# Prove that $Y = \frac{X_1+X_2*X_3}{\sqrt{1+X_1^2}}$ obeys normal distribution

given that $X_1, X_2, X_3$ are independent and identically distributed, $X_1 \sim N(0,1)$.

I tried to calculate the cumulative distribution function of Y: \begin{align} P(Y\leq y) &= \underset{\frac{x_1+x_2x_3}{\sqrt{1+x_1^2}}\leq y}{\iiint}f(x_1)f(x_2)f(x_3)\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3\\ &= \int_{-\infty}^\infty \mathrm{d}x_1 \underset{x_2x_3\leq y\sqrt{1+x_1^2}-x_1}{\iint}f(x_1)f(x_2)f(x_3)\mathrm{d}x_2\mathrm{d}x_3\\ &= \cdots \end{align} and then derive the probability density function of $Y$. But I can't figure out the above integral.

Prove that $Y = \frac{X_1+X_2*X_3}{\sqrt{1+X_1^2}}$ obeys normal distribution
Here is a quick Monte Carlo check of the empirical pdf of $Y$: