Inv. Laplace $\frac{1}{s}\frac{1}{\frac{\sqrt{sAB}}{A} \sinh \sqrt{sAB} \frac{C}{\sqrt{sCD}} \sinh \sqrt{s CD}+\cosh{\sqrt{sAB}} \cosh {\sqrt{sCD}} }$

Question

What would be the inverse Laplace of the following function?

$\frac{1}{s}\frac{1}{\frac{\sqrt{sAB}}{A} \sinh \sqrt{sAB} \frac{C}{\sqrt{sCD}} \sinh \sqrt{s CD}+\cosh{\sqrt{sAB}} \cosh {\sqrt{sCD}} }$ where $A,B,C,D>0$

I have approached the problem by using the residue method. First, there's a simple zero pole at $s=0$.

In order to find the analytic roots of the denominator, I set $\sqrt{\frac{C B}{A D}} \sinh \sqrt{s AB} \sinh \sqrt{s CD} + \cosh {\sqrt{s AB}} \cosh {\sqrt{s CD}}=0 $. All the roots are negative real numbers (solved it numerically in Matlab)

By setting $s=-s$ in order to make them positive real, we get the following graph.

Plot of Denominator=$f(s)$

and after some hyperbolic trigonometric identities the expression becomes:

$\left(1+\sqrt{\frac{C B}{A D}}\right) \cos\left[\sqrt{s} \left( \sqrt{AB} + \sqrt{CD} \right) \right] + \left(1-\sqrt{\frac{C B}{A D}}\right) \cos\left[\sqrt{s} \left( \sqrt{AB} - \sqrt{CD} \right) \right] = 0$

What would be the next step to solve this trigonometric equation in the real domain? Or, is this a better method to approach this inverse laplace transform - by using the definition and directly integrating?