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Topic: Rubi 4.5 released
Replies: 21   Last Post: Jun 29, 2014 4:29 AM

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 clicliclic@freenet.de Posts: 1,235 Registered: 4/26/08
Re: Rubi 4.5 released
Posted: Jun 22, 2014 8:00 AM

"Nasser M. Abbasi" schrieb:
>
> On 6/21/2014 10:06 AM, clicliclic@freenet.de wrote:

> >
>
> > Integration Example #75 from Chapter 5 of the Timofeev Test Suite ran
> > into a 25-second timeout on Version 4.4. Does Rubi 4.5 resurface
> > eventually, and what is the final result, or error message?
> >

>
> Taking the problem from
>
> http://www.apmaths.uwo.ca/~arich/IndependentTestResults/TimofeevIntegrationProblems.pdf
> page 55:
>
> Rubi 4.5 solves this on my PC in about 5 seconds.
>
> num = (Cos[2 x] - 3 Tan[x]) Cos[x]^3;
> den = (Sin[x]^2 - Sin[2 x]) Sin[2 x]^(5/2);
> Int[num/den, x]
>
> (*The result is too large to post here as is, but after
> simplify, here it is:)
>
> Simplify[%]
>
> -((1/(240*Sqrt[Sin[2*x]]))*((495*Cos[x]*EllipticF[ArcSin[Sqrt[Tan[x/2]]], -1])/
> (Sqrt[Cot[x/2]]*Sqrt[Cos[x]*Sec[x/2]^2]) +
> (1/2)*Sin[x/2]^2*(990*I*Cot[x/2]^(3/2)*
> EllipticPi[-(2/(-1 + Sqrt[5])), I*ArcSinh[Sqrt[Tan[x/2]]], -1]*
> Sqrt[Cos[x]*Sec[x/2]^2] + 990*I*Cot[x/2]^(3/2)*
> EllipticPi[2/(1 + Sqrt[5]), I*ArcSinh[Sqrt[Tan[x/2]]], -1]*
> Sqrt[Cos[x]*Sec[x/2]^2] + (1/8)*Csc[x/2]^4*Sec[x/2]^2*
> (99*Cos[x] - 147*Cos[3*x] + 200*Cos[x]^2*Sin[x]))))
>

Yes, this is the integral I had in mind, where the misprinted COS(x)^2
in Timofeev's book has been corrected to COS(x)^3. Somehow Rubi misses
the fact that the antiderivative is elementary:

INT((COS(2*x) - 3*TAN(x))*COS(x)^3 /
((SIN(x)^2 - SIN(2*x))*SIN(2*x)^(5/2)), x) =
COS(x)/SQRT(SIN(2*x))*(1/20*COT(x)^2 - 5/24*COT(x) - 9/16) +
33/32*ATANH(SQRT(SIN(2*x))/(2*COS(x)))

> > I am curious how far Rubi 4.5 allows one to go with Martin's integrals:
> >
> > <http://mathforum.org/kb/message.jspa?messageID=6477150&tstart=0>
> > <http://mathforum.org/kb/message.jspa?messageID=6872137&tstart=0>
> >

>
> I looked at these links. The integrals you have there are definite
> ones, but Rubi only does indefinite. How should one apply these to
> Rubi?
>
> ---------- from one of the above links--------
>
> int(int((jxx(r*cos(p)-a/2, r*SIN(p), z) * jxx(r*COS(p)+a/2, r*SIN(p),
> z) + jxy(r*cos(p)-a/2, r*SIN(p), z) * jxy(r*COS(p)+a/2, r*SIN(p),
> z))*r, r, 0, inf), p, 0, 2*pi)
> ------------------------------
>

I propose to proceed as Newton and Leibniz would have done: let Rubi do
the inner indefinite integral, form the difference between the values of
the antiderivative at the two endpoints (taking limits as needed), and
repeat this procedure for the outer integral.

It may be easier to use the simplified problem from a later post of this

int(int(1/2*(hxz(r*cos(p)-a/2, r*sin(p), z) * vxz(r*cos(p)+a/2,
r*sin(p), z) + vxz(r*cos(p)-a/2, r*sin(p), z) * hxz(r*cos(p)+a/2,
r*sin(p), z))*r, r, 0, inf), p, 0, 2*pi)

where:

hxz(x,y,z) := 3*x*z/(x^2+y^2+z^2)^(5/2)

vxz(x,y,z) := x*(sqrt(x^2+y^2+z^2) - z)/((x^2+y^2)*sqrt(x^2+y^2+z^2))

Martin.