can your system handle 4th-root pseudo-elliptics?

[clicl...@freenet.de]

... as promised last year (on Tue, 15 Nov 2016 23:04:20 +0100 in the
thread "FriCAS web interface failure") here are some pseudo-elliptic
integration exercises involving fourth roots. Does your favorite
integrator arrive at the elementary results?

INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x)

INT((a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x)

INT((3 + x^2)/((1 + x^2)*(1 + 6*x^2 + x^4)^(1/4)), x)

INT((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x)

The integrals are taken from or based on papers by Euler dated to 26th
March 1777, which can be found as items E688 and E690 at the Euler
Archive <http://eulerarchive.maa.org>, or on pp. 98-117 and 127-131 of
the Nova Acta Academiae Scientiarum Imperialis Petropolitanae IX (1795
for 1791).

Martin.

[Nasser M. Abbasi]

Hello;

Quick check shows that Maple 2017.2 can't do any. Mathematica 11.1.1
can do the first two and Rubi 12.1 can do second one. Have not
tried Fricas.

But result contain special functions AppellF1 or EllipticF
complex number, which I assume you are looking for only
result with elementary functions.

Mathematica:
-------------

In[16]:= Integrate[(a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x]
Out[16]= (x*(1 - x^2)^(1/4)*(2*a*AppellF1[1/2, 1/4, 1, 3/2, x^2, x^2/2] +
b*x*AppellF1[1, 1/4, 1, 2, x^2, x^2/2]))/(4*(-1 + x^2)^(1/4))

In[17]:= Integrate[(a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x]
Out[17]= (1/2)*a*x*AppellF1[1/2, 1/4, 1, 3/2, -x^2, -(x^2/2)] +
(1/4)*b*x^2*AppellF1[1, 1/4, 1, 2, -x^2, -(x^2/2)]

Rubi
-----

In[13]:= Int[(a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x]
Out[13]= -((b*ArcTan[1 - Sqrt[2]*(1 + x^2)^(1/4)])/Sqrt[2]) + (b*ArcTan[1 + Sqrt[2]*(1 + x^2)^(1/4)])/
Sqrt[2] + (I*a*Sqrt[-x^2]*EllipticPi[-I, ArcSin[(1 + x^2)^(1/4)], -1])/x -
(I*a*Sqrt[-x^2]*EllipticPi[I, ArcSin[(1 + x^2)^(1/4)], -1])/x +
(b*Log[1 - Sqrt[2]*(1 + x^2)^(1/4) + Sqrt[1 + x^2]])/(2*Sqrt[2]) -
(b*Log[1 + Sqrt[2]*(1 + x^2)^(1/4) + Sqrt[1 + x^2]])/(2*Sqrt[2])


--Nasser

[clicl...@freenet.de]

"Nasser M. Abbasi" schrieb:

>
> On 8/28/2017 11:09 AM, clicl...@freenet.de wrote:
> >
> > ... as promised last year (on Tue, 15 Nov 2016 23:04:20 +0100 in the
> > thread "FriCAS web interface failure") here are some pseudo-elliptic
> > integration exercises involving fourth roots. Does your favorite
> > integrator arrive at the elementary results?
> >
> > INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x)
> >
> > INT((a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x)
> >
> > INT((3 + x^2)/((1 + x^2)*(1 + 6*x^2 + x^4)^(1/4)), x)
> >
> > INT((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x)
> >
> > The integrals are taken from or based on papers by Euler dated to
> > 26th March 1777, which can be found as items E688 and E690 at the
> > Euler Archive <http://eulerarchive.maa.org>, or on pp. 98-117 and
> > 127-131 of the Nova Acta Academiae Scientiarum Imperialis
> > Petropolitanae IX (1795 for 1791).
> >
>

