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Topic: Goursat pseudo-elliptics and the Wolfram Integrator
Replies: 5   Last Post: Dec 18, 2012 12:15 PM

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clicliclic@freenet.de

Posts: 995
Registered: 4/26/08
Re: Goursat pseudo-elliptics and the Wolfram Integrator
Posted: Dec 18, 2012 12:15 PM
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Waldek Hebisch schrieb:
>
> clicliclic@freenet.de wrote:

> > Example 2 (quartic radicand):
> >
> > Integrate[(k*x^2 - 1)/((a*k*x + b)*(b*x + a)
> > *Sqrt[(1 - x^2)*(1 - k^2*x^2)]), x]
> >
> > ... "Mathematica could not find a formula for your integral. Most likely
> > this means that no formula exists." Waouw! Here the elementary
> > antiderivative is:
> >
> > 2/(Sqrt[(a + b)*(a*k + b)]*Sqrt[(a - b)*(a*k - b)])
> > *ArcTanh[Sqrt[(a + b)*(a*k + b)]*Sqrt[(1 - x^2)*(1 - k^2*x^2)]
> > /(Sqrt[(a - b)*(a*k - b)]*(1 - x)*(1 - k*x))]
> >
> > The theory behind these integrals is given in: Edouard Goursat, Note sur
> > quelques int?grales pseudo-elliptiques, Bulletin de la Soci?t?
> > Math?matique de France 15 (1887), 106-120, on-line at:
> >
> > <http://www.numdam.org/item?id=BSMF_1887__15__106_1>
> >
> > This was written 125 years ago - apparently too recent for the "Risch"
> > integrator of Mathematica 8. I expect that FriCAS can do the second
> > integral too.

>
> FriCAS result:
>
> integrate((k*x^2 - 1)/((a*k*x + b)*(b*x + a)*sqrt((1 - x^2)*(1 - k^2*x^2))), x)
>


[FriCAS result snipped as it could not be reposted on aioe.org]

>
> There are two alternatives, one in terms of 'log', the other
> (shorter) in terms of 'atan'.
>


The large size of the logarithmic result seems to arise through a
rationalization of the argument denominator; the arc tangent result is
practically identical to what I gave (done manually, with Derive
support).

Would the automatic rational factorization of all polynomial
subexpressions in a final result be too expensive?

Martin.



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