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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
Goursat pseudo-elliptics and the Wolfram Integrator
Posted: Dec 14, 2012 4:47 PM
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I was curious how Mathematica would handle the pseudo-elliptic integrals
recently referred to on this newsgroup. So I plugged two examples into
the Wolfram Integrator (which presumably represents Mathematica 8):

<http://integrals.wolfram.com/index.jsp>

Example 1 (cubic radicand):

Integrate[(k*x^2 - 1)/((a*k*x + b)*(b*x + a)
*Sqrt[x*(1 - x)*(1 - k*x)]), x]

... eeeek! The Integrator replies in terms of incomplete elliptic F,
incomplete elliptic Pi, and the imaginary unit. But the antiderivative
just is:

2/(Sqrt[a*b]*Sqrt[(a + b)*(a*k + b)])
*ArcTan[Sqrt[a*b]*Sqrt[x*(1 - x)*(1 - k*x)]
/(Sqrt[(a + b)*(a*k + b)]*x)]

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 intégrales pseudo-elliptiques, 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>

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. How do Maple and Sympy behave?

Martin.




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