> Quick check shows that Maple 2017.2 can't do any. Mathematica 11.1.1
> can do the first two and Rubi 12.1 can do second one. Have not
> tried Fricas.
>
> But result contain special functions AppellF1 or EllipticF
> complex number, which I assume you are looking for only
> result with elementary functions.
>
> Mathematica:
> -------------
>
> In[16]:= Integrate[(a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x]
> Out[16]= (x*(1 - x^2)^(1/4)*(2*a*AppellF1[1/2, 1/4, 1, 3/2, x^2, x^2/2] +
> b*x*AppellF1[1, 1/4, 1, 2, x^2, x^2/2]))/(4*(-1 + x^2)^(1/4))
>
> In[17]:= Integrate[(a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x]
> Out[17]= (1/2)*a*x*AppellF1[1/2, 1/4, 1, 3/2, -x^2, -(x^2/2)] +
> (1/4)*b*x^2*AppellF1[1, 1/4, 1, 2, -x^2, -(x^2/2)]
>
> Rubi
> -----
>
> In[13]:= Int[(a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x]
> Out[13]= -((b*ArcTan[1 - Sqrt[2]*(1 + x^2)^(1/4)])/Sqrt[2]) + (b*ArcTan[1 + Sqrt[2]*(1 + x^2)^(1/4)])/
> Sqrt[2] + (I*a*Sqrt[-x^2]*EllipticPi[-I, ArcSin[(1 + x^2)^(1/4)], -1])/x -
> (I*a*Sqrt[-x^2]*EllipticPi[I, ArcSin[(1 + x^2)^(1/4)], -1])/x +
> (b*Log[1 - Sqrt[2]*(1 + x^2)^(1/4) + Sqrt[1 + x^2]])/(2*Sqrt[2]) -
> (b*Log[1 + Sqrt[2]*(1 + x^2)^(1/4) + Sqrt[1 + x^2]])/(2*Sqrt[2])
>

Hum, not at all breath-taking. Yes, being elementary functions, the
antiderivatives should better be written as such.

I have already tried these integrals on the web-interface version of
FriCAS: the 1st and 2nd do not return within 5 minutes, and the 3rd and
4th result in ">> Error detected within library code: integrate:
implementation incomplete (residue poly has multiple non-linear
factors)".

It seems that FriCAS, provided that many silly bugs like infinite
looping are eradicated, excels at square-root algebraics (although some
results may look awkward), but originally was not meant to deal with 3rd
and 4th roots without supplementary code.

Martin.

[oldk...@gmail.com]

Like your previous examples, FriCAS can solve the first two cases
in elementary functions when a=0, b=0 separately, then do a linear
combination. (with setSimplifyDenomsFlag(true), the result is not
too long and returns in a few seconds).

I know there are problems in FriCAS dealing with integrals with symbolic
parameters, probably because when computing resultants, the symbolic
parameters don't cancel out like numbers, resulting in very big polynomials.

Thanks for your examples and references.

[clicl...@freenet.de]

oldk...@gmail.com schrieb:

However, when the role of a parameter is trivial, such as that of a in
integrate(a/((3 - x^2)*(1 - 3*x^2)^(1/3)), x), one would expect that its
role remains so when it ends up in properly conceived resultants.

Could it be that integrate(b*x/((3 - x^2)*(1 - 3*x^2)^(1/3)), x) and
integrate(b*x/((2 - x^2)*(x^2 - 1)^(1/4)), x) are preprocessed to remove
or simplify the radical? This could explain why the computations are
here not obstructed by the trivial parameter b.

Martin.

[Richard Fateman]

On 8/28/2017 9:30 AM, Nasser M. Abbasi wrote:
>

given this problem:

INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x)

it seems to me to be easily decomposed as

a*INT(1/((2 - x^2)*(x^2 - 1)^(1/4)), x)

+b*INT(x/((2 - x^2)*(x^2 - 1)^(1/4)), x)

Next, assuming that the integrals of actual interest are
not indefinite, but definite between 2 specific constants,
e.g.
INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), {x,1,4})

then this and similar problems can be done by numerical quadrature.

Even if you think that the symbolic form is more general,
if you are going to evaluate it at some value for x, consider whether
it wouldn't be better to evaluate the integral rather than the
elliptic functions.

Of course doing a symbolic integral has a magical
quality to it, but where should system builders spend their time?

RJF

[Albert Rich]

The just released version 4.13 is now available on Rubi's website at http://www.apmaths.uwo.ca/~arich/. Unlike previous versions, Rubi 4.13 is able to find an elementary antiderivative for Martin's first integration exercise (but not yet the rest).

