Distribution of $X\cdot Y +a\cdot X$ for $X,Y$ standardnormal I am searching for the exact or asymptotic CDF of the rv $X\cdot Y +a\cdot X$
with the $X,Y$ independent standard normal rv's. Found nothing till now.
Any hints?
Thanks.
 A: We can compute the characteristic function:
\begin{align}
\mathbb E\left[\exp\left(it(XY+aX)\right)\right]&=
\frac 1{2\pi}\iint_{\mathbb R^2}\exp(itxy)\exp(itax)\exp\left(-(x^2+y^2)/2\right)\mathrm dx\mathrm dy\\
&=\frac 1{\sqrt{2\pi}}\int_{\mathbb R}\exp(itax)\exp(-x^2/2)
\left(\frac1{\sqrt{2\pi}}\int_{\mathbb R}\exp(itxy)
\exp\left(-y^2/2\right)\mathrm dy\right)\mathrm dx.
\end{align}
The term between the parenthesis is the characteristic function of standard normal distribution evaluated at $tx$, hence 
$$\mathbb E\left[\exp\left(it(XY+aX)\right)\right]=\frac 1{\sqrt{2\pi}}\int_{\mathbb R}\exp(itax)\exp\left(-x^2/2\right)
\exp\left(-t^2x^2/2\right)\mathrm dx.$$
Now, rearranging the exponential and using the substitution $u:=\sqrt{1+t^2}x$, we get the wanted characteristic function.
By similar computations, we may deduce the density of $XY+aX$: 
$$f(x)=\frac 1{2\pi}\int_{-\infty}^{+\infty}\frac 1{|s|}\exp\left(-\frac 12\left((x/s)^2+(s-a)^2\right)\right)\mathrm ds;$$
I am not sure the cumulative density function has a simple expression, but maybe this formula can help to find asymptotics.
