TITLE: Closed-forms for $\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}}$ and $\int_0^\infty\frac{dx}{\sqrt[3]{45\big(23+4\sqrt{33}\big)+\cosh x}}$
QUESTION [10 upvotes]: (This summarizes results for cube roots from here and here. The fourth root version is this post.)
Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}=\frac1{B\big(\tfrac{1}{3},\tfrac{1}{2}\big)}$ with beta function $B(a,b)$. Then we have the nice evaluations,
$$\begin{aligned}\frac{3}{5^{5/6}} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)\\
&=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+4x^3}}\\[1.7mm]
&=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{9+4\sqrt{5}}\,x}}\\[1.7mm]
&=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{9+\cosh x}}
\end{aligned}\tag1$$
and,
$$\begin{aligned}\frac{4}{7} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-27\big)\\
&=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+27x^3}}\\[1.7mm]
&=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{55+12\sqrt{21}}\,x}}\\[1.7mm]
&=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}}
\end{aligned}\tag2$$
Note the powers of fundamental units,
$$U_{5}^6 = \big(\tfrac{1+\sqrt{5}}{2}\big)^6=\color{blue}{9+4\sqrt{5}}$$
$$U_{21}^3 = \big(\tfrac{5+\sqrt{21}}{2}\big)^3=\color{blue}{55+12\sqrt{21}}$$
Those two instances can't be coincidence.

Question:
  Is it true this observation can be explained by, let $b=2a+1$, then,
  $$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt[3]{b+\cosh x}}$$

Example: We assume it is true and use one of Noam Elkies' results as,
$$\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6}; -a\big)  = \frac{6}{11^{11/12}\, U_{33}^{1/4}}
$$
where $a=\sqrt{11}\,(U_{33})^{3/2}$ with fundamental unit $U_{33}=23+4\sqrt{33}$. Since $b=2a+1=45\,U_{33}$, we then have the nice integral,
$$2^{1/3}\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{45\big(23+4\sqrt{33}\big)+\cosh x}}=\frac{6}{11^{11/12}\,U_{33}^{1/4}}=0.255802\dots$$
where $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}.\,$ So is it true in general?

REPLY [3 votes]: I've noticed that for 
$$
I = \int_0^\infty \frac{dx}{\sqrt[3]{b + \cosh(x)}}
$$
if we take a Mellin transform with respect to $b$, we get 
$$
\mathcal{M}_b[I](s) = \frac{\Gamma(\frac{1}{3}-s)\Gamma(s)}{\Gamma(\frac{1}{3})}\int_0^\infty \frac{\text{sech}^{-s}{x}}{\cosh^{1/3}(x)}\;dx
$$
$$
\mathcal{M}_a[I](s) = \frac{\sqrt{\pi}\Gamma(\frac{1}{3}-s)\Gamma(\frac{1}{6}-\frac{s}{2})\Gamma(s)}{2\Gamma(\frac{1}{3})\Gamma(\frac{2}{3}-\frac{s}{2})}, \; \Re(s)<\frac{1}{3}
$$
$$
I(s) = \mathcal{M}^{-1}_s\left[\frac{\sqrt{\pi}\Gamma(\frac{1}{3}-s)\Gamma(\frac{1}{6}-\frac{s}{2})\Gamma(s)}{2\Gamma(\frac{1}{3})\Gamma(\frac{2}{3}-\frac{s}{2})}\right](b)
$$
$$
I(s) = \frac{\Gamma(\frac{1}{6})^2}{2^{5/3}\Gamma(\frac{1}{3})}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{b \Gamma(\frac{2}{3})^2}{2^{2/3}\Gamma(\frac{1}{3})}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)
$$
$$
2^{1/3}\beta I(s) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{2\sqrt{\pi}b \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)
$$
so you are looking for $a$ and $b$ such that satisfy
$$
\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-a\big) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{2\sqrt{\pi}b \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)
$$
i.e. $a=4,b=9$ and $a=27,b=55$, and would need to prove $$
\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-a\big) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},(2a+1)^2\right) - \frac{2\sqrt{\pi}(2a+1) \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},(2a+1)^2\right)
$$
The residues if you subtract the two and plot are around machine epsilon. I hope this is helpful.