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Topic: A bug in Reduce package 'algint'?
Replies: 9   Last Post: Feb 1, 2013 6:02 PM

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clicliclic@freenet.de

Posts: 959
Registered: 4/26/08
Re: A bug in Reduce package 'algint'?
Posted: Jan 22, 2013 12:42 PM
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acer schrieb:
>
> Le lundi 21 janvier 2013 16:16:30 UTC-5, Axel Vogt a écrit :

> > On 21.01.2013 19:55, clicliclic wrote:
> > >
> > > Wanting to refresh my knowledge of the capabilties of the Reduce algebra
> > > system, I have recently browsed the website. The system comes with the
> > > 'algint' package by J. Davenport which boosts the integrator
> > > capabilities for algebraic functions. The package documentation
> > >
> > > <http://www.reduce-algebra.com/docs/algint.pdf>
> > >
> > > introduces the example integrand sqrt(sqrt(a^2 + x^2) + x)/x. A correct
> > > antiderivative for this is
> > >
> > > 2*(sqrt(sqrt(a^2 + x^2) + x)
> > > - sqrt(a)*atanh(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a))
> > > - sqrt(a)*atan(sqrt(sqrt(a^2 + x^2) + x)/sqrt(a)))
> > >
> > > The antiderivative printed in the documentation, however, is either very
> > > wrong or garbled beyond recognition.
> > >

> >
> > Maple 16 returns 2*sqrt(2*x)*hypergeom([-1/4, -1/4, 1/4],[1/2, 3/4],-a^2/x^2)
> >
> > May be right, but not that 'usefull' w.r.t. your result
> >

>
> Just because Axel's brought up Maple (and not wishing to side-track Martin), in Maple 16.02,
>
> expr := sqrt(sqrt(a^2 + x^2) + x)/x:
> p := u = sqrt(a^2 + x^2) + x:
> new := student[changevar](p, Int(expr,x), u):
> sol := eval(value(new),p):
> lprint(sol);
>
> -((2*a^2*((a^2+x^2)^(1/2)+x)^2+a^4+((a^2+x^2)^(1/2)+x)^4)

/((a^2+x^2)^(1/2)+x)^2)^(1/2)
*((a^2+x^2)^(1/2)+x)*4^(1/2)*(-((a^2+x^2)^(1/2)+x)^(1/2)
+a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))
+a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2)))
/(a^2+((a^2+x^2)^(1/2)+x)^2)
>
> other := expand(radnormal(sol),power):
> lprint(other):
>
> 2*((a^2+x^2)^(1/2)+x)^(1/2)-2*a^(1/2)*arctan(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))-2*a^(1/2)*arctanh(((a^2+x^2)^(1/2)+x)^(1/2)/a^(1/2))
>
> What should we expect from the system, automatically?
>


My outside viewpoint: if the integrand is given in terms of elementary
functions, an elementary antiderivative whenever one exists. Even if it
can be expressed more compactly in terms of higher functions. But then
an infinite 3F2 series is not really simple.

The Wolfram Integrator produces an elementary answer, but includes an
unnecessary extra factor. Perhaps this is arrived at by automatic
simplification of 3F2(-1/4, -1/4, 1/4; 1/2, 3/4; -a^2/x^2) ?

How does Maple arrive at 3F2 in the first place? Solution of an ODE?
Series expansion of the integrand?

Side-tracking won't harm my ramblings,

Martin.



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