Define the function $Q:\mathbb{C}^{2}\rightarrow\mathbb{C}$ to be the binary quadratic form,
$$Q{\left(z,w\right)}:=z^{2}+w^{2}.\tag{1a}$$
Also, define $P:\mathbb{C}^{4}\rightarrow\mathbb{C}$ to be the polynomial of degree $5$, in four variables,
$$\begin{align} P{\left(a,b,x,y\right)} &:=a\left[\left(a^{2}+1\right)\left(a^{2}+b^{2}+1\right)-4b^{2}\right]\\ &~~~~~+2\left[\left(a^{2}+1\right)\left(a^{2}+b^{2}+1\right)-2b^{2}\right]x\\ &~~~~~+a\left(a^{2}+b^{2}+1\right)x^{2}\\ &~~~~~+a\left(a^{2}+b^{2}+1\right)y^{2}.\tag{1b}\\ \end{align}$$
Note that $P{\left(a,b,x,y\right)}$ is obviously even in both $b$ and $y$.
Then, define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the double integral
$$\mathcal{I}{\left(a,b\right)}:=\int_{-\infty}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\frac{2^{4}xy\,P{\left(a,b,x,y\right)}}{Q{\left(a+x,1-y\right)}\,Q{\left(a+x,1+y\right)}\,Q{\left(x,b-y\right)}\,Q{\left(x,b+y\right)}}.\tag{1c}$$
It's not hard to show then that $\mathcal{I}{\left(a,b\right)}$ is even in the second parameter $b$:
$$\mathcal{I}{\left(a,-b\right)}=\mathcal{I}{\left(a,b\right)};~~~\small{\left(a,b\right)\in\mathbb{R}^{2}}.$$
Problem: Given the pair of real parameters $\left(a,b\right)\in\mathbb{R}\times\mathbb{R}_{\ge0}$, find a closed form expression for the double integral $\mathcal{I}{\left(a,b\right)}$ in terms of elementary functions.
The obstacle in the way of solving this problem appears to be tedium more than anything else. Integrating the rational integrand in $(1c)$ over $x$ should in principle yield a piecewise rational function. Thus, subsequent integration over $y$ should lead to a function that is at the very least piecewise elementary, if not simpler.
However, attempting to solve the problem by brute force with partial fraction decompositions quickly leads to large numbers of cumbersome expressions, rending the integral quite unmanageable without a program such as Mathematica.
It is my hope that there is some cleverly efficient approach to this integral that I'm just not seeing at the moment. Any advice here would be welcome. Cheers!