I am facing the following integral:
$$I = \int_{\mathbb{R}^3} d^3 x_3 \int_{\mathbb{R}^3} d^3 x_4 \frac{1}{\left| \vec{x}_3 \right| \left| \vec{x}_4 \right|} \frac{1}{\left| \vec{x}_{13} \right| \left| \vec{x}_{24} \right| \left| \vec{x}_{34} \right|} \frac{1}{( \left| \vec{x}_{13} \right| + \left| \vec{x}_{24} \right| + \left| \vec{x}_{34} \right| )^3} \tag{1}$$
with $\vec{x}_{ij}:=\vec{x}_i-\vec{x}_j$, and $\vec{x}_1=(1,0,0)$, $\vec{x}_2=(x_2,y_2,0)$ in the (Cartesian) coordinate system that I used so far. Since the $z$-component of $\vec{x}_1$ and $\vec{x}_2$ are null, I can see that cylinder coordinates could be adequate, but the problem does not simplify much.
Is there another, non-orthodox maybe, coordinate system, in which the integral would turn out to be easier? The reason I am hoping for a positive answer is that the integral is symmetric with respect to $1 \leftrightarrow 2$ as well as for $3 \leftrightarrow 4$, but I am not sure that this brings much.
Note: by the way, the integral is divergent at $\vec{x}_3 = \vec{x}_4 = (0,0,0)$, so I am not expecting to be able to compute the integral until the end. However I am hoping to be able to perform some more integrations (for example the integration over $z_5$ and $z_6$ or similar in another coordinate system) before being stuck, with the divergence being contained in the remaining integral(s).