An integral involving hyperbolic functions I am wondering if it is possible to obtain a closed-form formula for
$$
f(\alpha) = \frac{1}{{\sqrt{2 \pi } \; \alpha }} \int^\infty_{-\infty} x^2 \cosh(x) \; e^{-\frac{\sinh ^2(x)}{2 \alpha ^2}} \mathop{}\!\mathrm{d}x \; .
$$
This integral came up when I tried to calculate the second moment of the random variable
$$
\DeclareMathOperator\arsinh{arsinh}
X = \arsinh(Z \cdot \alpha)
$$
where $Z \sim \mathcal{N}(0, 1)$ is a normally distributed random variable and $\arsinh$ denotes the inverse hyperbolic sine.
Motivation/Context: The above came up as I was investigating whether $\arsinh$ could be a useful variance-stabilizing, non-saturating activation function for artificial neural networks. For more details, see this Reddit thread and the original research cited there.
 A: Using a change of variables, I get 
$$ f(\alpha) = \frac{1}{\sqrt{2\pi} \alpha} \int_0^\infty \frac{\text{arcsinh}^2(\sqrt{x})}{\sqrt{x}} e^{-x/(2\alpha^2)}\; dx $$
so this is basically the Laplace transform of $\text{arcsinh}^2(\sqrt{x})/\sqrt{x}$.  Neither Maple nor Wolfram Alpha find a closed-form solution for this.
A: The small-$\alpha$ asymptotics is $f(\alpha)=\alpha^2$. For larger $\alpha$, you can replace the exponential function $e^{-x/2\alpha^2}$ in Robert Israel's expression by a cutoff at $x=2\alpha^2$, and then the integral can be evaluated in closed form,
$$f(\alpha)\approx \frac{4 \alpha+2 \alpha \,{\rm arcsinh}^2\, (\alpha\sqrt{2})-2 \sqrt{4 \alpha^2+2} \,{\rm arcsinh}\,(\alpha\sqrt{2})}{\sqrt{\pi } \alpha}$$
This is not a bad approximation over a large range of $\alpha$, see plot (gold is exact, blue is approximate):


In response to a comment by Wolfgang, the large-$\alpha$ behaviour of the approximation can be improved by placing the cutoff at $x=b\alpha^2$, with $b$ a bit smaller than $2$. The integral for arbitrary $b$ evaluates to
$$f(\alpha)\approx \frac{2 \sqrt{b} (2+\,{\rm arcsinh}^2\,\sqrt{b})-4 \sqrt{b+1} \,{\rm arcsinh}\,\sqrt{b}}{\sqrt{2 \pi } \alpha}
$$
A quite good agreement with the exact result is obtained for $b=1.7$, see the plot below (again, gold is exact, blue is approximate)

for $\alpha\ll 1$ the asymptotic $f(\alpha)\rightarrow\alpha^2$ is accurate, see plot (the gold and blue curves are almost indistinguishable):

A: Using the standart $\mathrm{arcsin}^2(x)$ Taylor expansion
$$
\arcsin^2(x)=\sum_{n=1}^\infty \frac{2^{2n-1}x^{2n}}{n^2 \binom{2n}{n}}
$$
one can show that $f(\alpha)$ can be expressed as
$$
f(\alpha)=\frac{1}{2}\int_{0}^\infty \frac{\log(1+2t\alpha^2)}{t} e^{-t} dt
$$
Maybe it would be helpful.
