Can $\exp\left(-i\frac{a}{b}\ln(2\cosh(bx))\right)$ be simplified further than $(e^{bx}+e^{-bx})^{-ia/b}$?

Question

Can the following expression be further simplified $$\exp\left(-i\frac{a}{b}\ln(2\cosh(bx))\right)$$ where $a$, $b$ are constant and $x$ is a variable?

I was able to do following simplification :

$$\begin{align} \exp\left(\ln\left[(2\cosh(bx))^{-i \frac{a}{b}}\right]\right) &=(2\cosh(bx))^{-i \frac{a}{b}} \\ &=(e^{bx}+e^{-bx})^{-i \frac{a}{b}} \end{align}$$

Answer 1

Since $$\frac ab\log({2\cosh bx})$$ is always real, provided $a,b,x$ are real, then you can use the identity $$\exp(i\phi)=\cos\phi+i\sin\phi.$$ Afraid it wouldn't go further than that.