As Richard recommends, Rubi first reduces the problem to the two integrands
1/((2 - x^2) (x^2 - 1)^(1/4)) and x/((2 - x^2) (x^2 - 1)^(1/4)). The obvious substitution u=x^2 makes the latter integration relatively trivial.

Using partial fractions the former integrand can be expanded into two integrands of the form 1/((d+e x) (a+c x^2)^(1/4)). As noted by Martin, in 1777 Leonhard Euler provided the basis for integrating such pseudo-elliptic integrands in his paper indexed E688 in The Euler Archive at http://eulerarchive.maa.org/.

On page 104 of E688, the antiderivative of 1/((1+x) (2 x^2-1)^(1/4)) is given, expressed in modern notation, as

1/2 arctan((2 x^2-1)^(1/4)/(1+x-sqrt(2 x^2-1))) +
1/4*log((1+x+sqrt(2 x^2-1)-(2 x^2-1)^(1/4))/
(1+x+sqrt(2 x^2-1)+(2 x^2-1)^(1/4)))

This formula can be generalized for integrands of the form
1/((d+e x) (a+c x^2)^(1/4)) when c d^2+2 a e^2=0 and a<0 to

1/(2 (-a)^(1/4) e)*
arctan((-1-c x^2/a)^(1/4)/(1-c d x/(2 a e)-sqrt(-1-c x^2/a))) +
1/(4 (-a)^(1/4) e)*
log((1-c d x/(2 a e)+ sqrt(-1-c x^2/a)-(-1-c x^2/a)^(1/4))/
(1-c d x/(2 a e)+ sqrt(-1-c x^2/a)+(-1-c x^2/a)^(1/4)))

Since 2 d/((d^2-e^2 x^2) (a+c x^2)^(1/4)) equals

1/((d+e x) (a+c x^2)^(1/4)) + 1/((d-e x) (a+c x^2)^(1/4)),

using the above formula twice and combining the resulting arctan and log terms gives the antiderivative Rubi 4.13.1 uses for integrands of the form 1/((a+b x^2)^(1/4) (c+d x^2)) when b c-2 a d=0 and a<0. If a is not less than 0, piecewise constant extraction is used to make it so.

So for the antiderivative of (a+b x)/((2-x^2) (x^2-1)^(1/4)) (Martin's first integration exercise), Rubi 4.13.1 returns

-b*ArcTan[(-1+x^2)^(1/4)] +
b*ArcTanh[(-1+x^2)^(1/4)] -
a/(4*Sqrt[2])*
ArcTan[(2*Sqrt[2]*x*(-1+x^2)^(1/4))/(x^2-2*Sqrt[-1+x^2])] +
a/(8*Sqrt[2])*
Log[(x^2+2*Sqrt[2]*x*(-1+x^2)^(1/4)+2*Sqrt[-1+x^2])/
(x^2-2*Sqrt[2]*x*(-1+x^2)^(1/4)+2*Sqrt[-1+x^2])]

I'm curious to know if Leonhard and Martin consider it optimal...


Finally, I must take issue with Richard's statement:

> Even if you think that the symbolic form is more general,
> if you are going to evaluate it at some value for x, consider
> whether it wouldn't be better to evaluate the integral rather
> than the elliptic functions.
>
> Of course doing a symbolic integral has a magical quality
> to it, but where should system builders spend their time?

As Martin makes clear, these exercises are pseudo-elliptic integrals, meaning they can be expressed in terms of elementary, non-elliptic functions. And as the antiderivatives given above show, they are relatively simple expressions easy to evaluate numerically.

Thus in order to take full advantage of THE Fundamental Theorem of Calculus, I think system implementers should spend some time and effort finding antiderivatives that are real and continuous on the real line, at least where the integrand is real and continuous; and of course valid throughout the complex plane...

