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, 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?

Alternative formulation of the integral $(1)$ is $$ \int_0^\infty\frac{dt}{(1+t+t^{\,\alpha})^2}=\frac23, \quad \alpha=\frac{1+i\sqrt3}{2}. \tag{1a} $$