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.
 A: 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}
A: 
Let's define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the double integral
$$\mathcal{I}{\left(a,b,c\right)}:=\iint_{\left[-a,a\right]\times\left[-b,b\right]}\mathrm{d}x\,\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}.$$

Suppose $\left(a,b,c\right)\in\mathbb{R}_{>0}^{3}$. As you've already found, the double integral can be boiled down to the following single-variable integral:
$$\begin{align}
\mathcal{I}{\left(a,b,c\right)}
&=\iint_{\left[-a,a\right]\times\left[-b,b\right]}\mathrm{d}x\,\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}\\
&=\int_{-a}^{a}\mathrm{d}x\int_{-b}^{b}\mathrm{d}y\,\frac{1}{\left(x^{2}+y^{2}+c^{2}\right)^{3/2}}\\
&=\int_{-a}^{a}\mathrm{d}x\int_{-b}^{b}\mathrm{d}y\,\frac{d}{dy}\left[\frac{y}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+y^{2}+c^{2}}}\right]\\
&=\int_{-a}^{a}\mathrm{d}x\,\frac{2b}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}}\\
&=2\int_{0}^{a}\mathrm{d}x\,\frac{2b}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}};~~~\small{even\,symmetry}\\
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+b^{2}+c^{2}}}\\
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+q^{2}}};~~~\small{\left[q:=\sqrt{b^{2}+c^{2}}>0\right]}.\\
\end{align}$$
Consider the substitution $\frac{x}{\sqrt{x^{2}+q^{2}}}=t$. Then, $x=\frac{qt}{\sqrt{1-t^{2}}}\implies dx=dt\,\frac{q}{\left(1-t^{2}\right)^{3/2}}$ and
$$\begin{align}
\mathcal{I}{\left(a,b,c\right)}
&=4b\int_{0}^{a}\mathrm{d}x\,\frac{1}{\left(x^{2}+c^{2}\right)\sqrt{x^{2}+q^{2}}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{q}{\left(1-t^{2}\right)^{3/2}}\cdot\frac{1}{\left(\frac{qt}{\sqrt{1-t^{2}}}\right)^{2}+c^{2}}\cdot\frac{\sqrt{1-t^{2}}}{q};~~~\small{\left[x=\frac{qt}{\sqrt{1-t^{2}}}\right]}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{\left(1-t^{2}\right)}\cdot\frac{1}{\left(\frac{q^{2}t^{2}}{1-t^{2}}\right)+c^{2}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{q^{2}t^{2}+c^{2}\left(1-t^{2}\right)}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+q^{2}}}}\mathrm{d}t\,\frac{1}{c^{2}+\left(q^{2}-c^{2}\right)t^{2}}\\
&=4b\int_{0}^{\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}}\mathrm{d}t\,\frac{1}{c^{2}+b^{2}t^{2}}\\
&=\frac{4}{c}\int_{0}^{\frac{ab}{c\sqrt{a^{2}+b^{2}+c^{2}}}}\mathrm{d}u\,\frac{1}{1+u^{2}};~~~\small{\left[t=\frac{cu}{b}\right]}\\
&=\frac{4}{c}\arctan{\left(\frac{ab}{c\sqrt{a^{2}+b^{2}+c^{2}}}\right)}.\blacksquare\\
\end{align}$$

