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Jan 12, 2017, 6:57:39 PM1/12/17

to

Grrrmbl.

A state-of-the-art implementation of a Risch integrator

)version

Value = "FriCAS 1.3.0 compiled at Thu Sep 1 17:01:37 UTC 2016"

responds to

integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)

with

>> Error detected within library code:

catdef: division by zero

In his "Symbolic Integration Tutorial", Bronstein appears to believe in

the existence of a workable integration algorithm for algebraic

integrands involving a simple radical extension. Didn't he know what he

was talking about?

Martin.

Jan 13, 2017, 6:17:38 PM1/13/17

to

code he apparently assumed that he will be lucky. Or to put it

differetly: code he wrote is enough to "demostrate" that algorithm

works. But there remain "uninteresting" programming probles to

resolve. To the point: his code uses random integers. If

substituting those integers into one of denominators gives zero,

then you will see this error. Correct algorithm would retry

with different values. But his code just gives error. Recent

changes to FriCAS lead to change in random choice and for

some integrals (like this one) the problem became very visible.

One can try to hide the problem using more randomness. Or

one can do "uninteresting" programming and properly implement

retry.

BTW: This is FriCAS bug and proper place to report this is

FriCAS mailing list or Sourceforge bugtracke. And this

problem was was already reported (after relase of 1.3.0).

--

Waldek Hebisch

Jan 14, 2017, 5:41:18 PM1/14/17

to

anti...@math.uni.wroc.pl schrieb:

familar cube-root integrand. But to implement a loop over different

pseudo-random numbers should be easy, and its omission may be a mere

oversight. I will submit this integrand again to the next version.

Another thing has me wondering. Your antiderivatives of rational

functions can be discontinuous because the ATAN part is not always

rewritten to involve polynomial arguments only, as proposed by Rioboo.

For example, FriCAS 1.3.0 computes (expressed in Derive notation)

INT((2560*x^3 - 400*x^2 - 576*x - 84)/(320*x^4 + 80*x^3 - 12*x^2

+ 24*x + 9), x) = 2*LN(320*x^4 + 80*x^3 - 12*x^2 + 24*x + 9)

+ 2*SQRT(11)*ATAN(SQRT(11)*(10*x + 3)/(80*x^2 + 10*x - 9))

This is surprising especially since Bronstein provides pseudocode of

Rioboo's algorithm in his book (Ch. 2.8).

Martin.

Jan 14, 2017, 11:05:25 PM1/14/17

to

> Oh, I thought I was the first to try this particular variation of a

> familar cube-root integrand. But to implement a loop over different

> pseudo-random numbers should be easy, and its omission may be a mere

> oversight. I will submit this integrand again to the next version.

After fixing this "division by zero" bug, your first integral seems to
> familar cube-root integrand. But to implement a loop over different

> pseudo-random numbers should be easy, and its omission may be a mere

> oversight. I will submit this integrand again to the next version.

loop indefinitely. I think there is no symbolic result for

"integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

When a,b are integers, FriCAS can give symbolic results.

> Another thing has me wondering. Your antiderivatives of rational

> functions can be discontinuous because the ATAN part is not always

> rewritten to involve polynomial arguments only, as proposed by Rioboo.

my patch, FriCAS gives continuous result, same as SymPy.

I think this is a small programming mistake made by Manuel Bronstein.

Jan 15, 2017, 12:33:23 PM1/15/17

to

oldk...@gmail.com schrieb:

>

> [...] I think there is no symbolic result for

> "integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

> When a,b are integers, FriCAS can give symbolic results.

Wow - we could all become world-famous if this proves to be the first
> When a,b are integers, FriCAS can give symbolic results.

parametrized algebraic integrand with this property! Alas, my Derive

oracle says:

goursat5a((x + 1)/(x^2 + x + 1), x, a, 0, 0, b)

[false, false, false, true, true]

which means integrable :(. Might something else perhaps be amiss with

the current "demonstration-only" code of FriCAS?

>

> [...] For this example, I have post a small patch to fricas-devel.

> After my patch, FriCAS gives continuous result, same as SymPy.

Thanks, I myself obtain (up to some piecewise constant):
ATAN(SQRT(11)*(10*x + 3)/(80*x^2 + 10*x - 9)) =

- ATAN(SQRT(11)*(800*x^3 - 40*x^2 + 30*x + 57)/66)

- ATAN(SQRT(11)*(40*x - 7)/55)

Martin.

Jan 15, 2017, 8:10:41 PM1/15/17

to

> Wow - we could all become world-famous if this proves to be the first

> parametrized algebraic integrand with this property!

prime numbers, and FriCAS always gives result.

Jan 16, 2017, 12:55:01 PM1/16/17

to

oldk...@gmail.com schrieb:

>

> > > I think there is no symbolic result for

> > > "integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

> > > When a,b are integers, FriCAS can give symbolic results.

>

> > > I think there is no symbolic result for

> > > "integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

> > > When a,b are integers, FriCAS can give symbolic results.

>

by extensive observation:

Let f(p,x) be an algebraic function in both arguments; if, with respect

to x, f(p,x) possesses an elementary antiderivative for an infinite

number of complex parameter values p, then it possesses a single

symbolic elementary antiderivative F(p,x) that holds for all complex

values of p.

Proof (or disproof) of the principle is left to our top experts in

symbolic integration theory.

Martin.

Jan 17, 2017, 8:03:03 AM1/17/17

to

> You appeared to be contemplating the violation of a principle supported

> by extensive observation:

Let's say "integrate(f(p,x), x) = F(p,x)", and maybe F is recursive:

F(p,x) = G(F(p-1,x))

I don't think there are algorithms can expand recursive definition.

Jan 17, 2017, 2:18:06 PM1/17/17

to

http://koutschan.de/conf/ICMS16/davenport.pdf

starting from page 25. It seems that we need to limit

how parameter is choosen, otherwise there is a

couterexample. The claim appeared as a theorem, but

recently couterexample was discovered, so there

remain some work to find correct statement. IIUC

the case when you substitute integers is considerd

to be proved.

--

Waldek Hebisch

Jan 17, 2017, 2:24:34 PM1/17/17

to

oldk...@gmail.com wrote:

>

> After fixing this "division by zero" bug, your first integral seems to

> loop indefinitely. I think there is no symbolic result for

> "integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

> When a,b are integers, FriCAS can give symbolic results.

Actually, computation of algebraic integrals in FriCAS may
>

> After fixing this "division by zero" bug, your first integral seems to

> loop indefinitely. I think there is no symbolic result for

> "integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)".

> When a,b are integers, FriCAS can give symbolic results.

be quite slow and require a lot of memory -- I can not say

how much but in all slow cases that I looked at there were

indication of forward progress.

--

Waldek Hebisch

Jan 17, 2017, 6:46:29 PM1/17/17

to

anti...@math.uni.wroc.pl schrieb:

>

> Look at

>

> http://koutschan.de/conf/ICMS16/davenport.pdf

>

> starting from page 25. It seems that we need to limit

> how parameter is choosen, otherwise there is a

> couterexample. The claim appeared as a theorem, but

> recently couterexample was discovered, so there

> remain some work to find correct statement. IIUC

> the case when you substitute integers is considerd

> to be proved.

>

Thanks! I wasn't aware of Davenport's paper, and the "principle" was
> Look at

>

> http://koutschan.de/conf/ICMS16/davenport.pdf

>

> starting from page 25. It seems that we need to limit

> how parameter is choosen, otherwise there is a

> couterexample. The claim appeared as a theorem, but

> recently couterexample was discovered, so there

> remain some work to find correct statement. IIUC

> the case when you substitute integers is considerd

> to be proved.

>

distilled from my own and only my own experience :).

Martin.

Jan 18, 2017, 1:15:36 PM1/18/17

to

anti...@math.uni.wroc.pl schrieb:

>

> Look at

>

> http://koutschan.de/conf/ICMS16/davenport.pdf

>

> starting from page 25. It seems that we need to limit

> how parameter is choosen, otherwise there is a

> couterexample. The claim appeared as a theorem, but

> recently couterexample was discovered, so there

> remain some work to find correct statement. IIUC

> the case when you substitute integers is considerd

> to be proved.

>

I have now looked into this a bit. A better reference is Davenport's
> Look at

>

> http://koutschan.de/conf/ICMS16/davenport.pdf

>

> starting from page 25. It seems that we need to limit

> how parameter is choosen, otherwise there is a

> couterexample. The claim appeared as a theorem, but

> recently couterexample was discovered, so there

> remain some work to find correct statement. IIUC

> the case when you substitute integers is considerd

> to be proved.

>

conference paper <http://koutschan.de/conf/ICMS16/Integration.pdf>. His

counterexample:

INT(x/((x^2 - u^2)*SQRT(x^3 - x)), x) = ?

is taken from unpublished work by Masser and Zannier. The integral is

Goursat pseudo-elliptic if and only if u^2 = -1 or u^2 = 3 +- 2*SQRT(2),

the three cases corresponding to the points [u,v]:

[#i, 1 - #i]

[SQRT(2) + 1, 2 + SQRT(2)]

[SQRT(2) - 1, (2 - SQRT(2))*#i]

on the elliptic curve v^2 = u^3 - u. I guess they are just the points

of order four on the curve. Elementary evaluations of the corresponding

Goursat integrals are:

INT(x/((x^2 + 1)*SQRT(x^3 - x)), x) =

1/2*ATANH((x - 1)/SQRT(x^3 - x)) - 1/2*ATAN((x + 1)/SQRT(x^3 - x))

INT(x/((x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) =

(2 + SQRT(2))/4*ATANH((SQRT(2) - 1)*(x - 1)/SQRT(x^3 - x))

- (2 + SQRT(2))/4*ATAN((SQRT(2) - 1)*(x + 1)/SQRT(x^3 - x))

INT(x/((x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) =

(SQRT(2) - 2)/4*ATANH((1 + SQRT(2))*(x - 1)/SQRT(x^3 - x))

- (SQRT(2) - 2)/4*ATAN((1 + SQRT(2))*(x + 1)/SQRT(x^3 - x))

Thus, the result in Davenport's paper is unnecessarily complicated -

perhaps it can be replaced still?

FriCAS 1.3.0 solves the Goursat integration problems readily. Does the

system allow to compute some (preferably simple) points of order three

on the elliptic curve? According to Davenport, Masser and Zannier claim

those points to determine an infinite number of non-Goursat integrals

that are nonetheless elementary. It would be interesting to see if

FriCAS succeeds in evaluating them and what the antiderivatives look

like.

Martin.

Jan 22, 2017, 2:29:11 AM1/22/17

to

clicl...@freenet.de schrieb:

counterexample integral.

The three Goursat cases do indeed correspond to the points of order

four. I have also determined the points of order three and discovered

that FriCAS 1.3.0 returns the associated pair of integrals unevaluated.

Something must be very wrong here - with the counterexample, with my

point computations, or with the FriCAS integrator.

Next I have tried the most simple points from those of order six: the

corresponding two integrals are again non-elementary according to FriCAS

1.3.0. Same for order eight.

Be warned, however, that I never before determined torsion points on an

elliptic curve and that I had to work from scratch because Derive lacks

commands and libraries for elliptic-curve computations.

Martin.

Jan 24, 2017, 8:10:54 PM1/24/17

to

u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

so something is wrong. Theory behind example is simple, so

I believe it. I am not sure about points, but this is probably

correct. So it remains to check FriCAS :(

--

Waldek Hebisch

Jan 25, 2017, 12:46:52 AM1/25/17

to

anti...@math.uni.wroc.pl schrieb:

order-3 integrands.

I suppose you have tried the integral on Axiom as well - the problem

would have to be quite old then.

The points of order 3, 6 and 8 determined by me correspond to the

following three pairs of integrals:

INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

x) = ?

INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

x) = ?

So your result for order 3 is confirmed. I derived these x^2 - u^2 from

a recursive definition of the division polynomials and then checked the

points [u,v] by iterating the group operation.

I should mention that my statements about Goursat integrability referred

to Moebius mappings realizing the dihedral group D_2 (isomorphic to the

Klein four-group K_4) only; the special nature of the radicand x^3 - x

may open other possibilities (thus, x <-> 1/x swaps two roots and

preserves the remaining two). Compare Goursat pp. 113-118 for x^3 - 1.

Martin.

Jan 31, 2017, 11:40:13 AM1/31/17

to

clicl...@freenet.de schrieb:

>

> anti...@math.uni.wroc.pl schrieb:

pp. 118-119) and concludes that no special pseudo-elliptic cases arise.

I tend to believe him.

The substitution x <-> 1/x takes the integral into itself, with u

replaced by 1/u. Thus my pair of order-6 integrals is equivalent to the

order-3 one.

Meanwhile I can confirm that the order-3 integrals are elementary. The

antiderivatives take less than 1000 Bytes to write down.

Martin.

Jan 31, 2017, 6:26:17 PM1/31/17

to

There are a few points that I think are worth mentioning.

1. The Risch/ etc/ methods determine algebraic anti-derivatives.

It is a bit of a stretch sometimes to identify them with functions

with certain properties. It is typically possible to multiply or add

or otherwise muck around with these functions by, for example, use

of delta or step functions. Or adding strange constants.

The results of a subsequent differentiation need not be affected.

2. Either we are dealing with functions of a single variable (e.g. x)

or multiple variables / parameterized or multidimensional integral.

a. If multiple variables, the notion of a singularity becomes much

more complicated, and insisting on a continuous function much more

hazardous. Try to find useful theorems about what amounts to

functions of several complex variables. Not too many that I have found.

b. If a single variable, you are presumably being presented with a

function that is computable and continuous [yeah, well some version of

continuous for some version of integration] within the limits of

integration (yes, I know you might not know the limits. But you really

have a lot of nerve asking for the integral of a function you are not

willing to define adequately between the limits of interest.)

Continuing on this -- a continuous function of the kind being discussed

here can almost always be integrated by numerical methods. That is

given f(x), excellent methods exist to produce a program F(a,b)

which returns the integral of f from a to b (where a and b are

numerical constants).

Arbitrarily high precision methods are available, as are error

bounds. Chance of being fooled very low.

These methods avoid the problems encountered with symbolic methods which

include (i) the uselessness of an implicit proof that there is no

elementary expression for the antiderivative or (ii) there may be one

but we can't tell for sure because there is a bug in our complicated

program or (iii) because the procedure requires a sub-algorithm that is

not computable [zero equivalence] we can't say for sure about (i) or (ii).

3. You might argue (certainly I have) that symbolic results display

more information than (say) a table of numbers for various a,b. Yet

if the result is going to be run through a plotting program, not so

convincing. A partial symbolic result can be obtained by doing a

numerical-partial-fraction decomposition of some expressions --

any denominator that looks like a polynomial in one variable with

coefficients that can be resolved to floating-point numbers can be

factored into a product of linear factors, thereby requiring only

integration of <numerator stuff>/ (x-r)^n.

4. You might argue that you need the results for a high-dimension

multiple integration (calculation of Feynman diagrams was a major

selling point of integration in the Reduce system). Eh, then you

have multiple variables, but given limits, I think.

Now if you want to posit a problem that is to find an antiderivative

that has the fewest singularities, that may be useful. Another challenge

is to find an antiderivative that is, by some data-structure complexity

measure, "the simplest". This seems to be the objective of Rubi.

These problems are interesting in a computer-sciency computer algebra

systems algebraic-geometry context.

I'm not aware of any current clamor from actual or potential users of

computer algebra systems for solution of these problems. Are there

any Feynman diagram people still out there?

I'm not saying that Bronstein's work, or the programmers attempting to

solve the problems in this thread are "wrong", just that the context and

relevance to (say) scientific computation is easy to misconstrue.

Cheers.

RJF

On 1/31/2017 8:39 AM, clicl...@freenet.de wrote:

>

> clicl...@freenet.de schrieb:

>>

>> anti...@math.uni.wroc.pl schrieb:

>>>

>>> My calculations indicate that points of order 3 correspond to

>>> u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

>>> so something is wrong...

1. The Risch/ etc/ methods determine algebraic anti-derivatives.

It is a bit of a stretch sometimes to identify them with functions

with certain properties. It is typically possible to multiply or add

or otherwise muck around with these functions by, for example, use

of delta or step functions. Or adding strange constants.

The results of a subsequent differentiation need not be affected.

2. Either we are dealing with functions of a single variable (e.g. x)

or multiple variables / parameterized or multidimensional integral.

a. If multiple variables, the notion of a singularity becomes much

more complicated, and insisting on a continuous function much more

hazardous. Try to find useful theorems about what amounts to

functions of several complex variables. Not too many that I have found.

b. If a single variable, you are presumably being presented with a

function that is computable and continuous [yeah, well some version of

continuous for some version of integration] within the limits of

integration (yes, I know you might not know the limits. But you really

have a lot of nerve asking for the integral of a function you are not

willing to define adequately between the limits of interest.)

Continuing on this -- a continuous function of the kind being discussed

here can almost always be integrated by numerical methods. That is

given f(x), excellent methods exist to produce a program F(a,b)

which returns the integral of f from a to b (where a and b are

numerical constants).

Arbitrarily high precision methods are available, as are error

bounds. Chance of being fooled very low.

These methods avoid the problems encountered with symbolic methods which

include (i) the uselessness of an implicit proof that there is no

elementary expression for the antiderivative or (ii) there may be one

but we can't tell for sure because there is a bug in our complicated

program or (iii) because the procedure requires a sub-algorithm that is

not computable [zero equivalence] we can't say for sure about (i) or (ii).

3. You might argue (certainly I have) that symbolic results display

more information than (say) a table of numbers for various a,b. Yet

if the result is going to be run through a plotting program, not so

convincing. A partial symbolic result can be obtained by doing a

numerical-partial-fraction decomposition of some expressions --

any denominator that looks like a polynomial in one variable with

coefficients that can be resolved to floating-point numbers can be

factored into a product of linear factors, thereby requiring only

integration of <numerator stuff>/ (x-r)^n.

4. You might argue that you need the results for a high-dimension

multiple integration (calculation of Feynman diagrams was a major

selling point of integration in the Reduce system). Eh, then you

have multiple variables, but given limits, I think.

Now if you want to posit a problem that is to find an antiderivative

that has the fewest singularities, that may be useful. Another challenge

is to find an antiderivative that is, by some data-structure complexity

measure, "the simplest". This seems to be the objective of Rubi.

These problems are interesting in a computer-sciency computer algebra

systems algebraic-geometry context.

I'm not aware of any current clamor from actual or potential users of

computer algebra systems for solution of these problems. Are there

any Feynman diagram people still out there?

I'm not saying that Bronstein's work, or the programmers attempting to

solve the problems in this thread are "wrong", just that the context and

relevance to (say) scientific computation is easy to misconstrue.

Cheers.

RJF

On 1/31/2017 8:39 AM, clicl...@freenet.de wrote:

>

> clicl...@freenet.de schrieb:

>>

>> anti...@math.uni.wroc.pl schrieb:

>>>

>>> My calculations indicate that points of order 3 correspond to

>>> u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

Jan 31, 2017, 9:00:19 PM1/31/17

to

> Actually, Goursat considered the special radicand x^3 - x as well (on

> pp. 118-119) and concludes that no special pseudo-elliptic cases arise.

> I tend to believe him.

talking about?

Jan 31, 2017, 9:22:17 PM1/31/17

to

On Wednesday, February 1, 2017 at 7:26:17 AM UTC+8, Richard Fateman wrote:

> There are a few points that I think are worth mentioning.

>

> 1. The Risch/ etc/ methods determine algebraic anti-derivatives.

