How to evaluate $\int_{0}^\infty \frac{{x^2}}{e^{\beta {\big(\sqrt{x^2 + m^2}}- \nu\big)} + 1} dx$ $$I = \int_{0}^\infty \frac{{x^2}}{e^{\beta {\big(\sqrt{x^2 + m^2}}- \nu\big)} + 1} dx = ?$$
If a constant is added to the exponential in the denominator along with a square root in the exponent and we have a polynomial at the top, how to perform the integral?
 A: First, we get rid of one of the parameters:
$$I = \int_{0}^\infty \frac{{x^2}}{e^{\beta {\big(\sqrt{x^2 + m^2}}- \nu\big)} + 1} dx=m^3 \int_{0}^\infty \frac{{y^2}}{e^{A \sqrt{y^2 + 1}- B} + 1} dy$$
For convergence we need $A>0$.
Now we are trying to evaluate the integral:

$$J=\int_{0}^\infty \frac{{y^2}}{e^{A \sqrt{y^2 + 1}- B} + 1} dy$$

First, let's make a substitution:
$$y=\sinh v$$
$$J=\int_{0}^\infty \frac{\sinh^2 v \cosh v}{e^{A \cosh v- B} + 1} dv$$
Now we expand the denominator in geometric series:
$$J=\sum_{k=0}^\infty (-1)^k e^{B(k+1)} \int_{0}^\infty e^{-A(k+1) \cosh v} (\cosh^3 v-\cosh v) dv$$
We know that the integral can be expressed through modified Bessel functions:
$$\int_{0}^\infty e^{-A(k+1) \cosh v}  dv=K_0 \left( A(k+1)\right)$$
Using differentiation w.r.t. $A(k+1)$ under the integral sign, we obtain:
$$\int_{0}^\infty e^{-A(k+1) \cosh v}\cosh v  dv=K_1 \left( A(k+1)\right)$$
$$\int_{0}^\infty e^{-A(k+1) \cosh v}\cosh^3 v  dv=K_1 \left( A(k+1)\right)+\frac{K_0 \left( A(k+1)\right)}{ A(k+1)}+\frac{2K_1 \left( A(k+1)\right)}{ A^2(k+1)^2}$$
Finally, we have:

$$J=\sum_{k=1}^\infty (-1)^{k+1} e^{Bk} \left( \frac{K_0 \left( Ak\right)}{ Ak}+\frac{2K_1 \left( Ak\right)}{ A^2k^2} \right) $$

I doubt the series has a closed form, but it's another way to represent the integral.
It's clear from asymptotics that we need to have $A>B$ for the series to converge. Which may be a stronger condition than for the integral.