Albert

[clicl...@freenet.de]

Albert Rich schrieb:

>
> The just released version 4.13 is now available on Rubi's website at
> http://www.apmaths.uwo.ca/~arich/. Unlike previous versions, Rubi
> 4.13 is able to find an elementary antiderivative for Martin's first
> integration exercise (but not yet the rest).
>

> As [oldk1331] recommends, Rubi first reduces the problem to the two

I have the following antiderivative:

INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x) = SQRT(2)*a/4
*(ATANH(SQRT(2)*(x^2 - 1)^(1/4)/x) + ATAN(x/(SQRT(2)*(x^2 - 1)^(1/4))))
+ b*(ATANH(1/(x^2 - 1)^(1/4)) - ATAN((x^2 - 1)^(1/4)))

When the radicand x^2 - 1 is converted to 1 - x^2 or 1 + x^2, the
arguments and prefactors of the arc tangent functions become complex.
One would then prefer to write:

INT((a + b*x)/((2 - x^2)*(1 - x^2)^(1/4)), x)
= a/2*(ATANH(x*(1 - x^2)^(1/4)/(1 + SQRT(1 - x^2)))
+ ATAN((1 - SQRT(1 - x^2))/(x*(1 - x^2)^(1/4))))
+ SQRT(2)*b/2*(ATANH(SQRT(2)*(1 - x^2)^(1/4)/(1 + SQRT(1 - x^2)))
+ ATAN((1 - SQRT(1 - x^2))/(SQRT(2)*(1 - x^2)^(1/4))))

INT((a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x)

= - a/2*(ATANH((1 - SQRT(1 + x^2))/(x*(1 + x^2)^(1/4)))
- ATAN(x*(1 + x^2)^(1/4)/(1 + SQRT(1 + x^2))))
- SQRT(2)*b/2*(ATANH(SQRT(2)*(1 + x^2)^(1/4)/(1 + SQRT(1 + x^2)))
+ ATAN((1 - SQRT(1 + x^2))/(SQRT(2)*(1 + x^2)^(1/4))))

The fourth possible radical, (-1 - x^2)^(1/4), is complex everywhere on
the real line, but has again the simpler type of antiderivative.

The need to introduce more complex arc tangent arguments does not arise
for comparable square-root or cube-root pseudo-elliptics. The same kind
of problem, however, affects example integral 108 in Section 4 of
Timofeev's textbook.

Martin.

[clicl...@freenet.de]

clicl...@freenet.de schrieb:

>
> I have the following antiderivative:
>
> INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x) = SQRT(2)*a/4
> *(ATANH(SQRT(2)*(x^2 - 1)^(1/4)/x) + ATAN(x/(SQRT(2)*(x^2 - 1)^(1/4))))
> + b*(ATANH(1/(x^2 - 1)^(1/4)) - ATAN((x^2 - 1)^(1/4)))
>

In fact, Leonhard Euler derives this version too in §13 - §19 ("Alia
Resolutio ...") of E688. He goes on to establish the equivalence of the
two antiderivatives in §20 - §23 ("Demonstratio Consensus ...").

Martin.

[Albert Rich]

The just released version 4.13.2 of Rubi defines 2 elegant rules that produce optimal antiderivatives for all integrands of the form 1/((a+b x^2)^(1/4) (c+d x^2)) wrt x when b c-2 a d=0. They are available for viewing at http://www.apmaths.uwo.ca/~arich/EulerE688.pdf.

These 2 new rules enable Rubi 4.13.2 to find optimal antiderivatives for all 4 permutations of Martin's first 2 pseudo-elliptic integration exercises.

However, I am curious as to why the arguments of the arctan and arctanh are systematically, and I think unnecessarily, the reciprocals of each other in Martin's expression of the optimal antiderivatives. Could it be due to an artifact of Derive's simplifer?

Albert

[Richard Fateman]

On 9/18/2017 10:53 PM, Albert Rich wrote:
> As Martin makes clear, these exercises are pseudo-elliptic integrals, meaning they can be expressed in terms of elementary, non-elliptic functions. And as the antiderivatives given above show, they are relatively simple expressions easy to evaluate numerically.

There are two atans, an atanh, one log, some fourth roots and square
roots, but
there also seem to be common subexpressions.
This formula would be fairly easy to evaluate
numerically. However, the time to produce that formula symbolically
must also be considered. Possibly add in the time to convert it to
a numerical program (optimized to reduce the redundant computation).
This cost could be amortized across the number of times the
formula is evaluated. If it is evaluated many times, it might
be a winner. That's certainly one argument I have made.

If the process is followed to get one or a few points,
it is less clear. Also sometimes there is an issue, as
Albert points out, that the integration result may be
poorly behaved. I don't know if this correlates 100%
to misbehavior in the integrand.