> It is a bit of a stretch sometimes to identify them with functions

> with certain properties. It is typically possible to multiply or add

> or otherwise muck around with these functions by, for example, use

> of delta or step functions. Or adding strange constants.

> The results of a subsequent differentiation need not be affected.

For rational functions integration, there exists algorithms that
> There are a few points that I think are worth mentioning.

>

> 1. The Risch/ etc/ methods determine algebraic anti-derivatives.

> It is a bit of a stretch sometimes to identify them with functions

> with certain properties. It is typically possible to multiply or add

> or otherwise muck around with these functions by, for example, use

> of delta or step functions. Or adding strange constants.

> The results of a subsequent differentiation need not be affected.

produce continuous anti-derivatives. It's fairly simple, I see no

reason not to implement it. As for other kinds of integration,

I don't think there's an algorithm that always works, but I think

heuristics can solve some of them.

> b. If a single variable, you are presumably being presented with a

> function that is computable and continuous [yeah, well some version of

> continuous for some version of integration] within the limits of

> integration (yes, I know you might not know the limits. But you really

> have a lot of nerve asking for the integral of a function you are not

> willing to define adequately between the limits of interest.)

> Continuing on this -- a continuous function of the kind being discussed

> here can almost always be integrated by numerical methods. That is

> given f(x), excellent methods exist to produce a program F(a,b)

> which returns the integral of f from a to b (where a and b are

> numerical constants).

> Arbitrarily high precision methods are available, as are error

> bounds. Chance of being fooled very low.

if the integrand includes variables, then numeric integration

must compute over and over again for these variables.

Another one is precision, zero is zero, no matter how many

digits are computed, you can't know if it's really zero or just

a small number.

> 3. You might argue (certainly I have) that symbolic results display

> more information than (say) a table of numbers for various a,b. Yet

> if the result is going to be run through a plotting program, not so

> convincing. A partial symbolic result can be obtained by doing a

> numerical-partial-fraction decomposition of some expressions --

> any denominator that looks like a polynomial in one variable with

> coefficients that can be resolved to floating-point numbers can be

> factored into a product of linear factors, thereby requiring only

> integration of <numerator stuff>/ (x-r)^n.

better than a plot (you can't draw a 4D function).

> I'm not saying that Bronstein's work, or the programmers attempting to

> solve the problems in this thread are "wrong", just that the context and

> relevance to (say) scientific computation is easy to misconstrue.

>

> Cheers.

> RJF

Cheers.

Feb 1, 2017, 12:40:40 AM2/1/17

to

oldk...@gmail.com schrieb:

Bulletin de la Société Mathématique de France 15 (1887), 106-120,

on-line at:

<http://www.numdam.org/item?id=BSMF_1887__15__106_1>

Practically forgotten. Straightforward to implement in rule-based

integrators. Covers all pseudo-elliptic integrands that matter in

practice. Can be extended to cube roots.

Martin.

Feb 6, 2017, 10:59:34 PM2/6/17

to

clicl...@freenet.de wrote:

>

> anti...@math.uni.wroc.pl schrieb:

I have less practice in hand calculations than say Gauss

so I am not eager to them by hand, but in principle this

is doable. Currently FriCAS thinks that point are

of infinite order, so probably something is wrong

with FriCAS.

> I suppose you have tried the integral on Axiom as well - the problem

> would have to be quite old then.

It makes no sense: due to known bugs Axiom has no chance

to do them.

> The points of order 3, 6 and 8 determined by me correspond to the

> following three pairs of integrals:

>

> INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> x) = ?

>

> INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> x) = ?

>

> So your result for order 3 is confirmed. I derived these x^2 - u^2 from

> a recursive definition of the division polynomials and then checked the

> points [u,v] by iterating the group operation.

Points of order 3 are special: here second derivative of defining

function vanishes:

(1) -> f := sqrt(x^3 - x)

+------+

| 3

(1) \|x - x

Type: Expression(Integer)

(2) -> D(f, x, 2)

4 2

3x - 6x - 1

(2) -------------------

+------+

3 | 3

(4x - 4x)\|x - x

Type: Expression(Integer)

(3) -> radicalSolve(numer(%)::Expression(Integer)::Polynomial(Integer))

(3)

+-----------+ +-----------+ +---------+ +---------+

| +-+ | +-+ | +-+ | +-+

\|- 2\|3 + 3 \|- 2\|3 + 3 \|2\|3 + 3 \|2\|3 + 3

[x = --------------,x = - --------------,x = ------------,x = - ------------]

+-+ +-+ +-+ +-+

\|3 \|3 \|3 \|3

Type: List(Equation(Expression(Integer)))

I am not aware of comparable simple method for different orders.

> I should mention that my statements about Goursat integrability referred

> to Moebius mappings realizing the dihedral group D_2 (isomorphic to the

> Klein four-group K_4) only; the special nature of the radicand x^3 - x

> may open other possibilities (thus, x <-> 1/x swaps two roots and

> preserves the remaining two). Compare Goursat pp. 113-118 for x^3 - 1.

>

> Martin.

--

Waldek Hebisch

>

> anti...@math.uni.wroc.pl schrieb:

> >

> > clicl...@freenet.de wrote:

> > > I have now also looked into the elliptic-curve aspect of this

> > > counterexample integral.

> > >

> > > The three Goursat cases do indeed correspond to the points of order

> > > four. I have also determined the points of order three and discovered

> > > that FriCAS 1.3.0 returns the associated pair of integrals unevaluated.

> > > Something must be very wrong here - with the counterexample, with my

> > > point computations, or with the FriCAS integrator.

> >

> > My calculations indicate that points of order 3 correspond to

> > u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

> > so something is wrong. Theory behind example is simple, so

> > I believe it. I am not sure about points, but this is probably

> > correct. So it remains to check FriCAS :(

> >

>

> A simple theory should allow to read off simple antiderivatives for the

> order-3 integrands.

It allows to read of antiderivatives. Concerning simple:
> > clicl...@freenet.de wrote:

> > > I have now also looked into the elliptic-curve aspect of this

> > > counterexample integral.

> > >

> > > The three Goursat cases do indeed correspond to the points of order

> > > four. I have also determined the points of order three and discovered

> > > that FriCAS 1.3.0 returns the associated pair of integrals unevaluated.

> > > Something must be very wrong here - with the counterexample, with my

> > > point computations, or with the FriCAS integrator.

> >

> > My calculations indicate that points of order 3 correspond to

> > u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

> > so something is wrong. Theory behind example is simple, so

> > I believe it. I am not sure about points, but this is probably

> > correct. So it remains to check FriCAS :(

> >

>

> A simple theory should allow to read off simple antiderivatives for the

> order-3 integrands.

I have less practice in hand calculations than say Gauss

so I am not eager to them by hand, but in principle this

is doable. Currently FriCAS thinks that point are

of infinite order, so probably something is wrong

with FriCAS.

> I suppose you have tried the integral on Axiom as well - the problem

> would have to be quite old then.

to do them.

> The points of order 3, 6 and 8 determined by me correspond to the

> following three pairs of integrals:

>

> INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

>

> INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> x) = ?

>

> INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> x) = ?

>

> So your result for order 3 is confirmed. I derived these x^2 - u^2 from

> a recursive definition of the division polynomials and then checked the

> points [u,v] by iterating the group operation.

function vanishes:

(1) -> f := sqrt(x^3 - x)

+------+

| 3

(1) \|x - x

Type: Expression(Integer)

(2) -> D(f, x, 2)

4 2

3x - 6x - 1

(2) -------------------

+------+

3 | 3

(4x - 4x)\|x - x

Type: Expression(Integer)

(3) -> radicalSolve(numer(%)::Expression(Integer)::Polynomial(Integer))

(3)

+-----------+ +-----------+ +---------+ +---------+

| +-+ | +-+ | +-+ | +-+

\|- 2\|3 + 3 \|- 2\|3 + 3 \|2\|3 + 3 \|2\|3 + 3

[x = --------------,x = - --------------,x = ------------,x = - ------------]

+-+ +-+ +-+ +-+

\|3 \|3 \|3 \|3

Type: List(Equation(Expression(Integer)))

I am not aware of comparable simple method for different orders.

> I should mention that my statements about Goursat integrability referred

> to Moebius mappings realizing the dihedral group D_2 (isomorphic to the

> Klein four-group K_4) only; the special nature of the radicand x^3 - x

> may open other possibilities (thus, x <-> 1/x swaps two roots and

> preserves the remaining two). Compare Goursat pp. 113-118 for x^3 - 1.

>

> Martin.

Waldek Hebisch

Feb 6, 2017, 11:09:46 PM2/6/17

to

says that there are none and you believe Goursat? I do not

understand.

> Meanwhile I can confirm that the order-3 integrals are elementary. The

> antiderivatives take less than 1000 Bytes to write down.

>

> Martin.

Waldek Hebisch

Feb 7, 2017, 12:12:48 PM2/7/17

to

anti...@math.uni.wroc.pl schrieb:

>

> clicl...@freenet.de wrote:

> >

> > anti...@math.uni.wroc.pl schrieb:

> > >

> > > clicl...@freenet.de wrote:

> > > >

> > > > I have now also looked into the elliptic-curve aspect of this

> > > > counterexample integral.

> > > >

> > > > The three Goursat cases do indeed correspond to the points of order

> > > > four. I have also determined the points of order three and discovered

> > > > that FriCAS 1.3.0 returns the associated pair of integrals unevaluated.

> > > > Something must be very wrong here - with the counterexample, with my

> > > > point computations, or with the FriCAS integrator.

> > >

> > > My calculations indicate that points of order 3 correspond to

> > > u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

> > > so something is wrong. Theory behind example is simple, so

> > > I believe it. I am not sure about points, but this is probably

> > > correct. So it remains to check FriCAS :(

> > >

> >

> > A simple theory should allow to read off simple antiderivatives for

> > the order-3 integrands.

