# Closed-forms for $\int_0^\infty\frac{dx}{\sqrt{55+\cosh x}}$ and $\int_0^\infty\frac{dx}{\sqrt{45\big(23+4\sqrt{33}\big)+\cosh x}}$

(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?