Inverse Laplace Transform of $\frac{\omega n^2}{s((s+z\omega n)^2+\omega n^2(1-z^2))}$

I've to solve the inverse Laplace transform of the following function, and am unable to find a good starting point. Any hints in the right direction would be appreciated. Thanks.

$$\mathcal{L}^{-1}\left[\frac{\omega n^2}{s\left((s+z\omega n)^2)+\omega n^2(1-z^{2})\right)}\right]$$

Attempted Solution:

Breaking down the original function into Partial function as under. However, the process seems rather tedious. Is there some elegant way around it?

\begin{aligned}\mathcal{L}^{-1}\left[\frac{a}{s((s+b)^{2}+c^2)}\right]&=\mathcal{L}^{-1}\left[\frac{A}{s}+\frac{Bs+C}{(s+b)^{2}+c^{2}}\right] \end{aligned}

The command of Mathematica 13.1

InverseLaplaceTransform[\[Omega]*n^2/s/((s+ z*\[Omega]*n)^2 + \[Omega]*n^2*(1 - z^2)), s,x]


results in -\frac{n \omega z e^{x \left(\sqrt{n^2 \omega \left(z^2-1\right)}-n \omega z\right)}-n \omega z e^{x \left(-\sqrt{n^2 \omega \left(z^2-1\right)}-n \omega z\right)}+\sqrt{n^2 \omega \left(z^2-1\right)} e^{x \left(-\sqrt{n^2 \omega \left(z^2-1\right)}-n \omega z\right)}+\sqrt{n^2 \omega \left(z^2-1\right)} e^{x \left(\sqrt{n^2 \omega \left(z^2-1\right)}-n \omega z\right)}-2 \sqrt{n^2 \omega \left(z^2-1\right)}}{2 \left(\omega z^2-z^2+1\right) \sqrt{n^2 \omega \left(z^2-1\right)}}.

Addition. The command of Maple

inttrans['invlaplace'](Omega*n^2/(s*((Omega*n*z+s)^2+Omega*n^2*(-z^2+1))), s, x);


performs (1-exp(-x*z*Omega*n)*(cosh(x*sqrt(Omega*n^2*(z^2-1)))*n*(z^2-1)+sqrt(Omega*n^2*(z^2-1))*z*sinh(x*sqrt(Omega*n^2*(z^2-1))))/(n*(z^2-1)))/(Omega*z^2-z^2+1) which is the same after some transforms.

• @Moo: You are right. I fixed my answer. Oct 7, 2022 at 15:15