I have the following integral that needs to be evaluated.
$$ \int_{0}^{2\pi} \int_0^{\frac 1 {\sqrt{4 \cos^2 \theta + 9 \sin^2 \theta}}} \frac r 2 \ln (r^2 + \rho^2 - 2r \rho \cos (\theta - \phi)) dr d\theta $$
I tried using Mathematica but it gives the following output
-((\[Rho] (-2 E^(2 I \[Theta]) Sqrt[
26 - 5 E^(-2 I \[Theta]) -
5 E^(2 I \[Theta])] + (5 - 26 E^(2 I \[Theta]) +
5 E^(4 I \[Theta])) \[Rho] ArcTan[
Cot[\[Theta] - \[Phi]]] Cos[
2 \[Theta] - 2 \[Phi]] Csc[\[Theta] - \[Phi]] - (5 -
26 E^(2 I \[Theta]) + 5 E^(4 I \[Theta])) \[Rho] ArcTan[
Cot[\[Theta] - \[Phi]] - (
Sqrt[2] Csc[\[Theta] - \[Phi]])/(\[Rho] Sqrt[
13 - 5 Cos[2 \[Theta]]])] Cos[
2 \[Theta] - 2 \[Phi]] Csc[\[Theta] - \[Phi]] -
10 \[Rho] Cos[\[Theta] - \[Phi]] Log[\[Rho]] +
52 E^(2 I \[Theta]) \[Rho] Cos[\[Theta] - \[Phi]] Log[\[Rho]] \
- 10 E^(4 I \[Theta]) \[Rho] Cos[\[Theta] - \[Phi]] Log[\[Rho]] +
5 \[Rho] Cos[\[Theta] - \[Phi]] Log[(
5 \[Rho]^2 + 5 E^(4 I \[Theta]) \[Rho]^2 -
2 E^(2 I \[Theta]) (2 + 13 \[Rho]^2))/(
5 - 26 E^(2 I \[Theta]) + 5 E^(4 I \[Theta])) - (
4 \[Rho] Cos[\[Theta] - \[Phi]])/Sqrt[
26 - 5 E^(-2 I \[Theta]) - 5 E^(2 I \[Theta])]] -
26 E^(2 I \[Theta]) \[Rho] Cos[\[Theta] - \[Phi]] Log[(
5 \[Rho]^2 + 5 E^(4 I \[Theta]) \[Rho]^2 -
2 E^(2 I \[Theta]) (2 + 13 \[Rho]^2))/(
5 - 26 E^(2 I \[Theta]) + 5 E^(4 I \[Theta])) - (
4 \[Rho] Cos[\[Theta] - \[Phi]])/Sqrt[
26 - 5 E^(-2 I \[Theta]) - 5 E^(2 I \[Theta])]] +
5 E^(4 I \[Theta]) \[Rho] Cos[\[Theta] - \[Phi]] Log[(
5 \[Rho]^2 + 5 E^(4 I \[Theta]) \[Rho]^2 -
2 E^(2 I \[Theta]) (2 + 13 \[Rho]^2))/(
5 - 26 E^(2 I \[Theta]) + 5 E^(4 I \[Theta])) - (
4 \[Rho] Cos[\[Theta] - \[Phi]])/Sqrt[
26 - 5 E^(-2 I \[Theta]) -
5 E^(2 I \[Theta])]]) Sin[\[Theta] - \[Phi]])/(5 -
26 E^(2 I \[Theta]) + 5 E^(4 I \[Theta])))
Does closed form of the above integral exists? What approaches might I try to solve this integral? Is this possible to express this integral in terms of special functions?