Here's an example illustrating a hazard where the formula
is a loser. I don't know if Rubi gets this, but Mathematica gets
a formula with 64 distinct terms, each with atan, atanh, some with logs,
and 32 different sines and cosines ...
vs. the simple integrand below..

integrate(1/(x^64+1),x)



details
http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.25.7855

I think we have to keep an open mind on what is best, not knowing
in advance what people will do next with our results. :)
RJF

[clicl...@freenet.de]

Albert Rich schrieb:
>
> [...]

>
> However, I am curious as to why the arguments of the arctan and
> arctanh are systematically, and I think unnecessarily, the reciprocals
> of each other in Martin's expression of the optimal antiderivatives.
> Could it be due to an artifact of Derive's simplifer?
>

Oh, I just took reciprocal ATAN or ATANH arguments as needed to avoid
discontinuities in plots of the antiderivatives on Derive. Since
Derive's definition of ATANH in particular is unusual, the arguments
should be reconsidered for Rubi on Mathematica. ACOT and ACOTH may also
be used.

As Rubi is referring to Euler for this, you may also want to cite his
E695, "Integratio succincta formulae integralis maxime memorabilis ..."
and Legendre's "Traité des Fonctions Elliptiques", Tome I, Chapitre XXVI
(De l'intégrale ...) if dedicated rules are supplied for those integrals
too.

Martin.

PS: Might Rubi be interested in:

INT(x^2/((2 - x^2)*(x^2 - 1)^(3/4)), x) = SQRT(2)/2
*(ATANH(SQRT(2)*(x^2 - 1)^(1/4)/x) - ATAN(x/(SQRT(2)*(x^2 - 1)^(1/4))))

and friends as well?

[Albert Rich]

On Thursday, September 21, 2017 at 7:08:29 AM UTC-10, clicl...@freenet.de wrote:
> As Rubi is referring to Euler for this, you may also want to cite his
> E695, "Integratio succincta formulae integralis maxime memorabilis ..."
> and Legendre's "Traité des Fonctions Elliptiques", Tome I, Chapitre XXVI
> (De l'intégrale ...) if dedicated rules are supplied for those integrals
> too.
>
> Martin.
>
> PS: Might Rubi be interested in:
>
> INT(x^2/((2 - x^2)*(x^2 - 1)^(3/4)), x) = SQRT(2)/2
> *(ATANH(SQRT(2)*(x^2 - 1)^(1/4)/x) - ATAN(x/(SQRT(2)*(x^2 - 1)^(1/4))))
>
> and friends as well?

Well that was relatively easy... As http://www.apmaths.uwo.ca/~arich/EulerE688.pdf now shows, the pair of rules for integrands of the form x^2/((a+b x^2)^(3/4) (c+d x^2)) when b c-2 a d=0 mirror the pair for 1/((a+b x^2)^(1/4) (c+d x^2)). Thanks for your and Euler's help deriving these pseudo-elliptic integration rules.

With the addition of these 2 pair of terminal rules along with some reduction rules, the just released version 4.13.3 of Rubi is able find optimal antiderivatives for all integrands of the form x^m/((a+b x^2)^(n/4) (c+d x^2)^p) when b c-2 a d=0, m is an integer, p=1, and n=1 or 3. (Rules for all integer m, n and p will have to await a future version.)

As Martin pointed out, when m=0 and n=1 or when m=2 and n=3 these integrals are pseudo-elliptic. However for other even values of m, the optimal antiderivatives do seem to require a single, simple elliptic term. Do you concur?

As far as properly citing Euler, Legendre et al as the source of individual rules in Rubi, I will leave that up to math history scholars able to read the original texts.

Albert

[clicl...@freenet.de]

Albert Rich schrieb:

>
> As Martin pointed out, when m=0 and n=1 or when m=2 and n=3 these
> integrals are pseudo-elliptic. However for other even values of m,
> the optimal antiderivatives do seem to require a single, simple
> elliptic term. Do you concur?

I suppose this is most easily answered by means of a capable algebraic
Risch integrator.

