I have two integrals that I'd like to compare. I've made similar computations and found that they were equal, but this one has me stumped, and the input is too large for Wolframalpha (I'm fine with an estimate of the integrals). Are the following integrals equal (this would be an enormous help)?
The form of the integral is:
$\displaystyle\iiiint \limits_{\varphi,\theta,\phi,\beta \in \left[0, 2\pi\right]} \; \iint \limits_{x,y \in \left(-\infty, \infty\right)} \exp\left(if\left(x,y,\varphi,\theta,\phi,\beta\right)\right)\frac{dx\,dy}{{(1+x^2+y^2)^2}}\,d\varphi\,d\theta\,d\phi\,d\beta\,$.
For the first integral, $\displaystyle f(x,y,\varphi,\theta,\phi,\beta)=\sin{\varphi}+\frac{1}{\sqrt{1+x^2+y^2}}(\sin\theta+x\sin\theta-y\cos\theta)+\frac{\sqrt{2}}{\sqrt{1+x^2+y^2}}(\sin\phi-\sin\beta)-\sqrt{2}\frac{1-x^2-y^2}{1+x^2+y^2}\sin(\phi+\beta)\,$
and for the second integral
$\displaystyle f(x,y,\varphi,\theta,\phi,\beta)= \sin{\varphi}+\frac{1}{\sqrt{1+x^2+y^2}}(\sin\theta+x\sin\theta+y\cos\theta)-\frac{2\sqrt{2}}{(1+x^2+y^2)^{3/2}}\big((x^2-y^2)\sin(\theta+\varphi)+2xy\cos(\theta+\varphi)\big)+\frac{\sqrt{2}}{\sqrt{1+x^2+y^2}}(\sin\phi-\sin\beta)-\sqrt{2}\frac{1-x^2-y^2}{1+x^2+y^2}\sin(\phi+\beta)$
Not a full answer but, for the "second" integral, consider an equivalent variable swap $y\leftrightarrow -y$ such that
$$f_2 = \cdots + \frac{(x+1)\sin\theta \mathbf{-y}\cos\theta}{\sqrt{1+x^2+y^2}}+\cdots$$
that way, we have that the only real difference between the integrals is
$$f_1-f_2 = 2\sqrt{2}\sin^2(a)\cos (a)\sin(\theta+\varphi-2b)$$
under the simplifying substitution
$$\begin{cases}x = \tan (a) \cos (b) \\ y = \tan(a)\sin(b)\end{cases}\implies \frac{dx\:dy}{(1+x^2+y^2)^2} = \frac{1}{2}\sin(2a)\;da\:db$$
Since the only difference between the integrands is the single exponential factor above, it would be best try numerical integration with this change of variables provided, since the range is finite now in every variable.