>

> It allows to read of antiderivatives. Concerning simple:

> I have less practice in hand calculations than say Gauss

> so I am not eager to them by hand, but in principle this

> is doable. Currently FriCAS thinks that point are

> of infinite order, so probably something is wrong

> with FriCAS.

There can be no doubt that something is wrong with FriCAS since Gauss
> clicl...@freenet.de wrote:

> >

> > anti...@math.uni.wroc.pl schrieb:

> > >

> > > clicl...@freenet.de wrote:

> > > >

> > > > I have now also looked into the elliptic-curve aspect of this

> > > > counterexample integral.

> > > >

> > > > The three Goursat cases do indeed correspond to the points of order

> > > > four. I have also determined the points of order three and discovered

> > > > that FriCAS 1.3.0 returns the associated pair of integrals unevaluated.

> > > > Something must be very wrong here - with the counterexample, with my

> > > > point computations, or with the FriCAS integrator.

> > >

> > > My calculations indicate that points of order 3 correspond to

> > > u^2 = 1 +- 2*sqrt(3)/3. FriCAS says that integal is nonelementary,

> > > so something is wrong. Theory behind example is simple, so

> > > I believe it. I am not sure about points, but this is probably

> > > correct. So it remains to check FriCAS :(

> > >

> >

> > A simple theory should allow to read off simple antiderivatives for

> > the order-3 integrands.

>

> It allows to read of antiderivatives. Concerning simple:

> I have less practice in hand calculations than say Gauss

> so I am not eager to them by hand, but in principle this

> is doable. Currently FriCAS thinks that point are

> of infinite order, so probably something is wrong

> with FriCAS.

has provided me with a hand-written finite elementary evaluation of one

order-3 integral. And I find his evaluation to hold up.

>

> > I suppose you have tried the integral on Axiom as well - the problem

> > would have to be quite old then.

>

> It makes no sense: due to known bugs Axiom has no chance

> to do them.

>

u^2 appears as factors of the so-called division polynomials for the

elliptic curve v^2 = u^3 - u:

divpol(3, x, 0, -1, 0)

3*x^4 - 6*x^2 - 1

1/3*(3*x^2 + 2*SQRT(3) - 3)*(3*x^2 - 2*SQRT(3) - 3)

divpol(8, x, 0, -1, 0)

8*(x^6 - 5*x^4 - 5*x^2 + 1)*(x^24 - 68*x^22 - 1694*x^20 + 3276*x^18

- 19601*x^16 + 63352*x^14 - 86436*x^12 + 63352*x^10 - 19601*x^8 +

3276*x^6 - 1694*x^4 - 68*x^2 + 1)*SQRT(x^3 - x)

8*(x^2 + 1)*(x^2 + 2*SQRT(2) - 3)*(x^2 - 2*SQRT(2) - 3)*(x^2 +

2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*(x^2 - 2*SQRT(10*SQRT(2) +

14) + 4*SQRT(2) + 5)*(x^2 + 2*SQRT(41*SQRT(2) + 58) +

4*SQRT(10*SQRT(2) + 14) - 6*2^(3/4) - 8*SQRT(2) - 10*2^(1/4) -

11)*(x^2 + 2*SQRT(41*SQRT(2) + 58) - 4*SQRT(10*SQRT(2) + 14) +

6*2^(3/4) - 8*SQRT(2) + 10*2^(1/4) - 11)*(x^2 - 2*SQRT(41*SQRT(2) +

58) + 4*SQRT(10*SQRT(2) + 14) + 6*2^(3/4) - 8*SQRT(2) + 10*2^(1/4) -

11)*(x^2 - 2*SQRT(41*SQRT(2) + 58) - 4*SQRT(10*SQRT(2) + 14) -

6*2^(3/4) - 8*SQRT(2) - 10*2^(1/4) - 11)*(x^4 + 2*x^2*(5 - 4*SQRT(2))

+ 1)*(x^8 + 4*x^6*(8*SQRT(2) - 11) + 2*x^4*(51 - 32*SQRT(2)) +

4*x^2*(8*SQRT(2) - 11) + 1)*SQRT(x^3 - x)

Ugh. Note that the order-4 factors are repeated for order 8.

Martin.

Feb 7, 2017, 12:13:01 PM2/7/17

to

anti...@math.uni.wroc.pl schrieb:

>

> clicl...@freenet.de wrote:

> >

> > clicl...@freenet.de schrieb:

> > >

> > > I should mention that my statements about Goursat integrability

> > > referred to Moebius mappings realizing the dihedral group D_2

> > > (isomorphic to the Klein four-group K_4) only; the special nature

> > > of the radicand x^3 - x may open other possibilities (thus, x <->

> > > 1/x swaps two roots and preserves the remaining two). Compare

> > > Goursat pp. 113-118 for x^3 - 1.

> > >

> >

> > Actually, Goursat considered the special radicand x^3 - x as well

> > (on pp. 118-119) and concludes that no special pseudo-elliptic cases

> > arise. I tend to believe him.

>

> Hmm, you checked that for order 3 one gets special case, Goursat

> says that there are none and you believe Goursat? I do not

> understand.

>

Goursat claims ("... ne fournissent pas d'intégrales pseudo-elliptiques
> > > referred to Moebius mappings realizing the dihedral group D_2

> > > (isomorphic to the Klein four-group K_4) only; the special nature

> > > of the radicand x^3 - x may open other possibilities (thus, x <->

> > > 1/x swaps two roots and preserves the remaining two). Compare

> > > Goursat pp. 113-118 for x^3 - 1.

> > >

> >

> > Actually, Goursat considered the special radicand x^3 - x as well

> > (on pp. 118-119) and concludes that no special pseudo-elliptic cases

> > arise. I tend to believe him.

>

> Hmm, you checked that for order 3 one gets special case, Goursat

> says that there are none and you believe Goursat? I do not

> understand.

>

nouvelles") that for the special radicand x^3 - x his method of Moebius

mapping cannot detect pseudo-elliptic integrands beyond those already

found from equation (5) on p. 111 of his paper.

For x^3 - 1, however, the situation is different (pp. 113-118).

Martin.

Feb 11, 2017, 12:37:57 AM2/11/17

to

clicl...@freenet.de wrote:

>

> There can be no doubt that something is wrong with FriCAS since Gauss

> has provided me with a hand-written finite elementary evaluation of one

> order-3 integral. And I find his evaluation to hold up.

<snip>
>

> There can be no doubt that something is wrong with FriCAS since Gauss

> has provided me with a hand-written finite elementary evaluation of one

> order-3 integral. And I find his evaluation to hold up.

> > > The points of order 3, 6 and 8 determined by me correspond to the

> > > following three pairs of integrals:

> > >

> > > INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> > > x) = ?

> > >

> > > INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> > > x) = ?

FriCAS had a silly bug due to which it treated correctly computed
> > > following three pairs of integrals:

> > >

> > > INT(x/((3*x^2 + 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((3*x^2 - 2*SQRT(2) - 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 + 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 - 2*SQRT(2) + 3)*SQRT(x^3 - x)), x) = ?

> > >

> > > INT(x/((x^2 + 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> > > x) = ?

> > >

> > > INT(x/((x^2 - 2*SQRT(10*SQRT(2) + 14) + 4*SQRT(2) + 5)*SQRT(x^3 - x)),

> > > x) = ?

integral as nonelementary. For finding the bug it did not help

that point that you gave in integral for order 3 and 6 were

wrong (you wrote SQRT(2) while SQRT(3) is correct).

> Thanks for the details. As noted above, the special denominators x^2 -

> u^2 appears as factors of the so-called division polynomials for the

> elliptic curve v^2 = u^3 - u:

one corresponds to u^2 = (1 + 2*sqrt(-1))/5. FriCAS have

now computed integrals for orders 3, 5, 6, 8. The results

are rather lengthy, so instead of posting them here I have

put them at:

http://www.math.uni.wroc.pl/~hebisch/fricas/p3

(3 above means order 3, replace 3 by 5, 6, 8 for higher order).

--

Waldek Hebisch

Feb 11, 2017, 7:32:59 AM2/11/17

to

anti...@math.uni.wroc.pl schrieb:

When I copy expressions from Derive, such roots consist of two

characters looking like "V3": one non-ASCII square-root character

immediately followed by the ASCII integer. In the above, I regularized

them manually to SQRT(2), unthinkingly mistaking "3" for "2", probably

doing it once and then using copy-and-paste. (The alternative is to

extract expressions from session files stored on disk; they are regular

but consist of character blocks lacking spaces and exhibiting arbitrary

line-breaks preceded by "~".)

