$\int_{0}^{2\pi}\frac{e^{ir\left ( \cos \phi +\sin \phi \right )}}{\cos^2\phi+a^2\sin^2\phi}\,\mathrm d\phi$ $$\int_{0}^{2\pi}\frac{e^{ir\left ( \cos \phi +\sin \phi \right )}}{\cos^2\phi+a^2\sin^2\phi}\,\mathrm d\phi$$
I've spent many days working on this problem. I haven't been able to take this integral
First idea:
Note that the function is even with respect to $r$, and in what follows consider the integral only up to $\pi.$
But I haven't come to anything. It's a dead end again!
Then I thought I could prove that the integral is not expressed in elementary functions, but it didn't work either.
I've done a differential equation, but then there's another integral, which also doesn't take.
 A: I doubt there's a simple closed form. What I get (easily) is a series using Bessel functions.
Taking the following known generating function $$\sum_{n\in\mathbb{Z}}J_n(z)t^n=\exp\left[\frac{z}{2}\left(t-\frac1t\right)\right]\qquad(z,t\in\mathbb{C},\ t\neq0)$$ at $z=r\sqrt2$ and $t=\exp[i(\phi+\pi/4)]$, we get $$e^{ir(\cos\phi+\sin\phi)}=\sum_{n\in\mathbb{Z}}J_n(r\sqrt2)e^{in(\phi+\pi/4)}.$$
The following one (basically a sum of two geometric series) $$\sum_{n\in\mathbb{Z}}z^{|n|}e^{int}=\frac{1-z^2}{1-2z\cos t+z^2}\qquad(|z|<1,t\in\mathbb{R})$$ at $z=(a-1)/(a+1)$ (assuming $a>0$) and $t=2\phi$, becomes $$\frac1{\cos^2\phi+a^2\sin^2\phi}=\frac1a\sum_{n\in\mathbb{Z}}\left(\frac{a-1}{a+1}\right)^{|n|}e^{2in\phi}.$$
Now it's "easy" to integrate the product:
\begin{align*}
\int_0^{2\pi}\frac{e^{ir(\cos\phi+\sin\phi)}\,d\phi}{\cos^2\phi+a^2\sin^2\phi}
&=\frac{2\pi}{a}\sum_{n\in\mathbb{Z}}J_{-2n}(r\sqrt2)e^{i(-2n)\pi/4}\left(\frac{a-1}{a+1}\right)^{|n|}\\&=\frac{2\pi}{a}\left[1+2\sum_{n=1}^\infty(-1)^n\left(\frac{a-1}{a+1}\right)^{2n}J_{4n}(r\sqrt2)\right].\end{align*}
