Is there any explanation based on algebraic number theory that the integral 
$$
\int\_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}\tag{1}
$$
has a closed form? Analytic proof of this integral is given in this [MSE post](https://math.stackexchange.com/questions/2030017/proof-int-infty-infty-fracdx-leftexe-xeix-sqrt3-right2), however this proof does not explain why a similar looking integral
$$
\int\_{-\infty}^\infty\frac{e^{ix\sqrt{3}}\ dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^3}=\frac{\sqrt{3}}{8\pi}\int\_{0}^\infty\frac{dx}{\left(1+\frac{x^3}{1^3}\right)\left(1+\frac{x^3}{2^3}\right)\left(1+\frac{x^3}{3^3}\right)\ldots}
$$
probably does not have a closed form. Is it possible that $(1)$ is related to [Eisenstein integers](https://en.wikipedia.org/wiki/Eisenstein_integer)?


Alternative formulation of the integral $(1)$ is
$$
\int\_0^\infty\frac{dt}{(1+t+t^{\,\alpha})^2}=\frac23, \quad \alpha=\frac{1+i\sqrt3}{2}. \tag{1a}
$$



Number theoretic interpretation of the integral $\int_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{ix\sqrt{3}}\right)^2}=\frac{1}{3}$?
The following formula gives a parametric extension of $(1)$ for $|a|$ sufficiently small
\begin{align}
\small\int\_{-\infty}^\infty\frac{dx}{\left(e^x+e^{-x}+e^{a+ix\sqrt{3}}\right)^2}+e^a\int\_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{ix\sqrt{3}}\right)^2}+e^a\int\_{-\infty}^\infty\frac{dx}{\left(e^{a+x}+e^{-x}+e^{-ix\sqrt{3}}\right)^2}=1
\end{align}
This means that $(1)$ is not an isolated result.
In view of this one might be very sceptical that any number theoretic interpretation of the integral exists.


