This question was also asked here and here.

I have faced some difficulties to do the following integral:

$$ I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\int_{0}^{\infty}dr~r^2\frac{3x^2y^2\cos(u r \sin\theta \cos\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{r^2}{2}} \tag{1}, $$

where $x$, $y$, and $u$ are real positive constants. I tried at least two ways to solve this integral:

- First attempt:

I began to solve the $r$ integral first. By using Mathematica, then

$$ I=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2(1-u^2\sin^2\theta\cos^2\phi)\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}\mathrm e^{-\frac{u^2}{2}\sin^2\theta\cos^2\phi} \tag{2}. $$

After that, I looked for a solution for $\phi$ integral. My best attempt was:

$$ I_\phi(x,y,u,\theta)=\frac{2}{B}\left[B\left(\frac{1}{2}\right)F_1\left(\frac{1}{2},1,-;1;\nu,-\frac{a}{2}\right)-aB\left(\frac{1}{2}\right)F_1\left(\frac{3}{2},1,-;2;\nu,-\frac{a}{2}\right)\right], $$

where $B=x^2\sin^2\theta+x^2y^2\cos^2\theta$, $a=u^2\sin^2\theta$, and $\nu=\frac{x^2-y^2}{x^2+x^2y^2\cot^2\theta}$. In this way, the final results it's something like that:

$$ I= \int_{0}^{\pi} \mathrm d \theta~3x^2y^2\sin\theta \cos^2\theta~ I_\phi(x,y,u,\theta). \tag{3}. $$

Eq. $(3)$ cannot be further simplied in general and is the nal result.

- Second attempt:

To avoid the hypergeometric function $F_1$, I tried to start with the $\phi$ integral. In this case, my initial problem is an integral something like that:

$$ \int_{0}^{2\pi} \mathrm d \phi \frac{\cos(A \cos\phi)}{a^2\cos^2\phi+b^2\sin^2\phi}. \tag{4} $$

This integral $(4)$ can be solved by series (see Vincent's answer and Jack's answer). However those solutions, at least for me, has not a closed form. This is my final step on this second attempt :(

What is the point? It turns out that someone has managed to solve the integral $(1)$, at least the integral in $r$ and $\phi$. The final resuls found by this person was:

$$ I_G=\frac{12 \pi x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2 \exp\left(-\frac{u^2}{2}\frac{x^2k^2}{(1-x^2)(1-k^2)}\right)}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}, $$

where, I belive, $k=\sqrt{1-x^2}\cos\theta$. As you can see in this following code performed in Mathematica

```
IG[x_, y_, u_] :=
Sqrt[Pi/2] NIntegrate[(12 Pi x y)/(1 - x^2)^(3/2)
(v^2 Exp[-(u^2 x^2 v^2)/(2 (1 - x^2) (1 - v^2))])/(Sqrt[1 - v^2] Sqrt[1 - v^2 (1 - y^2)/(1 - x^2)]), {v, 0, Sqrt[1 - x^2]}]
IG[.3, .4, 1]
** 4.53251 **
I[x_, y_, u_] :=
NIntegrate[(r^2 Sin[a] Cos[
u r Sin[a] Cos[b]] 3 x^2 y^2 Cos[a]^2 Exp[-r^2/
2])/((y^2 Cos[b]^2 + x^2 Sin[b]^2) Sin[a]^2 +
x^2 y^2 Cos[a]^2), {r, 0, Infinity}, {a, 0, Pi}, {b, 0, 2 Pi}]
I[.3, .4, 1]
** 4.53251 **
```

the integrals $I$ and $I_G$ are equals. Indeed, since that they emerge from the same physical problem.

**So, my question is: what are the steps applied for that integral $I$ gives the integral $I_G$?**

**Edit**

Since my question was not solved yet, I think it is because it is a tough question, I will show a particular case of the integral $I$, letting $u=0$. I hope with this help you help me.

In this case, the $r$ integral in $(1)$ is trivial and the integral takes the form:

$$ I_P=\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta~\sin\theta\frac{3x^2y^2\cos^2\theta}{(y^2\cos\phi+x^2\sin^2\phi)\sin^2\theta+x^2y^2\cos^2\theta}. \tag{5} $$

The $\phi$ integral can be integrated with the help of Eq. 3.642.1 in Gradstein and Ryzhik's tables of integrals. Thereby, the $I_P$ takes the for:

$$ I_P=3xy\int_{0}^{\pi}d\theta\frac{\sin\theta\cos^2\theta}{\sqrt{1+(x^2-1)\cos^2\theta}\sqrt{1+(y^2-1)\cos^2\theta}}. \tag{6}$$

Now the change of variable $k=\sqrt{1-x^2}\cos\theta$ bring expression $(6)$ to the form

$$ I_P= \frac{(const) x~y}{(1-x^2)^{3/2}}\int_{0}^{\sqrt{1-x^2}} \mathrm dk \frac{k^2}{\sqrt{1-k^2}\sqrt{1-k^2\frac{1-y^2}{1-x^2}}}. $$

Did you notice how $I_G$ and $I_P$ are similar? Do you think a similar approach can be applied to my original problem? Please, let me know.

I've solved this problem applying the Schwinger proper-time substitution: $$\frac{1}{q^2}=\int_{0}^{\infty}\mathrm{d\xi}~\mathrm{e^{-q^2\xi}}. $$