For comparison, here are my direct and factored division polynomials for

orders 5,6,7:

divpol(5, x, 0, -1, 0)

5*x^12 - 62*x^10 - 105*x^8 + 300*x^6 - 125*x^4 + 50*x^2 + 1

(x^2 + 2*SQRT(2*SQRT(5) + 5) - 2*SQRT(5) - 3)*

(x^2 - 2*SQRT(2*SQRT(5) + 5) - 2*SQRT(5) - 3)*

(x^2 + 2*SQRT(5 - 2*SQRT(5)) + 2*SQRT(5) - 3)*

(x^2 - 2*SQRT(5 - 2*SQRT(5)) + 2*SQRT(5) - 3)*(5*x^4 - 2*x^2 + 1)

divpol(6, x, 0, -1, 0)

2*(x^12 - 22*x^10 - 165*x^8 + 92*x^6 - 185*x^4 + 90*x^2 - 3)*(3*x^4 -

6*x^2 - 1)*SQRT(x^3 - x)

2*(x^2 + 2*SQRT(3) + 3)*(x^2 - 2*SQRT(3) + 3)*

(x^2 + 2*SQRT(2)*3^(3/4) - 4*SQRT(3) + 4*12^(1/4) - 7)*

(x^2 - 2*SQRT(2)*3^(3/4) - 4*SQRT(3) - 4*12^(1/4) - 7)*

(x^4 + 2*x^2*(4*SQRT(3) - 7) + 1)*(3*x^4 - 6*x^2 - 1)*SQRT(x^3 - x)

divpol(7, x, 0, -1, 0)

7*x^24 - 308*x^22 - 2954*x^20 + 19852*x^18 - 35231*x^16

+ 82264*x^14 - 111916*x^12 + 42168*x^10 + 15673*x^8

- 14756*x^6 + 1302*x^4 - 196*x^2 - 1

The last one exceeds my factorization capabilities. And here is the

order-3 antiderivative provided by Gauss:

INT(x/((3*x^2 + 2*SQRT(3) - 3)*SQRT(x^3 - x)), x) =

12^(3/8)*SQRT(SQRT(3) + 1)/36*ATANH(SQRT(SQRT(3) + 1)*(SQRT(2)*3^(3/4)

*x^3 + 3*(3 - SQRT(3))*x^2 - SQRT(2)*3^(3/4)*x + 3*SQRT(3) - 5)

/(12^(3/8)*(SQRT(2)*3^(3/4)*x - SQRT(3) + 1)*SQRT(x^3 - x))) +

12^(3/8)*SQRT(SQRT(3) + 1)/36*ATAN(SQRT(SQRT(3) + 1)*(SQRT(2)*3^(3/4)

*x^3 + 3*(SQRT(3) - 3)*x^2 - SQRT(2)*3^(3/4)*x - 3*SQRT(3) + 5)

/(12^(3/8)*(SQRT(2)*3^(3/4)*x + SQRT(3) - 1)*SQRT(x^3 - x)))

Martin.

Feb 11, 2017, 10:58:05 PM2/11/17

to

clicl...@freenet.de wrote:

>

> divpol(7, x, 0, -1, 0)

>

> 7*x^24 - 308*x^22 - 2954*x^20 + 19852*x^18 - 35231*x^16

> + 82264*x^14 - 111916*x^12 + 42168*x^10 + 15673*x^8

> - 14756*x^6 + 1302*x^4 - 196*x^2 - 1

>

> The last one exceeds my factorization capabilities.

This one is irreducible. One possible u = x^2 in terms
>

> divpol(7, x, 0, -1, 0)

>

> 7*x^24 - 308*x^22 - 2954*x^20 + 19852*x^18 - 35231*x^16

> + 82264*x^14 - 111916*x^12 + 42168*x^10 + 15673*x^8

> - 14756*x^6 + 1302*x^4 - 196*x^2 - 1

>

> The last one exceeds my factorization capabilities.

of radicals is below:

-1

*

+---+ +-+ +-+ +---+ +-+

((- 13\|- 1 \|3 + 9)\|7 - 35\|- 1 \|3 + 21)

*

+---------------------------------------------------+2

3| +---+ +-+ +-+ +---+ +-+

\|(1524\|- 1 \|3 + 3196)\|7 - 4032\|- 1 \|3 - 8456

+

+---+ +-+

(- 42\|- 1 \|3 + 42)

*

+---------------------------------------------------+

3| +---+ +-+ +-+ +---+ +-+

\|(1524\|- 1 \|3 + 3196)\|7 - 4032\|- 1 \|3 - 8456

+

+-+

- 336\|7 + 924

/

126

--

Waldek Hebisch

May 16, 2017, 12:55:31 PM5/16/17

to

clicl...@freenet.de schrieb:

>

> A state-of-the-art implementation of a Risch integrator

>

> )version

>

> Value = "FriCAS 1.3.0 compiled at Thu Sep 1 17:01:37 UTC 2016"

>

> responds to

>

> A state-of-the-art implementation of a Risch integrator

>

> )version

>

> Value = "FriCAS 1.3.0 compiled at Thu Sep 1 17:01:37 UTC 2016"

>

> responds to

>

> integrate((1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)), x)

>

>

> with

>

> >> Error detected within library code:

> catdef: division by zero

>

Meanwhile, the FriCAS version has advanced to 1.3.2, the web interface
>

> >> Error detected within library code:

> catdef: division by zero

>

is running version 1.3.1, and the division by zero no longer occurs when

the integral is submitted to the web interface. Its computation,

however, exceeds the 5-minute timeout imposed by the interface (Proxy

Error - Reason: Error reading from remote server). Does FriCAS now

manage to return a correct antiderivative eventually?

Martin.

May 31, 2017, 12:55:53 PM5/31/17

to

clicl...@freenet.de schrieb:

Martin.

Jun 15, 2022, 9:39:36 AM6/15/22

to

Okay, why does the case u=i returns nonelementary

result that has weierstrassPInverse(4, 0, x) on latest

HEAD of master?

Also I cannot get these results on latest HEAD of

master I compiled today.

http://www.math.uni.wroc.pl/~hebisch/fricas/p3

not even talking about crazy

http://www.math.uni.wroc.pl/~hebisch/fricas/p5

http://www.math.uni.wroc.pl/~hebisch/fricas/p6

http://www.math.uni.wroc.pl/~hebisch/fricas/p8

This looks like a regression from 1.3.1!

Also, Masser and Zannier published very nice one

integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

It does not work, timeouts after 8 hours and looks

like even crashes because it exists (no warnings) as

if )quit happened.

result that has weierstrassPInverse(4, 0, x) on latest

HEAD of master?

Also I cannot get these results on latest HEAD of

master I compiled today.

http://www.math.uni.wroc.pl/~hebisch/fricas/p3

not even talking about crazy

http://www.math.uni.wroc.pl/~hebisch/fricas/p5

http://www.math.uni.wroc.pl/~hebisch/fricas/p6

http://www.math.uni.wroc.pl/~hebisch/fricas/p8

This looks like a regression from 1.3.1!

Also, Masser and Zannier published very nice one

integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

It does not work, timeouts after 8 hours and looks

like even crashes because it exists (no warnings) as

if )quit happened.

Jun 16, 2022, 2:16:00 AM6/16/22

to

??????? ???????????? schrieb:

>

> Okay, why does the case u=i returns nonelementary

> result that has weierstrassPInverse(4, 0, x) on latest

> HEAD of master?

Good grief! However:
> Okay, why does the case u=i returns nonelementary

> result that has weierstrassPInverse(4, 0, x) on latest

> HEAD of master?

integrate(x/(x^2+1)/(x^3-x)^(1/2), x)

still works on the web interface running version 1.3.7, the

instantaneous result being:

(log(((4*x^2+8*x+(-4))*(x^3+(-1)*x)^(1/2)+(x^4+8*x^3+2*x^2+(-8)*x+1))

/(x^4+2*x^2+1))

+2*atan(((x^2+(-2)*x+(-1))*(x^3+(-1)*x)^(1/2))/(2*x^3+(-2)*x)))/8

which is reasonably compact. As mentioned earlier in this thread, it

reduces to (in Derive notation):

1/2*ATANH((x - 1)/SQRT(x^3 - x)) - 1/2*ATAN((x + 1)/SQRT(x^3 - x))

>

> Also I cannot get these results on latest HEAD of

> master I compiled today.

> http://www.math.uni.wroc.pl/~hebisch/fricas/p3

> not even talking about crazy

> http://www.math.uni.wroc.pl/~hebisch/fricas/p5

> http://www.math.uni.wroc.pl/~hebisch/fricas/p6

> http://www.math.uni.wroc.pl/~hebisch/fricas/p8

>

> This looks like a regression from 1.3.1!

>

> Also, Masser and Zannier published very nice one

>

> integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

>

> It does not work, timeouts after 8 hours and looks

> like even crashes because it exists (no warnings) as

> if )quit happened.

I fear that the "master" branch you are using must be regarded a work
> master I compiled today.

> http://www.math.uni.wroc.pl/~hebisch/fricas/p3

> not even talking about crazy

> http://www.math.uni.wroc.pl/~hebisch/fricas/p5

> http://www.math.uni.wroc.pl/~hebisch/fricas/p6

> http://www.math.uni.wroc.pl/~hebisch/fricas/p8

>

> This looks like a regression from 1.3.1!

>

> Also, Masser and Zannier published very nice one

>

> integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

>

> It does not work, timeouts after 8 hours and looks

> like even crashes because it exists (no warnings) as

> if )quit happened.

in progress. One should perhaps better stick with version 1.3.7 until

an official 1.3.8 arrives.

Martin.

Jun 16, 2022, 7:59:27 AM6/16/22

to

You are right, I just checked HEAD of master again,

and u=i is giving what you said. Strange.

This still of course does not work:

http://www.math.uni.wroc.pl/~hebisch/fricas/p3

BTW, I cannot compile 1.3.2, compiler does no like this

old code no longer. Do you maybe have it precompiled?

and u=i is giving what you said. Strange.

This still of course does not work:

http://www.math.uni.wroc.pl/~hebisch/fricas/p3

BTW, I cannot compile 1.3.2, compiler does no like this

old code no longer. Do you maybe have it precompiled?

Jun 16, 2022, 8:21:24 AM6/16/22

to

Actually I was notified on a way to solve p3, p5, p8!

Instead of integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

use

setSimplifyDenomsFlag(true)

integrate(x/((x^2 - ((1 + 2*sqrt(-1))/5))*sqrt(x^3 - x)), x)

See:

https://github.com/fricas/fricas/pull/92

Case closed!

Instead of integrate(x/(x^2-1/5-2*%i/5)/(x^3-x)^(1/2), x)

use

setSimplifyDenomsFlag(true)

integrate(x/((x^2 - ((1 + 2*sqrt(-1))/5))*sqrt(x^3 - x)), x)

See:

https://github.com/fricas/fricas/pull/92

Case closed!

Jun 16, 2022, 9:23:27 AM6/16/22

to

??????? ???????????? schrieb:

>

> You are right, I just checked HEAD of master again,

> and u=i is giving what you said. Strange.

