TITLE: On $\big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\small 161+72\sqrt{5}$ and $\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{161+72\sqrt{5}\,x}}$
QUESTION [14 upvotes]: (This summarizes scattered results from here, here, here and elsewhere. See also this older post.)

I. Cubic

Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}= \frac{1}{48^{1/4}\,K(k_3)}$. 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.

II. Quartic

Define $\gamma= \tfrac{\sqrt{2\pi}}{\Gamma^2\big(\tfrac14\big)}= \frac{1}{2\sqrt2\,K(k_1)}=\frac1{2L}$ with lemniscate constant $L$. Then we have the nice,
$$\begin{aligned}\frac{2}{3^{3/4}} &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-3\big)\\
&=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+3x^4}}\\[1.7mm]
&\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{7+4\sqrt{3}}\,x}}\\[1.7mm]
&=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}
\end{aligned}\tag3$$
and,
$$\begin{aligned}\frac{3}{5}&=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-80\big)\\
&=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+80x^4}}\\[1.7mm]
&\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{161+72\sqrt{5}}\,x}}\\[1.7mm]
&=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{161+\cosh x}}
\end{aligned}\tag4$$
with $a=161$ given by Noam Elkies in this comment. (For $4$th roots, I just assumed the equality using the blue radicals based on the ones for cube roots.) Note again the powers of fundamental units,
$$U_{3}^2 = \big(2+\sqrt3\big)^2=\color{blue}{7+4\sqrt{3}}$$
$$U_{5}^{12} = \big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\color{blue}{161+72\sqrt{5}}$$
Just like for the cube roots version, these can't be coincidence.

Questions:

Is it true these observations can be explained by, let $b=2a+1$, then,
$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt[3]{b+\cosh x}}$$
$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{b+\sqrt{b^2-1}\,x}}=2^{1/4}\int_0^\infty\frac{dx}{\sqrt[4]{b+\cosh x}}$$

REPLY [3 votes]: Too long for a comment : In general, for strictly positive values of n we have 

$$\begin{align}
\sqrt[n]2\int_0^\infty\frac{dx}{\sqrt[n]{\cosh2t~+~\cosh x}}
~&=~
\int_0^1\frac{dx}{\sqrt{1-x}\cdot\sqrt[n]{x^{n-1}~+~x^n\cdot\sinh^2t}}
\\\\
~&=~
\int_{-1}^1\frac{dx}{\sqrt[n]{(1-x^2)^{n-1}}\cdot\sqrt[n]{\cosh2t~+~x\cdot\sinh2t}}
\end{align}$$