Inverse Laplace transform of $F(s)=\frac{1}{s\sqrt{s+b\sqrt{as}\tanh{(\sqrt{as})}}}$ using complex integration I want to find the inverse Laplace transform of 
$$F(s)=\frac{1}{s\sqrt{s+b\sqrt{as}\tanh{(\sqrt{as})}}}$$
I tried to use the Bromwich integral
$$f(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{1}{s\sqrt{s+b\sqrt{as}\tanh{(\sqrt{as})}}}e^{st}\,ds$$ 
My progress so far has been stunted by the fact that we have a branch point at s=0. The contour may be like this. But I don't know how to perform the integration.

Any help is appreciated. 
 A: The best I can do is reduce the expression for the ILT to a real integral.  Based on your diagram, I get the following expression for the contour integral about the Bromwich contour:
$$\int_{c-i \infty}^{c+ i \infty} ds \, \frac{e^{s t}}{s^{3/2} \sqrt{1+ a b \frac{\tanh{\sqrt{a s}}}{\sqrt{a s}}}} + e^{i \pi} \int_{\epsilon}^R dx \, \frac{e^{-x t}}{e^{i 3 \pi/2} x^{3/2} \sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\ + e^{-i \pi} \int_R^{\epsilon} dx \, \frac{e^{-x t}}{e^{-i 3 \pi/2} x^{3/2} \sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} + i \epsilon^{-1/2} \int_{\pi}^{-\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}} $$
A few observations.  Note how I rewrote the integrand of the ILT.  The function inside the square root in the denominator is now an even function of $s$, so that there are no further branch points except the factor of $s^{3/2}$ in the denominator.  Also note that the integrals appear singular in $\epsilon$; I will show that the singularities cancel, as they so often do in ILT computations.
By Cauchy's theorem, the contour integral is zero.  Thus, we have as $R \to \infty$:
$$\int_{c-i \infty}^{c+ i \infty} ds \, \frac{e^{s t}}{s^{3/2} \sqrt{1+ a b \frac{\tanh{\sqrt{a s}}}{\sqrt{a s}}}} = -i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{e^{-x t}}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\+ i \epsilon^{-1/2} \int_{-\pi}^{\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}} $$
To see how the singularities in $\epsilon$ cancel, rewrite the RHS as
$$-i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{e^{-x t}-1}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} -i 2 \int_{\epsilon}^{\infty} dx \, x^{-3/2} \frac{1}{\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} \\+ i \epsilon^{-1/2} \int_{\pi}^{-\pi} d\phi \, e^{-i \phi/2} \frac{e^{\epsilon e^{i \phi} t}}{\sqrt{1+ a b \frac{\tanh{\sqrt{a \epsilon e^{i \phi}}}}{\sqrt{a \epsilon e^{i \phi}}}}}$$
Note that the first integral is finite as $\epsilon \to 0$.  In this limit, the second integral approaches
$$- i 2 \epsilon^{-1/2} \int_{1}^{\infty} dy \, y^{-3/2} = -i 4  \epsilon^{-1/2}$$
The third integral approaches
$$i \epsilon^{-1/2} \int_{-\pi}^{\pi} d\phi \, e^{-i \phi/2} = +i 4 \epsilon^{-1/2}$$
Obviously, these cancel.  Higher order expansions in $\epsilon$ vanish as $\epsilon \to 0$, so the ILT may now be written as

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+ i \infty} ds \, \frac{e^{s t}}{s^{3/2} \sqrt{1+ a b \frac{\tanh{\sqrt{a s}}}{\sqrt{a s}}}} = \frac1{\pi} \int_0^{\infty} dx \, \frac{1-e^{-x t}}{x^{3/2}\sqrt{1+ a b \frac{\tan{\sqrt{a x}}}{\sqrt{a x}}}} $$

This is as far as I can take it.  The integral on the RHS is well-defined for all values of $t$ but I imagine the presence of the $\tan$ function may give a numerical integrator some fits.  Good luck!
ADDENDUM
The root in the denominator may also give rise to other branch points in the complex plane.  Calculating these branch point locations seems to me to be quite difficult and maybe it is just as well to assume that $a$ and $b$ are defined such that there are no such branch points in the left half-plane (if that is even possible).
