Distribution of $\frac{Z_1^2 + Z_2^2}{Z_1+Z_2}$ where $Z_1, Z_2$ are standard normals Let $Z_1, Z_2 \sim \mathcal{N}(0,1)$ be i.i.d random variables. I wish to find the distribution of
\begin{align}
 \frac{Z_1^2 + Z_2^2}{Z_1+Z_2} \,.
\end{align}
It is well known that $W = Z_1^2 + Z_2^2 \sim \chi^2_2$ and $Q = Z_1 + Z_2 \sim N(0,2)$. If $W$ and $Q$ were independent, then we might hope to invert this result relating to $t$-distributions, but of course they are not independent since
\begin{align}
Q^2 = W + 2Z_1 Z_2 \,.
\end{align}
Any thoughts on how to proceed are welcome!

One way to proceed is to transform to polar coordinates $(Z_1,Z_2) \mapsto (R,\Theta)$ such that
$$Z_1=R\cos\Theta \quad, \quad Z_2=R\sin\Theta$$
Then,
$$\frac{Z_1^2+Z_2^2}{Z_1+Z_2}=\frac{R}{\cos\Theta+\sin\Theta}=\frac{R}{\sqrt 2\sin\left(\Theta+\frac{\pi}4\right)}$$
Now $R$ and $\Theta$ are independently distributed with $R$ having a Rayleigh distribution and $\Theta$ having a uniform distribution on $(0,2\pi)$. Independence of $R$ and $\Theta$ implies the independence of $R$ and $\sin\left(\Theta+\frac{\pi}4\right)$. So it is theoretically possible to derive this distribution.
