<s>[INST] Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$ I am trying to solve this:
$$\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh x+\frac{ix}{2}\right)\,dx$$

I don't have much ideas about the problem. I thought of writing $\cos x=\dfrac{e^{ix}+e^{-ix}}{2}$ but couldn't proceed after that. 
$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{ -(ab\cosh x-iac \sinh x)-\frac{x}{2}}+e^{ -(ab\cosh x+iac \sinh x)+\frac{x}{2}}\right)\,dx$$
$\displaystyle 
\begin{aligned}
ab\cosh x-iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)-\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\
&=a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)\\
\end{aligned}$
Similarly,
$\displaystyle 
\begin{aligned}
ab\cosh x+iac\sinh x &=a\sqrt{b^2+c^2}\left(\frac{b}{\sqrt{b^2+c^2}}\cos(ix)+\frac{c}{\sqrt{b^2+c^2}}\sin(ix)\right)\\
&=a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)\\
\end{aligned}$
where $\alpha=\arcsin\left(\dfrac{b}{\sqrt{b^2+c^2}}\right)$
So the integral can be simplified to:
$$\Large \frac{1}{2}\int_0^{\infty} \left(e^{-a\sqrt{b^2+c^2}\sin\left(\alpha-ix\right)-\frac{x}{2}}+e^{-a\sqrt{b^2+c^2}\sin\left(\alpha+ix\right)+\frac{x}{2}}\right)\,dx$$
A solution without using contour integration is appreciated. Thanks!
 [/INST] *

*Using parity properties, the integral can be rewritten as
$$I=\frac12\int_{-\infty}^{\infty}e^{-ab\cosh x+iac\sinh x-\frac{x}{2}}dx$$

*Note that $b\cosh x-ic\sinh x=\sqrt{b^2+c^2}\cosh\left(x-i\phi\right)$ with 
$$\cos\phi=\frac{b}{\sqrt{b^2+c^2}},\qquad\sin\phi=\frac{c}{\sqrt{b^2+c^2}}.$$ Hence, shifting $x$ by $i\phi$ [for that we need to assume that $\phi\in(-\pi/2,\pi/2)$], we obtain
$$I=\frac{e^{-i\phi/2}}{2}\int_{-\infty}^{\infty}e^{-a\sqrt{b^2+c^2}\cosh x-\frac{x}{2}}dx$$

*The last integral can be expressed in terms of the Macdonald function $K_{\nu}(r)$, which has integral representation
$$K_{\nu}(r)=\frac12\int_{-\infty}^{\infty}e^{-r\cosh x\pm\nu x}dx.$$
In addition, for $\nu=\frac12$ this function reduces to an elementary one: 
$\displaystyle K_{1/2}(x)=\sqrt{\frac{\pi}{2x}}e^{-x}$.
Therefore the final result is
$$\boxed{\displaystyle I= 
\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}\;\exp\left\{-a\sqrt{b^2+c^2}-\frac{i}2\arctan\frac{c}{b}\right\}}$$
</s>