Evaluate $\int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}$ How do you evaluate this integral?

$$\int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}$$

Thanks for any help.

$$
? \equiv \int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}
$$
With $\quad\displaystyle{y = \sqrt{{a^{2} \over 2} + z^{2}\,}\ \tan\left(\theta\right)}$
\begin{align}
?
&=
\int
{\cos\left(\theta\right)
 \over
a^{2}\sin^{2}\left(\theta\right)/2 + z^{2}}\,{\rm d}\theta
=
{1 \over z^{2}}\ {z \over 2^{-1/2}a}\int
{2^{-1/2}a\cos\left(\theta\right)/z
 \over
\left\lbrack 2^{-1/2}a\sin\left(\theta\right)/z \right\rbrack^{2} + 1}\,{\rm d}\theta
\\[3mm]&=
{\sqrt{2} \over az}\,\arctan\left(a\sin\left(\theta\right) \over \sqrt{2}z\right)
=
{\sqrt{2} \over az}\,
\arctan\left({a \over \sqrt{2}z}\,
            {\tan\left(\theta\right) \over \sqrt{\tan^{2}\left(\theta\right) + 1}}\right)
\\[3mm]&={\large
{\sqrt{2} \over az}\,
\arctan\left({a \over \sqrt{2}z}\,
            {y \over \sqrt{y^{2} + a^{2}/2 + z^{2}}}\right)}
\end{align}
$+$ some constant !!!.
