I am trying to compute the inverse Laplace transform of $$F(s)=\frac{1}{\sqrt{[(s+\lambda_3)^2-(\lambda_1-\lambda_2)^2][(s-\lambda_3)^2-(\lambda_1+\lambda_2)^2]}},$$ where we can assume that $\lambda_3>\lambda_2>\lambda_1$.
If the polynomial was of order 2, the inverse Laplace transform would involve a modified Bessel function. However, here I am stuck.
By playing with the integral contour in the complex plane, I managed to rewrite the inverse Laplace transform as $$ f(t)=\int_0^{2\pi}\frac{d\theta}{2\pi}\left(\frac{e^{(\lambda_3+a \cos\theta)t}}{\sqrt{(a \cos\theta+2\lambda_3)^2-b^2}}-\frac{e^{(-\lambda_3+b \cos\theta)t}}{\sqrt{(b \cos\theta-2\lambda_3)^2-a^2}}\right), $$ with $a=\lambda_2+\lambda_3$ and $b=\lambda_2-\lambda_3$. The arguments of the square roots are always positive.
Is there a way to compute this integral ? I would expect some Bessel function to come out of it, but I have no idea how to proceed.
A non-trivial thing is that the polynomial is invariant under exchange of the $\lambda_i$'s, and so should be $f(t)$. This is clearly not explicit in my integral representation, so maybe there is a better representation to use...