I am looking for a closed form expression for the following integral. One approach that comes to my mind was to use Laplace transform, using the following, I came with the function $f(t)$. What can be the Laplace transform for this function.
$$\int^\infty_0f(r) dr = A\int^\infty_0 \frac{r}{1+Cr^4} e^{-Br^2} dr$$
If I consider $t= r^2$ and $s = B$
$$\int^\infty_0f(r) dr = \frac{A}{2}\int^\infty_0 \frac{1}{1+Ct^2} e^{-st} dt$$
This means now $$f(t) = \frac{1}{1 + Ct^2}$$ I cannot find $\mathcal{L}_{f(t)} (s)$ since it is again complicated. Any assistance would be highly appreciated.
PS: I am concerned with the closed form expression of the integral, if I can get it without Laplace transform, that is suffice for me. I will appreciate any help.