>
> As far as properly citing Euler, Legendre et al as the source of
> individual rules in Rubi, I will leave that up to math history
> scholars able to read the original texts.
>

They will find both the Latin original and an English translation of
E695 at the Euler Archive:

<http://eulerarchive.maa.org/docs/originals/E695.pdf>
<http://eulerarchive.maa.org/docs/translations/E695en.pdf>

Martin.

[Albert Rich]

On Friday, September 22, 2017 at 6:58:04 AM UTC-10, clicl...@freenet.de wrote:
> Albert Rich schrieb:
> >
> > As Martin pointed out, when m=0 and n=1 or when m=2 and n=3 these
> > integrals are pseudo-elliptic. However for other even values of m,
> > the optimal antiderivatives do seem to require a single, simple
> > elliptic term. Do you concur?
>
> I suppose this is most easily answered by means of a capable algebraic
> Risch integrator.

For the antiderivative of x^2/((2-x^2)*(x^2-1)^(1/4)) Rubi 4.13.3 gets

ArcTan[x/(Sqrt[2]*(x^2-1)^(1/4))]/Sqrt[2] +
ArcTanh[x/(Sqrt[2]*(x^2-1)^(1/4))]/Sqrt[2] -
2*(1-x^2)^(1/4)/(x^2-1)^(1/4)*EllipticE[ArcSin[x]/2,2]

involving only a single EllipticE function. If this was actually a pseudo-elliptic integral, that would imply EllipticE[ArcSin[x]/2,2] could be expressed in terms of elementary functions. But surely that is not the case(?). Ergo the integral is elliptic.

Albert

[clicl...@freenet.de]

Albert Rich schrieb:

>
> On Friday, September 22, 2017 at 6:58:04 AM UTC-10, clicl...@freenet.de wrote:
> >
> > Albert Rich schrieb:
> > >
> > > As Martin pointed out, when m=0 and n=1 or when m=2 and n=3 these
> > > integrals are pseudo-elliptic. However for other even values of
> > > m, the optimal antiderivatives do seem to require a single, simple
> > > elliptic term. Do you concur?
> >
> > I suppose this is most easily answered by means of a capable
> > algebraic Risch integrator.
>
> For the antiderivative of x^2/((2-x^2)*(x^2-1)^(1/4)) Rubi 4.13.3 gets
>
> ArcTan[x/(Sqrt[2]*(x^2-1)^(1/4))]/Sqrt[2] +
> ArcTanh[x/(Sqrt[2]*(x^2-1)^(1/4))]/Sqrt[2] -
> 2*(1-x^2)^(1/4)/(x^2-1)^(1/4)*EllipticE[ArcSin[x]/2,2]
>
> involving only a single EllipticE function. If this was actually a

> seudo-elliptic integral, that would imply EllipticE[ArcSin[x]/2,2]
> could be expressed in terms of elementary functions. But surely that
> is not the case(?). Ergo the integral is elliptic.
>

I too believe that E(phi, k) := INT(SQRT(1 - k^2*SIN(p)^2), p, 0, phi)
cannot be expressed in terms of elementary function unless k^2 = 0 or
k^2 = 1, but I cannot name a source for this. The Digital Library of
Mathematical Functions at <http://dlmf.nist.gov/> may be a good point
to start digging. For specific values of k^2, however, an on-line proof
can be ordered at:
<http://axiom-wiki.newsynthesis.org/FriCASIntegration?root=FriCAS>
Scroll down to the bottom of the page, enter:

\begin{axiom}
setSimplifyDenomsFlag(true)
integrate(your_elliptic_integrand, your_integration_variable)
\end{axiom}

into the grey text box, and hit the Preview Button. After the screen
has been updated, inspect the result: if your integral is returned
unevaluated, FriCAS claims that it cannot be expressed in terms of
elementary functions.

Martin.

PS: Use lower-case sqrt(), sin(), etc.

[Albert Rich]

According to Mathematica, 1/(2*(1-x^2)^(1/4)) is the derivative of Elliptic[ArcSin[x]/2,2]. At the FriCASIntegration website

\begin{axiom}
setSimplifyDenomsFlag(true)
integrate(1/(1-x^2)^(1/4), x)
\end{axiom}

is unable to find a closed-form antiderivative. So apparently FriCAS concurs it is elliptic.