> This still of course does not work:

> http://www.math.uni.wroc.pl/~hebisch/fricas/p3

For:
> You are right, I just checked HEAD of master again,

> and u=i is giving what you said. Strange.

> This still of course does not work:

> http://www.math.uni.wroc.pl/~hebisch/fricas/p3

integrate(x/((x^2 - (1 - 2*sqrt(3)/3))*sqrt(x^3 - x)), x)

on the web interface running version 1.3.7 I get the instantaneous

result (hopefully I didn't mess it up):

(4*(6*3^(1/2)+9)^(1/4)*atan(((((27*x^6+(-171)*x^4+117*x^2+(-5))*3^(1/2)

+((-81)*x^6+315*x^4+(-171)*x^2+9))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1

/4))^3+(((-216)*x^6+216*x^4)*3^(1/2)+(324*x^6+(-360)*x^4+36*x^2))*((6*3

^(1/2)+9)^(1/4))^2+(((-270)*x^5+144*x^3+(-18)*x)*3^(1/2)+(324*x^5

+(-396)*x^3))*(x^3+(-1)*x)^(1/2)*(6*3^(1/2)+9)^(1/4)+((216*x^5+(-216)*x

^3)*3^(1/2)+(486*x^7+(-486)*x^5+18*x^3+(-18)*x)))*((((768*x^9+(-288)*x

^7+(-96)*x^5+(-96)*x^3+(-32)*x)*3^(1/2)+((-1296)*x^9+288*x^7+576*x^5

+(-32)*x^3+(-48)*x))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1/4))^3+

(((-432)*x^11+336*x^9+672*x^7+(-864)*x^5+272*x^3+16*x)*3^(1/2)+(864*x

^11+(-1440)*x^9+1152*x^7+(-1152)*x^5+544*x^3+32*x))*((6*3^(1/2)+9)^(1

/4))^2+((216*x^10+(-1080)*x^8+1104*x^6+(-1104)*x^4+600*x^2+8)*3^(1/2)

+((-216)*x^10+648*x^8+1584*x^6+(-2448)*x^4+936*x^2+8))*(x^3+(-1)*x)^(1

/2)*(6*3^(1/2)+9)^(1/4)+((864*x^10+(-1728)*x^6+1536*x^4+(-672)*x^2)*3

^(1/2)+(108*x^12+(-2376)*x^10+2916*x^8+(-2160)*x^6+2484*x^4+(-1224)*x^2

+(-4))))/(27*x^12+(-162)*x^10+297*x^8+(-108)*x^6+(-99)*x^4+(-18)*x^2

+(-1)))^(1/2)+((((-54)*x^6+(-198)*x^4+198*x^2+(-10))*3^(1/2)+(162*x^6

+342*x^4+(-378)*x^2+18))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1/4))^3

+(((-432)*x^5+144*x^3)*3^(1/2)+(324*x^5+(-504)*x^3+36*x))*(x^3+(-1)*x)

^(1/2)*(6*3^(1/2)+9)^(1/4)))/(972*x^7+(-324)*x^5+(-684)*x^3+36*x))+(4

*(6*3^(1/2)+9)^(1/4)*atan(((((27*x^6+(-171)*x^4+117*x^2+(-5))*3^(1/2)

+((-81)*x^6+315*x^4+(-171)*x^2+9))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1

/4))^3+((216*x^6+(-216)*x^4)*3^(1/2)+((-324)*x^6+360*x^4+(-36)*x^2))

*((6*3^(1/2)+9)^(1/4))^2+(((-270)*x^5+144*x^3+(-18)*x)*3^(1/2)+(324*x^5

+(-396)*x^3))*(x^3+(-1)*x)^(1/2)*(6*3^(1/2)+9)^(1/4)+(((-216)*x^5+216

*x^3)*3^(1/2)+((-486)*x^7+486*x^5+(-18)*x^3+18*x)))*(((((-768)*x^9+288

*x^7+96*x^5+96*x^3+32*x)*3^(1/2)+(1296*x^9+(-288)*x^7+(-576)*x^5+32*x^3

+48*x))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1/4))^3+(((-432)*x^11+336*x

^9+672*x^7+(-864)*x^5+272*x^3+16*x)*3^(1/2)+(864*x^11+(-1440)*x^9+1152

*x^7+(-1152)*x^5+544*x^3+32*x))*((6*3^(1/2)+9)^(1/4))^2+(((-216)*x^10

+1080*x^8+(-1104)*x^6+1104*x^4+(-600)*x^2+(-8))*3^(1/2)+(216*x^10

+(-648)*x^8+(-1584)*x^6+2448*x^4+(-936)*x^2+(-8)))*(x^3+(-1)*x)^(1/2)

*(6*3^(1/2)+9)^(1/4)+((864*x^10+(-1728)*x^6+1536*x^4+(-672)*x^2)*3^(1

/2)+(108*x^12+(-2376)*x^10+2916*x^8+(-2160)*x^6+2484*x^4+(-1224)*x^2

+(-4))))/(27*x^12+(-162)*x^10+297*x^8+(-108)*x^6+(-99)*x^4+(-18)*x^2

+(-1)))^(1/2)+((((-54)*x^6+(-198)*x^4+198*x^2+(-10))*3^(1/2)+(162*x^6

+342*x^4+(-378)*x^2+18))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1/4))^3

+(((-432)*x^5+144*x^3)*3^(1/2)+(324*x^5+(-504)*x^3+36*x))*(x^3+(-1)*x)

^(1/2)*(6*3^(1/2)+9)^(1/4)))/(972*x^7+(-324)*x^5+(-684)*x^3+36*x))+((6

*3^(1/2)+9)^(1/4)*log((((768*x^9+(-288)*x^7+(-96)*x^5+(-96)*x^3+(-32)

*x)*3^(1/2)+((-1296)*x^9+288*x^7+576*x^5+(-32)*x^3+(-48)*x))*(x^3+(-1)

*x)^(1/2)*((6*3^(1/2)+9)^(1/4))^3+(((-432)*x^11+336*x^9+672*x^7+(-864)

*x^5+272*x^3+16*x)*3^(1/2)+(864*x^11+(-1440)*x^9+1152*x^7+(-1152)*x^5

+544*x^3+32*x))*((6*3^(1/2)+9)^(1/4))^2+((216*x^10+(-1080)*x^8+1104*x^6

+(-1104)*x^4+600*x^2+8)*3^(1/2)+((-216)*x^10+648*x^8+1584*x^6+(-2448)

*x^4+936*x^2+8))*(x^3+(-1)*x)^(1/2)*(6*3^(1/2)+9)^(1/4)+((864*x^10

+(-1728)*x^6+1536*x^4+(-672)*x^2)*3^(1/2)+(108*x^12+(-2376)*x^10+2916

*x^8+(-2160)*x^6+2484*x^4+(-1224)*x^2+(-4))))/(27*x^12+(-162)*x^10+297

*x^8+(-108)*x^6+(-99)*x^4+(-18)*x^2+(-1)))+(-1)*(6*3^(1/2)+9)^(1/4)

*log(((((-768)*x^9+288*x^7+96*x^5+96*x^3+32*x)*3^(1/2)+(1296*x^9+(-288)

*x^7+(-576)*x^5+32*x^3+48*x))*(x^3+(-1)*x)^(1/2)*((6*3^(1/2)+9)^(1/4))

^3+(((-432)*x^11+336*x^9+672*x^7+(-864)*x^5+272*x^3+16*x)*3^(1/2)+(864

*x^11+(-1440)*x^9+1152*x^7+(-1152)*x^5+544*x^3+32*x))*((6*3^(1/2)+9)^(1

/4))^2+(((-216)*x^10+1080*x^8+(-1104)*x^6+1104*x^4+(-600)*x^2+(-8))*3

^(1/2)+(216*x^10+(-648)*x^8+(-1584)*x^6+2448*x^4+(-936)*x^2+(-8)))*(x^3

+(-1)*x)^(1/2)*(6*3^(1/2)+9)^(1/4)+((864*x^10+(-1728)*x^6+1536*x^4

+(-672)*x^2)*3^(1/2)+(108*x^12+(-2376)*x^10+2916*x^8+(-2160)*x^6+2484*x

^4+(-1224)*x^2+(-4))))/(27*x^12+(-162)*x^10+297*x^8+(-108)*x^6+(-99)*x^4

+(-18)*x^2+(-1))))))/24

A much shorter version appears earlier in this thread.

>

> BTW, I cannot compile 1.3.2, compiler does no like this

> old code no longer. Do you maybe have it precompiled?

Martin.

Jun 16, 2022, 9:24:09 AM6/16/22

to

??????? ???????????? schrieb:

In view of the potential consequences, the developers should make this

the default setting.

Martin.

Jun 16, 2022, 8:17:24 PM6/16/22

to

I did know about it (and did try it), but %i was the problem.

Jun 19, 2022, 2:31:01 AM6/19/22

to

??????? ???????????? schrieb:

>

> I did know about it (and did try it), but %i was the problem.

I have no luck either when I submit the order-5 integrand with sqrt(-1)
> I did know about it (and did try it), but %i was the problem.

in place of %i to the FriCAS web interface currently running version

1.3.7:

unparse(integrate(x/(x^2 - 1/5 - 2*sqrt(-1)/5)/sqrt(x^3 - x),

x)::InputForm)

There are 9 exposed and 6 unexposed library operations named -

having 2 argument(s) but none was determined to be applicable.

Use HyperDoc Browse, or issue

)display op -

to learn more about the available operations. Perhaps

package-calling the operation or using coercions on the arguments

will allow you to apply the operation.

Cannot find a definition or applicable library operation named -

with argument type(s)

Polynomial(Fraction(Integer))

AlgebraicNumber

Perhaps you should use "@" to indicate the required return type,

or "$" to specify which version of the function you need.

Grrrmbl. Will version 1.3.8 do better?

Martin.

Jun 19, 2022, 2:57:32 AM6/19/22

to

No, you need both. Then it works.

