The integral under discussion is of the form $$\int_0^{\infty} \frac{p^2 dp}{\mathrm{exp}(a\sqrt{p^2+b^2}) \pm 1}$$, which can be transformed using the substitution ##x(p) = a\sqrt{p^2 + b^2}##. This leads to a new integral expression that involves the term $$\frac{1}{a^3} \int_0^{\infty} \frac{\sqrt{x^2-(b/a)^2}}{e^x \pm 1} dx$$. While Wolfram Alpha is not providing useful results for this integral, references to similar integrals can be found in Schaum's Outline of Mathematical Handbook of Formulas and Tables, specifically formulas 18.81 and 18.82. It is advised to double-check the algebra for the variable change to ensure the bounds of the integral are correct.