Albert

[clicl...@freenet.de]

Albert Rich schrieb:

>
> According to Mathematica, 1/(2*(1-x^2)^(1/4)) is the derivative of
> Elliptic[ArcSin[x]/2,2]. At the FriCASIntegration website
>
> \begin{axiom}
> setSimplifyDenomsFlag(true)
> integrate(1/(1-x^2)^(1/4), x)
> \end{axiom}
>
> is unable to find a closed-form antiderivative. So apparently FriCAS
> concurs it is elliptic.
>

While integrals returned unevaluated by FriCAS are thereby declared
non-integrable in elementary terms, occasional error exits or endless
computations have no such implication. One may then only rephrase the
integral and try again.

There have also been bugs causing integrals to be mistakenly returned
unevaluated; thus INT(x/((3*x^2 + 2*SQRT(3) - 3)*SQRT(x^3 - x)), x) was
declared non-integrable in elementary terms by FriCAS version 1.3.0 and
earlier.

Martin.

[Sam Blake]

5 years later... if you split the first two integrals into a sum of two integrals, then Mathematica 13.2 can compute these:

In[7391]:= Integrate[a/((2 - x^2) (x^2 - 1)^(1/4)), x] +
Integrate[(b x)/((2 - x^2) (x^2 - 1)^(1/4)), x]

Out[7391]= (a (-ArcTan[(Sqrt[2] (-1 + x^2)^(1/4))/x] +
ArcTanh[x/(Sqrt[2] (-1 + x^2)^(1/4))]))/(2 Sqrt[2]) +
b (-ArcTan[(-1 + x^2)^(1/4)] + ArcTanh[(-1 + x^2)^(1/4)])

In[7392]:= Integrate[a/((2 + x^2) (1 + x^2)^(1/4)), x] +
Integrate[(b x)/((2 + x^2) (1 + x^2)^(1/4)), x]

Out[7392]= (b (ArcTan[(-1 + Sqrt[1 + x^2])/(Sqrt[2] (1 + x^2)^(1/4))] -
ArcTanh[(Sqrt[2] (1 + x^2)^(1/4))/(1 + Sqrt[1 + x^2])]))/Sqrt[2] +
a (-(1/4) ArcTan[(-x + (1 + x^2)^(1/4))/(1 + x^2)^(1/4)] +
1/4 ArcTan[(x + (1 + x^2)^(1/4))/(1 + x^2)^(1/4)] +
1/4 ArcTanh[(2 x (1 + x^2)^(1/4))/(x^2 + 2 Sqrt[1 + x^2])])

and Mathematica gets the last two integrals in around 8 seconds each:

In[7393]:= Integrate[(3 + x^2)/((1 + x^2) (1 + 6 x^2 + x^4)^(1/4)), x] // Timing

Out[7393]= {8.00801,
ArcTan[(-1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTan[(1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTanh[(-1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTanh[(1 + x)/(1 + 6 x^2 + x^4)^(1/4)]}

In[7394]:= Integrate[(3 - x^2)/((1 - x^2) (1 - 6 x^2 + x^4)^(1/4)), x] // Timing

Out[7394]= {8.30879,
ArcTan[(I + x)/(1 - 6 x^2 + x^4)^(1/4)] -
ArcTan[(1 - 6 x^2 + x^4)^(1/4)/(-I + x)] +
ArcTanh[(I + x)/(1 - 6 x^2 + x^4)^(1/4)] +
ArcTanh[(1 - 6 x^2 + x^4)^(1/4)/(-I + x)]}

the general form of these integrals takes a bit longer and times-out (10 seconds) before IntegrateAlgebraic can compute it:

In[7397]:=
Integrate`IntegrateAlgebraic[((b^2 - 2 a b x + a^2 x^2) (b^2 +
2 a b x + a^2 x^2))/((b^2 + a^2 x^2) (b^4 + 6 a^2 b^2 x^2 +
a^4 x^4)^(3/4)), x] // Timing

Out[7397]= {14.2513,
ArcTan[(b - a x)/(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)]/a +
ArcTan[(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)/(b + a x)]/a -
ArcTanh[(b - a x)/(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)]/a +
ArcTanh[(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)/(b + a x)]/a}