inverse Laplace transform by finding residues of essential singularities

Question

I want to find the inverse Laplace transform of $$F(s)=\exp\Big(-\sqrt{2s}\tanh(\sqrt{2s})\Big).$$ Despite the square roots, $F$ doesn't have any branch points since $$\sqrt{2s}\tanh(\sqrt{2s})=\frac{\sqrt{2s}\sinh(\sqrt{2s})}{\cosh(\sqrt{2s})}$$ and both the numerator and denominator are seen to be holomorphic by looking at their power series. Hence there are only isolated essential singularities at the zeros of $\cosh(\sqrt{2s})$, namely $$s=-\frac{\pi^2}{8},-\frac{9\pi^2}{8},-\frac{25\pi^2}{8},...$$ Assuming that we have sufficient decay of the modulus of $F$ away from the singularities in the left half-plane, we can use the residue theorem to write the inverse Laplace transform of $F$ as $$f(t)=\sum_{n=1}^\infty\exp\Big(-\frac{(2n-1)^2\pi^2}{8}t\Big)Res\Big(F,-\frac{(2n-1)^2\pi^2}{8}\Big)$$ where $Res\Big(F,-\frac{(2n-1)^2\pi^2}{8}\Big)$ is the residue of $F$ at the essential singularity $-\frac{(2n-1)^2\pi^2}{8}$. However, I don't know how to calculate these essential singularity residues.