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

 Messages: [ Previous | Next ]
 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: Rubi 4.5 released
Posted: Jun 23, 2014 7:17 AM

"Nasser M. Abbasi" schrieb:
>
> On 6/22/2014 7:01 AM, clicliclic@freenet.de wrote:
>

> > 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))
> >

>
> This is an impossible integrand :) Can't even do the internal one.
> Gave up waiting. Tried Rubi 4.5 and Mathematica 9.01.
>
> 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])
> Clear[r, p, a, z];
> integrand = 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;
>
> integrand = FullSimplify[integrand]
>
> (48*r*z*(a - 2*r*Cos[p])*(a + 2*r*Cos[p])*(-(1/((2*z)/(a^2 + 4*(r^2 + z^2)
> - 4*a*r*Cos[p])^2 + 1/(a^2 + 4*(r^2 + z^2) - 4*a*r*Cos[p])^(3/2)))
> - 1/((2*z)/(a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^2 +
> 1/(a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^(3/2))))/
> ((a^2 + 4*(r^2 + z^2) - 4*a*r*Cos[p])^(5/2)*
> (a^2 + 4*(r^2 + z^2) + 4*a*r*Cos[p])^(5/2))
>
> Int[integrand, r]
> .... no answer after 1/2 hr. Aborted.
>
> Please give a simpler one :)
>

Some assistance to Rubi may be in order. First, all coordinates and
variables are meant to be real, so a,p,r,z should better be declared
real. As mentioned in the original thread, one should further declare
z>0 to preclude the appearance of a pole in vxz(x,y,z) at x^2+y^2 = 0.
Independent of such declarations, the indefinite inner integral:

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)

is fully equivalent to:

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^2 + 4*a*c*r + 4*r^2)
*(a^2 - 4*a*c*r + 4*r^2))/((a^2 + 4*a*c*r + 4*r^2)
*(a^2 - 4*a*c*r + 4*r^2)*(4*z^2 + a^2 + 4*a*c*r + 4*r^2)^(5/2)
*(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)
+ 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)

where the abbreviation c = COS(p) is used; one should therefore declare
-1 < c < 1 as well. Now, since the radicands 4*z^2 + a^2 +- 4*a*c*r +
4*r^2 are positive for -1 < c < 1, the inner integral also equals:

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.

Martin.