The problem I am stuck with is as follows:
$$\int^\infty_0\frac{\exp(-A\sqrt{e^t-1})}{1 + (e^t-1)B} dt$$
considering $g = \sqrt{e^t-1}$, I get $dg= \frac{e^t}{2\sqrt{e^t-1}}$ and $e^t = g^2-1$
$$\int^\infty_0\frac{\exp(-A\sqrt{e^t-1})}{1 + (e^t-1)B} dt$$
$$\int^\infty_0\frac{2g \ \exp{(-Ag)}}{(g^2-1)(1+g^2B)} dg$$
This the maximum I can get, it still is too complex. Any help will be highly appreciated.
I got the answer out of Mathematica which couldn't do it directly, but could after the trick I used below. If $$ I(A,B)=\int^\infty_0\frac{\exp(-A\sqrt{e^t-1})}{1 + (e^t-1)B} dt $$
To get the integration to work we can first take a Mellin transform of the integrand with respect to $A$,
$$ \mathcal{M}_A[I(A,B)](s)=\Gamma(s)\int^\infty_0\frac{(e^t-1)^{-s/2}}{1 + (e^t-1)B} dt $$ this integrates to $$ \mathcal{M}_A[I(A,B)](s)= \frac{(1-B^{s/2})\pi \csc(\pi s/2)\Gamma(s)}{1-B} $$ then taking the inverse Mellin transform of both sides $$ \mathcal{M}^{-1}[\mathcal{M}_A[I(A,B)](s)](A)= \mathcal{M}^{-1}\left[\frac{(1-B^{s/2})\pi \csc(\pi s/2)\Gamma(s)}{1-B}\right](A) $$ $$ I(A,B)= \mathcal{M}^{-1}\left[\frac{(1-B^{s/2})\pi \csc(\pi s/2)\Gamma(s)}{1-B}\right](A) $$
gives the answer $$ (B-1)I(A,B) = -2 \text{Ci}\left(\frac{A}{\sqrt{B}}\right) \cos \left(\frac{A}{\sqrt{B}}\right)-2 \text{Si}\left(\frac{A}{\sqrt{B}}\right) \sin \left(\frac{A}{\sqrt{B}}\right)+\pi\sin \left(\frac{A}{\sqrt{B}}\right)+2 \text{Ci}(A) \cos (A)+2 \text{Si}(A) \sin (A)-\pi \sin (A) $$ where $\text{Ci}$ and $\text{Si}$ are the Cos and Sine integral functions. This expression seems to be correct numerically for various positive values of $A$ and $B$.