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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