Solving the Integral $int_0^{infty} , mathrm{arcsinh} left(frac{a}{sqrt{x^2+y^2}} right) , mathrm{cos}(b , x) ,mathrm{d}x$

Is there any possibily to solve the following integral

$$\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$$

with $a>0$, $y>0$ and $-\pi/2<\mathrm{arg}(b)<0$.

I assume, the result is connected to Bessel and Struve functions. Thank you.

Edit:
Using integration by parts with

$$\int\mathrm{cos}(b \, x)\,\mathrm{d}x=\frac{\mathrm{sin}(b \, x)}{b}$$

$$\frac{\mathrm{d}}{\mathrm{d}x}( \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right))= - \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}}$$

and the limits

$$\lim_{x\to0} \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0$$
$$\lim_{x\to\infty} \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0$$

Gives the integral

$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x$$

if that makes anything simpler...

Edit2:

Mathematica tells me that the last integrand can be presented as the product of three G-functions. Inhere it is said that the integral of the product of three G-functions can be computed under certain restrictions. Sadly it is not mentioned which restrictions. Does anybody know anything about this?

It would be:

$$\frac{1}{x^2+y^2+a^2} = \frac{1}{\sqrt{\pi} \, \sqrt{y^2+a^2}} \,\mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right]$$

$$\frac{1}{x^2+y^2} = \frac{1}{y^2} \,\mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right]$$

$$\mathrm{sin}(b\,x)= \sqrt{\pi} \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{ 0\} \right\},\tfrac{x^2 \, b^2}{4}\right]$$

which finally results in

$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, \mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right] \, \mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right] \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{0 \} \right\},\tfrac{x^2 \, b^2}{4}\right] \, \mathrm{d}x$

The $\mathrm{MeijerG}$ are defined according to Mathematica syntax. Or (I hope I converted this correctly)

$$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}\frac{1}{2}\\ 0 \end{matrix}&\frac{x^2}{y^2+a^2}\end{array}\right) \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}0\\ 0 \end{matrix}&\frac{x^2}{y^2}\end{array}\right)\, G^{1,0}_{0,2}\left(\begin{array}{c|c}\begin{matrix}-\\\frac{1}{2},\, 0\end{matrix}&\frac{x^2 \,b^2}{4}\end{array}\right) \, \mathrm{d}x$$