Difficult trigonometric integral

Here is an outline of the approach I have taken to solve this integral.

First rewrite the integral $(1)$ in Cartesian variables:

$$I=\int_{-\infty}^{\infty} \mathrm{d}^3v~ \frac{3x^2y^2v_z^2}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}\cos(uv_x)\exp\left(-\frac{v_x^2}{2}-\frac{v_y^2}{2}-\frac{v_z^2}{2}\right). $$

Now use the following substitution

$$ \frac{1}{y^2v_x^2+x^2v_y^2+x^2y^2v_z^2}=\int_{0}^{\infty}d\tau~\mathrm{ e^{-(y^2v_x^2+x^2v_y^2+x^2y^2v_z^2)\tau}},$$

such that

$$ I=\int_{0}^{\infty}d\tau\int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-v_x^2(\tau y^2+1/2)-v_y^2(\tau x^2+1/2)-v_z^2(\tau x^2y^2+1/2)}}. $$

The $(v_x,v_y,v_z)$ can be evaluated with the help of Mathematica. The results gives

$$ \int_{-\infty}^{\infty} \mathrm{d}^3v~3x^2y^2v_z^2\cos(uv_x) \mathrm{e^{-\alpha v_x^2-\beta v_y^2-\gamma v_z^2}}=\frac{3\pi^{3/2}}{2x^2y^2}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}.$$

Thereby,

$$ I= -\frac{3~\mathrm{const}}{x^2y^2~\mathrm{const}}\int_{0}^{\infty}\mathrm{d\tau}\frac{\exp\left(-\frac{u^2}{4y^2}\frac{1}{\tau+1/2y^2}\right)}{\left(\tau+1/2x^2y^2\right)^{3/2}\left(\tau+1/2x^2\right)^{1/2}\left(\tau+1/2y^2\right)^{1/2}}. $$

Now, performing the substitution $\tau=\frac{1-x^2}{2x^2y^2k^2}-\frac{1}{2x^2y^2}$ gives us

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

which is the desired integral unless of a constant. ;)

Here is an outline for a possible approach for the first integral: expand $e^{a\cos^2x}=\sum_ka^2\frac{\cos^{2k}x}{k!}$. So, focus on $J_k:=\int_0^{\pi/2}\frac{\cos^{2k}x}{b^2\cos^2x+c^2\sin^2}dx$. Notice that $I_k:=\int_0^{2\pi}\frac{\cos^{2k}x}{b^2\cos^2x+c^2\sin^2}dx=4J_k$. This allows for contour integration over the circle $\vert z\vert=1$, using complex analysis, by letting $z=e^{ix}$: $$I_k=\frac1{4^{n-1}i}\int_{\vert z\vert=1}\frac{(z^2+1)^{2n}}{(\alpha z^2+\beta)(\beta z^2+\alpha)z^{2n-1}}\,dz$$ where $\alpha=b+c$ and $\beta=b-c$.

I would leave the details to you.