$$\frac1{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} ds \, \frac{e^{-\sqrt{a s}}}{c s+s^{3/2}} \cos{\sqrt{b s}} \, e^{s t}$$
where $a, b$ and $c$ constant. To evaluate this, I used Cauchy's theorem and considered the contour integral, but I think I have miscalculation, I could not find my mistake.
This is an inverse Laplace transform. We assume $t \gt 0$; the ILT is zero when $t \lt 0$.
To evaluate, we consider the contour integral
$$\oint_C \frac{dz}{z} \frac{e^{-\sqrt{a z}}}{c+\sqrt{z}} \cos{\sqrt{b z}} e^{t z} $$
where $C$ is a Bromwich contour which is detoured above and below the negative real axis and around the origin, as we define a branch cut of the square root there. We define the branch cut by setting $-1=e^{i \pi}$ above the cut and $-1=e^{-i \pi}$ below the cut.
I will leave the details to the reader for now, but I get as a final result after applying Cauchy's theorem
$$\frac{c}{i 2 \pi} \int_{\gamma-i \infty}^{\gamma+i \infty} \frac{ds}{s} \frac{e^{-\sqrt{a s}}}{c+\sqrt{s}} \cos{\sqrt{b s}} e^{t s} = \\1-\frac1{\pi} \int_{-\infty}^{\infty} \frac{dx}{1+x^2} \left [\frac{\sin{\left (\sqrt{a} c x\right )}}{x} + \cos{\left (\sqrt{a} c x\right )} \right ] \cosh{\left (\sqrt{b} c x\right )} e^{-t c^2 x^2}$$
I will work on evaluating the integral on the RHS as time permits. I suspect the final answer involves error functions of complex arguments, but I cannot say for sure right now.