Transform integral to elliptic or hypergeometric forms?

Question

I have two integrals that I suspect can be expressed as elliptic integrals:

$$int_0^{2pi} dphi' frac{1}{( 1 + alpha^2 + beta^2 - 2 beta cos(phi') )^{3/2}} $$
$$int_0^{2pi} dphi' frac{cos(phi')}{( 1 + alpha^2 + beta^2 - 2 beta cos(phi') )^{3/2}}$$

Is there a way to have Mathematica attempt to transform these into elliptic integrals?

Akku14 · Accepted Answer

When doing the indefinite integrals and taking the limits, you get the results very fast.

int1[p_, a_, b_] = Integrate[1/(1 + a^2 + b^2 - 2 b Cos[p])^(3/2), p, 
           Assumptions -> 
               a [Element] Reals && b [Element] Reals && 0 <= p <= 2 Pi]

lim1t = Limit[int1[p, a, b], p -> 2 Pi, Direction -> 1, 
           Assumptions -> a [Element] Reals && b [Element] Reals]

(*    (4 EllipticE[-((4 b)/(a^2 + (-1 + b)^2))])/(Sqrt[
        a^2 + (-1 + b)^2] (a^2 + (1 + b)^2))    *)

lim1b = Limit[int1[p, a, b], p -> 0, Direction -> -1, 
           Assumptions -> a [Element] Reals && b [Element] Reals]

(*    0    *)

int2[p_, a_, b_] = Integrate[Cos[p]/(1 + a^2 + b^2 - 2 b Cos[p])^(3/2), p, 
                     Assumptions -> a [Element] Reals && b [Element] Reals]

lim2t = Limit[int2[p, a, b], p -> 2 Pi, Direction -> 1, 
            Assumptions -> a [Element] Reals && b [Element] Reals]

(*    (2 (1 + a^2 + b^2) EllipticE[-((4 b)/(a^2 + (-1 + b)^2))] - 
       2 (a^2 + (1 + b)^2) EllipticK[-((4 b)/(a^2 + (-1 + b)^2))])/(Sqrt[
       a^2 + (-1 + b)^2] b (a^2 + (1 + b)^2))    *)

lim2b = Limit[int2[p, a, b], p -> 0, Direction -> -1, 
            Assumptions -> a [Element] Reals && b [Element] Reals]

(*    0    *)

So the definite integrals are lim1t and lim2t.

Peeter Joot · Answer

I was able to do this after a manual transformation (related to the original integrals by a factor of $2 (alpha^2 + (1-beta)^2)^{3/2}$) :

$Assumptions = 
  beta >= 0 && alpha [Element] Reals && (1 - beta)^2 + alpha^2 != 0;

i1 = Integrate[ ( 
    1 + 4 beta Sin[u]^2/((1 - beta)^2 + alpha^2))^(-3/2), {u, 0, Pi}]

i2 = Integrate[ 
  Cos[2 u] ( 1 + 4 beta Sin[u]^2/((1 - beta)^2 + alpha^2))^(-3/2), {u,
    0, Pi}]

and then waiting a very long time (around an hour).  The results comes back as not very illuminating sums of EllipticE and EllipticK functions.

Perhaps Mathematica would have given me a similar result for Integrate[]'s of the original integrands?  However, I was too impatient, and Aborted the evaluation, thinking that it was hung.

user64494 · Answer

In version 12.0
Integrate[1/(1 + [Alpha]^2 + [Beta]^2 - 2 [Beta] Cos[[Phi]])^(3/
 2), {[Phi], 0, 2*Pi}, GenerateConditions -> False]

$$frac{4 Eleft(-frac{4 beta }{alpha ^2+(beta -1)^2}right)}{sqrt{alpha ^2+(beta -1)^2} left(alpha ^2+(beta +1)^2right)} $$
Integrate[Cos[[Phi]]/(1 + [Alpha]^2 + [Beta]^2 - 2 [Beta] Cos[[Phi]])^(3/
 2), {[Phi], 0, 2*Pi}, GenerateConditions -> False]

$$frac{2 left(alpha ^2+beta ^2+1right) Eleft(-frac{4 beta }{alpha ^2+(beta -1)^2}right)-2 left(alpha ^2+(beta +1)^2right) Kleft(-frac{4 beta }{alpha ^2+(beta -1)^2}right)}{beta  sqrt{alpha ^2+(beta -1)^2} left(alpha ^2+(beta +1)^2right)} $$

Transform integral to elliptic or hypergeometric forms?

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