Jun 19, 2022, 3:35:28 AM6/19/22

to

setSimplifyDenomsFlag(true)

r=integrate(x/((x^2 - ((1 + 2*sqrt(-1))/5))*sqrt(x^3 - x)), x);

unparse(r::InputForm)

"((4^(1/4))^3*((-2)*(-1)^(1/2)+11)^(1/4)*log(((((75669959187507629394531250*x

^18+(-669388100504875183105468750)*x^16+1261942088603973388671875000*x^14+959

26225185394287109375000*x^12+(-1562852412462234497070312500)*x^10+86302682757

3776245117187500*x^8+(-33326447010040283203125000)*x^6+(-41164457798004150390

625000)*x^4+3458932042121887207031250*x^2+31664967536926269531250)*(-1)^(1/2)

+((-52386894822120666503906250)*x^18+29103830456733703613281250*x^16+11501833

79650115966796875000*x^14+(-2308748662471771240234375000)*x^12+10368414223194

12231445312500*x^10+378187745809555053710937500*x^8+(-28655678033828735351562

5000)*x^6+28498470783233642578125000*x^4+2047047019004821777343750*x^2+(-5774

1999626159667968750)))*(4^(1/4))^2*(((-2)*(-1)^(1/2)+11)^(1/4))^2+((488944351

673126220703125000*x^17+3352761268615722656250000000*x^15+(-13653188943862915

039062500000)*x^13+12353062629699707031250000000*x^11+(-882893800735473632812

50000)*x^9+(-3137230873107910156250000000)*x^7+748246908187866210937500000*x^

5+7510185241699218750000000*x^3+(-5058944225311279296875000)*x)*(-1)^(1/2)+(1

676380634307861328125000000*x^17+(-6798654794692993164062500000)*x^15+4079192

876815795898437500000*x^13+8519738912582397460937500000*x^11+(-95494091510772

70507812500000)*x^9+1992255449295043945312500000*x^7+412732362747192382812500

000*x^5+(-104635953903198242187500000)*x^3+1281499862670898437500000*x)))*(x^

3+(-1)*x)^(1/2)+(((128056854009628295898437500*x^19+(-50058588385581970214843

7500)*x^17+9313225746154785156250000*x^15+1659616827964782714843750000*x^13+(

-1897476613521575927734375000)*x^11+522278249263763427734375000*x^9+139370560

646057128906250000*x^7+(-64030289649963378906250000)*x^5+33043324947357177734

37500*x^3+152736902236938476562500*x)*(-1)^(1/2)+(23283064365386962890625000*

x^19+(-535510480403900146484375000)*x^17+1732259988784790039062500000*x^15+(-

1598149538040161132812500000)*x^13+(-192597508430480957031250000)*x^11+860676

169395446777343750000*x^9+(-301808118820190429687500000)*x^7+6645917892456054

687500000*x^5+5342066287994384765625000*x^3+(-141561031341552734375000)*x))*4

^(1/4)*(((-2)*(-1)^(1/2)+11)^(1/4))^3+((5820766091346740722656250*x^20+(-3492

45965480804443359375000)*x^18+2005835995078086853027343750*x^16+(-32074749469

75708007812500000)*x^14+960472971200942993164062500*x^12+14509409666061401367

18750000*x^10+(-1016844063997268676757812500)*x^8+131219625473022460937500000

*x^6+19775703549385070800781250*x^4+(-2406537532806396484375000)*x^2+(-186264

5149230957031250))*(-1)^(1/2)+((-32014213502407073974609375)*x^20+46566128730

7739257812500000*x^18+(-554719008505344390869140625)*x^16+(-16838312149047851

56250000000)*x^14+3611529245972633361816406250*x^12+(-19629001617431640625000

00000)*x^10+(-26958063244819641113281250)*x^8+226378440856933593750000000*x^6

+(-32397918403148651123046875)*x^4+(-268220901489257812500000)*x^2+1024454832

0770263671875))*(4^(1/4))^3*((-2)*(-1)^(1/2)+11)^(1/4)))/(1600000*x^20+(-3200

000)*x^18+4160000*x^16+(-3584000)*x^14+2380800*x^12+(-1193984)*x^10+476160*x^

8+(-143360)*x^6+33280*x^4+(-5120)*x^2+512))+((-1)*(4^(1/4))^3*((-2)*(-1)^(1/2

)+11)^(1/4)*log(((((75669959187507629394531250*x^18+(-66938810050487518310546

8750)*x^16+1261942088603973388671875000*x^14+95926225185394287109375000*x^12+

(-1562852412462234497070312500)*x^10+863026827573776245117187500*x^8+(-333264

47010040283203125000)*x^6+(-41164457798004150390625000)*x^4+34589320421218872

07031250*x^2+31664967536926269531250)*(-1)^(1/2)+((-5238689482212066650390625

0)*x^18+29103830456733703613281250*x^16+1150183379650115966796875000*x^14+(-2

308748662471771240234375000)*x^12+1036841422319412231445312500*x^10+378187745

809555053710937500*x^8+(-286556780338287353515625000)*x^6+2849847078323364257

8125000*x^4+2047047019004821777343750*x^2+(-57741999626159667968750)))*(4^(1/

4))^2*(((-2)*(-1)^(1/2)+11)^(1/4))^2+((488944351673126220703125000*x^17+33527

61268615722656250000000*x^15+(-13653188943862915039062500000)*x^13+1235306262

9699707031250000000*x^11+(-88289380073547363281250000)*x^9+(-3137230873107910

156250000000)*x^7+748246908187866210937500000*x^5+7510185241699218750000000*x

^3+(-5058944225311279296875000)*x)*(-1)^(1/2)+(1676380634307861328125000000*x

^17+(-6798654794692993164062500000)*x^15+4079192876815795898437500000*x^13+85

19738912582397460937500000*x^11+(-9549409151077270507812500000)*x^9+199225544

9295043945312500000*x^7+412732362747192382812500000*x^5+(-1046359539031982421

87500000)*x^3+1281499862670898437500000*x)))*(x^3+(-1)*x)^(1/2)+((((-12805685

4009628295898437500)*x^19+500585883855819702148437500*x^17+(-9313225746154785

156250000)*x^15+(-1659616827964782714843750000)*x^13+189747661352157592773437

5000*x^11+(-522278249263763427734375000)*x^9+(-139370560646057128906250000)*x

^7+64030289649963378906250000*x^5+(-3304332494735717773437500)*x^3+(-15273690

2236938476562500)*x)*(-1)^(1/2)+((-23283064365386962890625000)*x^19+535510480

403900146484375000*x^17+(-1732259988784790039062500000)*x^15+1598149538040161

132812500000*x^13+192597508430480957031250000*x^11+(-860676169395446777343750

000)*x^9+301808118820190429687500000*x^7+(-6645917892456054687500000)*x^5+(-5

342066287994384765625000)*x^3+141561031341552734375000*x))*4^(1/4)*(((-2)*(-1

)^(1/2)+11)^(1/4))^3+(((-5820766091346740722656250)*x^20+34924596548080444335

9375000*x^18+(-2005835995078086853027343750)*x^16+320747494697570800781250000

0*x^14+(-960472971200942993164062500)*x^12+(-1450940966606140136718750000)*x^

10+1016844063997268676757812500*x^8+(-131219625473022460937500000)*x^6+(-1977

5703549385070800781250)*x^4+2406537532806396484375000*x^2+1862645149230957031

250)*(-1)^(1/2)+(32014213502407073974609375*x^20+(-46566128730773925781250000

0)*x^18+554719008505344390869140625*x^16+1683831214904785156250000000*x^14+(-

3611529245972633361816406250)*x^12+1962900161743164062500000000*x^10+26958063

244819641113281250*x^8+(-226378440856933593750000000)*x^6+3239791840314865112

3046875*x^4+268220901489257812500000*x^2+(-10244548320770263671875)))*(4^(1/4

))^3*((-2)*(-1)^(1/2)+11)^(1/4)))/(1600000*x^20+(-3200000)*x^18+4160000*x^16+

(-3584000)*x^14+2380800*x^12+(-1193984)*x^10+476160*x^8+(-143360)*x^6+33280*x

^4+(-5120)*x^2+512))+((-1)*(-1)^(1/2)*(4^(1/4))^3*((-2)*(-1)^(1/2)+11)^(1/4)*

log((((((-75669959187507629394531250)*x^18+669388100504875183105468750*x^16+(

-1261942088603973388671875000)*x^14+(-95926225185394287109375000)*x^12+156285

2412462234497070312500*x^10+(-863026827573776245117187500)*x^8+33326447010040

283203125000*x^6+41164457798004150390625000*x^4+(-3458932042121887207031250)*

x^2+(-31664967536926269531250))*(-1)^(1/2)+(52386894822120666503906250*x^18+(

-29103830456733703613281250)*x^16+(-1150183379650115966796875000)*x^14+230874

8662471771240234375000*x^12+(-1036841422319412231445312500)*x^10+(-3781877458

09555053710937500)*x^8+286556780338287353515625000*x^6+(-28498470783233642578

125000)*x^4+(-2047047019004821777343750)*x^2+57741999626159667968750))*(4^(1/

4))^2*(((-2)*(-1)^(1/2)+11)^(1/4))^2+((488944351673126220703125000*x^17+33527

61268615722656250000000*x^15+(-13653188943862915039062500000)*x^13+1235306262

9699707031250000000*x^11+(-88289380073547363281250000)*x^9+(-3137230873107910

156250000000)*x^7+748246908187866210937500000*x^5+7510185241699218750000000*x

^3+(-5058944225311279296875000)*x)*(-1)^(1/2)+(1676380634307861328125000000*x

^17+(-6798654794692993164062500000)*x^15+4079192876815795898437500000*x^13+85

19738912582397460937500000*x^11+(-9549409151077270507812500000)*x^9+199225544

9295043945312500000*x^7+412732362747192382812500000*x^5+(-1046359539031982421

87500000)*x^3+1281499862670898437500000*x)))*(x^3+(-1)*x)^(1/2)+(((2328306436

5386962890625000*x^19+(-535510480403900146484375000)*x^17+1732259988784790039

062500000*x^15+(-1598149538040161132812500000)*x^13+(-19259750843048095703125

0000)*x^11+860676169395446777343750000*x^9+(-301808118820190429687500000)*x^7

+6645917892456054687500000*x^5+5342066287994384765625000*x^3+(-14156103134155

2734375000)*x)*(-1)^(1/2)+((-128056854009628295898437500)*x^19+50058588385581

9702148437500*x^17+(-9313225746154785156250000)*x^15+(-1659616827964782714843

750000)*x^13+1897476613521575927734375000*x^11+(-522278249263763427734375000)

