Closed form solution of the normal density of a superfluid for the Bogoliubov spectrum I've been trying to solve the following definite integral
$$
\int_0^\infty dx\, x^4\, \frac{e^{\sqrt{x^4+2 x^2}/Tp}}{\left(e^{\sqrt{x^4+2 x^2}/Tp}-1\right)^2}\, ,\quad Tp = \frac{T}{Un}
$$
This is the density of the normal part of a superfluid. However, so far I could not find any solution. I'd prefer an exact one but a good approximation would be also nice. (I do have the results for the limiting cases where one sets $\sqrt{x^4+2 x^2}\approx \sqrt{2}x$ and the case $\sqrt{x^4+2 x^2}\approx x^2 + 1$, so I am interested in the exact result or at least an approximation which is higher order than the limiting cases.)
(@Alex Trounev :) This is a follow up question of Closed form solution to normal fluid density integral in the two fluid model)
 A: Since the integral depends only one parameter $Tp$, a good approach is to evaluate it numerically, since all you need is the dependence on this parameter. One could also try to evaluate analytically limiting cases: $Tp \ll 1$ and $Tp \gg 1$.
It is possible that by a clever substitution this integral can be reduced to one of the integrals in Gradshtein&Ryzhik, but there is a high chance that the solution is given in terms of special functions, which is often not better than just looking at the limiting cases. There is a fine difference between an exact result and a result that one can understand/interpret.
Update
I suggest using substitution $$y = \sqrt{x^4+2x^2}/(Tp) \Leftrightarrow x^2 = \sqrt{1+y^2(Tp)^2}-1$$, which reduces the integral to
$$\int_0^{+\infty}dx x^4 \frac{e^{\sqrt{x^4+2x^2}/Tp}}{(e^{\sqrt{x^4+2x^2}/Tp}-1)^2} =
\frac{(Tp)^2}{2}\int_0^{+\infty}dy \frac{y(\sqrt{1+y^2(Tp)^2}-1)^3}{\sqrt{1+y^2(Tp)^2}}\frac{e^y}{(e^y-1)^2}
$$
One can now obtain the required limiting cases by expanding $\sqrt{1+y^2(Tp)^2}$ in powers of $(Tp)^2$ (for small $Tp$) and by expanding $Tp\sqrt{\frac{1}{(Tp)^2}+y^2}$ in powers of $1/(Tp)^2$ (for $Tp \gg 1$). It is also easier to see whether the integral is reducible to any of those is the integral tables.
