Any neat way to solve the integral $\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$? Straight to the point: given the integral
$$\iint_Q \frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$$
where $Q=[-a,a]\times[-b,b]$, can you think of any neat way to solve it?
At a first glance it looked quite innocent. No need to say that I’ve changed my mind.
EDIT:
Integrating first w.r.t. y, using the integral
$$\int \frac{1}{\left(\xi^2+\alpha^2\right)^{3/2}}\,d\xi = \frac{\xi}{\alpha^2\sqrt{\xi^2+\alpha^2}} + \text{constant}, $$
one gets to
$$ 2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx $$
and here I get pretty stuck to be honest, so if someone could give me any hints that would be appreciated. Still, I think there should be some nicer way to approach this from the start.
Since I was asked for the source of this problem: I simply asked myself a basic physics question, and trying to find an answer led to this integral.

Substitute $x = \sqrt{b^2+z^2}\sinh t$ after integrating over $y$
\begin{align}
&\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy \\
= &2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx \\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\sinh^2 t +z^2}dt\\
 =&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\cosh^2 t -b^2}dt\\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac{d(\tanh t)}{
z^2+b^2\tanh^2 t }\\
 =&\frac 4z \tan^{-1} \left(\frac bz \tanh t\right)\bigg|_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}} }\\
=&\frac 4z \tan^{-1} \left(\frac{ab}{z\sqrt{a^2+b^2+z^2}}\right)
\end{align}