*x^9+(-139370560646057128906250000)*x^7+64030289649963378906250000*x^5+(-3304

332494735717773437500)*x^3+(-152736902236938476562500)*x))*4^(1/4)*(((-2)*(-1

)^(1/2)+11)^(1/4))^3+((32014213502407073974609375*x^20+(-46566128730773925781

2500000)*x^18+554719008505344390869140625*x^16+1683831214904785156250000000*x

^14+(-3611529245972633361816406250)*x^12+1962900161743164062500000000*x^10+26

958063244819641113281250*x^8+(-226378440856933593750000000)*x^6+3239791840314

8651123046875*x^4+268220901489257812500000*x^2+(-10244548320770263671875))*(-

1)^(1/2)+(5820766091346740722656250*x^20+(-349245965480804443359375000)*x^18+

2005835995078086853027343750*x^16+(-3207474946975708007812500000)*x^14+960472

971200942993164062500*x^12+1450940966606140136718750000*x^10+(-10168440639972

68676757812500)*x^8+131219625473022460937500000*x^6+1977570354938507080078125

0*x^4+(-2406537532806396484375000)*x^2+(-1862645149230957031250)))*(4^(1/4))^

3*((-2)*(-1)^(1/2)+11)^(1/4)))/(1600000*x^20+(-3200000)*x^18+4160000*x^16+(-3

584000)*x^14+2380800*x^12+(-1193984)*x^10+476160*x^8+(-143360)*x^6+33280*x^4+

(-5120)*x^2+512))+(-1)^(1/2)*(4^(1/4))^3*((-2)*(-1)^(1/2)+11)^(1/4)*log((((((

-75669959187507629394531250)*x^18+669388100504875183105468750*x^16+(-12619420

88603973388671875000)*x^14+(-95926225185394287109375000)*x^12+156285241246223

4497070312500*x^10+(-863026827573776245117187500)*x^8+33326447010040283203125

000*x^6+41164457798004150390625000*x^4+(-3458932042121887207031250)*x^2+(-316

64967536926269531250))*(-1)^(1/2)+(52386894822120666503906250*x^18+(-29103830

456733703613281250)*x^16+(-1150183379650115966796875000)*x^14+230874866247177

1240234375000*x^12+(-1036841422319412231445312500)*x^10+(-3781877458095550537

10937500)*x^8+286556780338287353515625000*x^6+(-28498470783233642578125000)*x

^4+(-2047047019004821777343750)*x^2+57741999626159667968750))*(4^(1/4))^2*(((

-2)*(-1)^(1/2)+11)^(1/4))^2+((488944351673126220703125000*x^17+33527612686157

22656250000000*x^15+(-13653188943862915039062500000)*x^13+1235306262969970703

1250000000*x^11+(-88289380073547363281250000)*x^9+(-3137230873107910156250000

000)*x^7+748246908187866210937500000*x^5+7510185241699218750000000*x^3+(-5058

944225311279296875000)*x)*(-1)^(1/2)+(1676380634307861328125000000*x^17+(-679

8654794692993164062500000)*x^15+4079192876815795898437500000*x^13+85197389125

82397460937500000*x^11+(-9549409151077270507812500000)*x^9+199225544929504394

5312500000*x^7+412732362747192382812500000*x^5+(-104635953903198242187500000)

*x^3+1281499862670898437500000*x)))*(x^3+(-1)*x)^(1/2)+((((-23283064365386962

890625000)*x^19+535510480403900146484375000*x^17+(-17322599887847900390625000

00)*x^15+1598149538040161132812500000*x^13+192597508430480957031250000*x^11+(

-860676169395446777343750000)*x^9+301808118820190429687500000*x^7+(-664591789

2456054687500000)*x^5+(-5342066287994384765625000)*x^3+1415610313415527343750

00*x)*(-1)^(1/2)+(128056854009628295898437500*x^19+(-500585883855819702148437

500)*x^17+9313225746154785156250000*x^15+1659616827964782714843750000*x^13+(-

1897476613521575927734375000)*x^11+522278249263763427734375000*x^9+1393705606

46057128906250000*x^7+(-64030289649963378906250000)*x^5+330433249473571777343

7500*x^3+152736902236938476562500*x))*4^(1/4)*(((-2)*(-1)^(1/2)+11)^(1/4))^3+

(((-32014213502407073974609375)*x^20+465661287307739257812500000*x^18+(-55471

9008505344390869140625)*x^16+(-1683831214904785156250000000)*x^14+36115292459

72633361816406250*x^12+(-1962900161743164062500000000)*x^10+(-269580632448196

41113281250)*x^8+226378440856933593750000000*x^6+(-32397918403148651123046875

)*x^4+(-268220901489257812500000)*x^2+10244548320770263671875)*(-1)^(1/2)+((-

5820766091346740722656250)*x^20+349245965480804443359375000*x^18+(-2005835995

078086853027343750)*x^16+3207474946975708007812500000*x^14+(-9604729712009429

93164062500)*x^12+(-1450940966606140136718750000)*x^10+1016844063997268676757

812500*x^8+(-131219625473022460937500000)*x^6+(-19775703549385070800781250)*x

^4+2406537532806396484375000*x^2+1862645149230957031250))*(4^(1/4))^3*((-2)*(

-1)^(1/2)+11)^(1/4)))/(1600000*x^20+(-3200000)*x^18+4160000*x^16+(-3584000)*x

^14+2380800*x^12+(-1193984)*x^10+476160*x^8+(-143360)*x^6+33280*x^4+(-5120)*x

^2+512)))))/80"

--Nasser

Jun 19, 2022, 3:41:34 AM6/19/22

to

Let me try again

setSimplifyDenomsFlag(true)

r=integrate(x/(x^2 - 1/5 - 2*sqrt(-1)/5)/sqrt(x^3 - x),x);

There are 9 exposed and 6 unexposed library operations named -

having 2 argument(s) but none was determined to be applicable.

Use HyperDoc Browse, or issue

)display op -

to learn more about the available operations. Perhaps

package-calling the operation or using coercions on the arguments

will allow you to apply the operation.

Cannot find a definition or applicable library operation named -

with argument type(s)

Polynomial(Fraction(Integer))

AlgebraicNumber

Perhaps you should use "@" to indicate the required return type,

or "$" to specify which version of the function you need.

So I get same error as you do on the pre-release 1.3.8 Fricas
having 2 argument(s) but none was determined to be applicable.

Use HyperDoc Browse, or issue

)display op -

to learn more about the available operations. Perhaps

package-calling the operation or using coercions on the arguments

will allow you to apply the operation.

Cannot find a definition or applicable library operation named -

with argument type(s)

Polynomial(Fraction(Integer))

AlgebraicNumber

Perhaps you should use "@" to indicate the required return type,

or "$" to specify which version of the function you need.

--Nasser

Jun 19, 2022, 9:15:44 AM6/19/22

to

But that is the same integral, Nasser.

Jun 19, 2022, 9:56:31 AM6/19/22

to

--

Waldek Hebisch

Jun 19, 2022, 4:49:49 PM6/19/22

to

On 6/19/2022 8:15 AM, Валерий Заподовников wrote:

> But that is the same integral, Nasser.

You are right. But they are written differently. Since one gave an error
> But that is the same integral, Nasser.

and not the second, I assumed they are different and I thought I copied

the wrong one from somewhere else.

This looks like bug in Fricas parser in this case?

(1) -> setSimplifyDenomsFlag(true)

(2) -> r=integrate(x/((x^2 - ((1 + 2*sqrt(-1))/5))*sqrt(x^3 - x)), x);

Type: Equation(Expression(Integer))

(3) -> r2=integrate(x/(x^2 - 1/5 - 2*sqrt(-1)/5)/sqrt(x^3 - x),x);

There are 9 exposed and 6 unexposed library operations named -

having 2 argument(s) but none was determined to be applicable.

Use HyperDoc Browse, or issue

)display op -

to learn more about the available operations. Perhaps

package-calling the operation or using coercions on the arguments

will allow you to apply the operation.

Cannot find a definition or applicable library operation named -

with argument type(s)

Polynomial(Fraction(Integer))

AlgebraicNumber

Perhaps you should use "@" to indicate the required return type,

or "$" to specify which version of the function you need.

--Nasser
having 2 argument(s) but none was determined to be applicable.

Use HyperDoc Browse, or issue

)display op -

to learn more about the available operations. Perhaps

package-calling the operation or using coercions on the arguments

will allow you to apply the operation.

Cannot find a definition or applicable library operation named -

with argument type(s)

Polynomial(Fraction(Integer))

AlgebraicNumber

Perhaps you should use "@" to indicate the required return type,

or "$" to specify which version of the function you need.

Jun 20, 2022, 3:10:35 AM6/20/22

to

"Nasser M. Abbasi" schrieb:

resolve:

Polynomial(Fraction(Integer)) - AlgebraicNumber

by automatically downgrading to Polynomial(AlgebraicNumber), assuming

that is a valid FriCAS type. Presumably, the integrand would then again

end up as Expression(Integer) and thus be palatable to the integrator.

Martin.

Jun 21, 2022, 1:17:21 PM6/21/22