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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: 988
Registered: 4/26/08
Re: Rubi 4.5 released
Posted: Jun 24, 2014 2:52 AM
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Albert Rich schrieb:
>
> On Monday, June 23, 2014 1:17:12 AM UTC-10, clicl...@freenet.de wrote:
>

> > INT(192*r*z^2*(a^2 - 4*c^2*r^2)*(16*z^4*(a^2 + 4*r^2)
> > + (8*z^2 + a^2 + 4*r^2)*(a^4 + 8*a^2*r^2*(1 - 2*c^2) + 16*r^4))
> > /((a^4 + 8*a^2*r^2*(1 - 2*c^2) + 16*r^4)*((4*z^2 + a^2)^2
> > + 8*r^2*(4*z^2 + a^2*(1 - 2*c^2)) + 16*r^4)^(5/2)), r)
> > + INT(48*r*z*(4*c^2*r^2 - a^2)/((a^2 - 4*a*c*r + 4*r^2)
> > *(4*z^2 + a^2 + 4*a*c*r + 4*r^2)^(5/2)), r)
> > + INT(48*r*z*(4*c^2*r^2 - a^2)/((a^2 + 4*a*c*r + 4*r^2)
> > *(4*z^2 + a^2 - 4*a*c*r + 4*r^2)^(5/2)), r)
> >
> > and served in this form should be found digestible by Rubi.

>
> Yes, Rubi can digest it and returns an antiderivative with a leaf
> count of 4850. Mathematica returns one with a leaf count of 7479.
>


The next step is to extract the definite integral from r=0 to r=inf. If
the antiderivative is continuous for real a, z > 0, -1 < c < 1, the
result should equal:

- 3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) - 4*c^4 + c^2 + 1)*LN(-
2*z*SQRT(z^2 + a^2*c*(c - SQRT(c^2 - 1)))*(4*z^2 + a^2) + 8*z^4 -
2*a^2*z^2*(2*c*SQRT(c^2 - 1) - 2*c^2 - 1) + a^4*c*(SQRT(c^2 - 1)*(2*c^2
- 1) - 2*c^3 + 2*c))/(16*(z^2 + a^2*c*(c - SQRT(c^2 - 1)))^(5/2)) +
3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) + 4*c^4 - c^2 - 1)*LN(-
2*z*SQRT(z^2 + a^2*c*(SQRT(c^2 - 1) + c))*(4*z^2 + a^2) + 8*z^4 +
2*a^2*z^2*(2*c*SQRT(c^2 - 1) + 2*c^2 + 1) - a^4*c*(SQRT(c^2 - 1)*(2*c^2
- 1) + 2*c^3 - 2*c))/(16*(z^2 + a^2*c*(SQRT(c^2 - 1) + c))^(5/2)) +
3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) - 4*c^4 + c^2 + 1)*LN(- 2*SQRT(z^2
+ a^2*c*(c - SQRT(c^2 - 1)))*SQRT(4*z^2 + a^2) + 4*z^2 - a^2*(c*SQRT(c^2
- 1) - c^2 - 1))/(8*(z^2 + a^2*c*(c - SQRT(c^2 - 1)))^(5/2)) -
3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) + 4*c^4 - c^2 - 1)*LN(- 2*SQRT(z^2
+ a^2*c*(SQRT(c^2 - 1) + c))*SQRT(4*z^2 + a^2) + 4*z^2 + a^2*(c*SQRT(c^2
- 1) + c^2 + 1))/(8*(z^2 + a^2*c*(SQRT(c^2 - 1) + c))^(5/2)) +
3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) - 4*c^4 + c^2 + 1)*LN(-
2*z*SQRT(z^2 + a^2*c*(c - SQRT(c^2 - 1))) + 2*z^2 - a^2*c*SQRT(c^2 -
1))/(16*(z^2 + a^2*c*(c - SQRT(c^2 - 1)))^(5/2)) - 3*a^2*z*(c*SQRT(c^2 -
1)*(4*c^2 + 1) - 4*c^4 + c^2 + 1)*LN(a*(SQRT(c^2 - 1) - 2*c) -
2*SQRT(z^2 + a^2*c*(c - SQRT(c^2 - 1))))/(16*(z^2 + a^2*c*(c - SQRT(c^2
- 1)))^(5/2)) - 3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) - 4*c^4 + c^2 +
1)*LN(a*(2*c - SQRT(c^2 - 1)) - 2*SQRT(z^2 + a^2*c*(c - SQRT(c^2 -
1))))/(16*(z^2 + a^2*c*(c - SQRT(c^2 - 1)))^(5/2)) - 3*a^2*z*(c*SQRT(c^2
- 1)*(4*c^2 + 1) + 4*c^4 - c^2 - 1)*LN(- 2*z*SQRT(z^2 + a^2*c*(SQRT(c^2
- 1) + c)) + 2*z^2 + a^2*c*SQRT(c^2 - 1))/(16*(z^2 + a^2*c*(SQRT(c^2 -
1) + c))^(5/2)) + 3*a^2*z*(c*SQRT(c^2 - 1)*(4*c^2 + 1) + 4*c^4 - c^2 -
1)*LN(a*(SQRT(c^2 - 1) + 2*c) - 2*SQRT(z^2 + a^2*c*(SQRT(c^2 - 1) +
c)))/(16*(z^2 + a^2*c*(SQRT(c^2 - 1) + c))^(5/2)) + 3*a^2*z*(c*SQRT(c^2
- 1)*(4*c^2 + 1) + 4*c^4 - c^2 - 1)*LN(- 2*SQRT(z^2 + a^2*c*(SQRT(c^2 -
1) + c)) - a*(SQRT(c^2 - 1) + 2*c))/(16*(z^2 + a^2*c*(SQRT(c^2 - 1) +
c))^(5/2)) - z*(z*SQRT(4*z^2 + a^2)*(96*c^2*z^10 - 24*a^2*z^8*(c^4 -
4*c^2 - 3) + 2*a^4*z^6*(108*c^6 - 101*c^4 - 22*c^2 + 15) +
a^6*z^4*(262*c^6 - 249*c^4 - 64*c^2 + 3) + 8*a^8*c^2*z^2*(c^2 -
1)*(14*c^2 + 3) + a^10*c^2*(c^2 - 1)*(13*c^2 + 3)) - 256*c^2*z^12 +
64*a^2*z^10*(5*c^4 - 6*c^2 - 4) - 16*a^4*z^8*(58*c^6 - 71*c^4 - 6*c^2 +
11) - 4*a^6*z^6*(332*c^6 - 303*c^4 - 83*c^2 + 10) - a^8*z^4*(698*c^6 -
493*c^4 - 176*c^2 + 3) - 2*a^10*c^2*z^2*(71*c^4 - 44*c^2 - 19) -
3*a^12*c^2*(3*c^4 - 2*c^2 - 1))/(2*(z^4 + 2*a^2*c^2*z^2 +
a^4*c^2)^2*(4*z^2 + a^2)^(3/2)*(4*z^2 + a^2*(1 - c^2))^2)

which is real for -1 < c < 1 although the subexpression SQRT(c^2 - 1) is
imaginary (Rubi may prefer real ATANs to my complex LNs). Now c should
be replaced by COS(p) and this result be integrated from p=0 to p=2*pi;
for symmetry reasons it suffices to integrate from p=0 to p=pi and to
double the result, whereby the singularity of SQRT(c^2 - 1) at c^2 = 1
is avoided.

Martin.



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