How to prove $_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-2^7\phi^9\big)=\large \frac{3}{5^{5/6}}\,\phi^{-1}\,$ with golden ratio $\phi$?

Question

(Note: This is the case $a=\frac16$ of ${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}.\,$ There is also $a=\frac13$ and $a=\frac14$.)

After investigating $a=\frac13$ and $a=\frac14$, I wondered if there was for $a=\frac16$. And happily there was,

$$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{\tfrac{125}3}x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{\tfrac{125}{3}})=\frac{2}{3^{5/6}}$$ $$\frac{1}{\color{red}{432}^{1/4}\,K(k_3)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[6]{x^5+\color{blue}{2^7\phi^9}\, x^6}}=\,_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\color{blue}{2^7\phi^9})=\frac{3}{5^{5/6}}\phi^{-1}$$

The first was found by computer search and, from previous posts, the denominator with $K(k_3)$ was enough to give me a clue that $\tau=\frac{1+3\sqrt{-3}}2$ was involved. After fiddling around with some equations, a third conjecture can be made, that there is an infinite family of algebraic numbers $\alpha$ and $\beta$ such that,
$$_2F_1\left(\frac16,\frac16;\frac23;-\alpha\right)=\beta$$

Conjecture: "Let $\tau = \frac{1+p\sqrt{-3}}{2}$ with integer $p>1$. Then $\alpha$ is the root of an analogous quadratic, $$16\cdot\color{red}{432}\,\alpha(1+\alpha)=-j(\tau)$$ with j-function $j(\tau)$. And if odd $p=3k\pm1$ is a prime, then $\alpha$ and $\beta^6$ are algebraic numbers of degree $k$."

$$\begin{array}{|c|c|c|c|c|} \hline p&\tau&\alpha(\tau)&\beta(\tau)&\text{Deg}\\ \hline 3&\frac{1+3\sqrt{-3}}2&\frac{125}3& \large\frac2{3^{5/6}} &1\\ 5&\frac{1+5\sqrt{-3}}2&2^7\phi^9& \large\frac3{5^{5/6}}\phi^{-1} &2\\ 7&\frac{1+7\sqrt{-3}}2&\Big(\frac{129 + 29\sqrt{21}}2\Big)^3& \large\frac47 \frac1{U_{21}^{1/2}} &2\\ 11&\frac{1+11\sqrt{-3}}2& x_1 & \large\frac6{11} x_2 &4 \\ 13&\frac{1+13\sqrt{-3}}2& y_1 & \large\frac7{13} y_2 &4 \\ \hline \end{array}$$ $U_{21}=\frac{5+\sqrt{21}}2$ is a fundamental unit, while $x_i,y_i$ are roots of quartics which are rather tedious to write down. And so on.

Q: How do we prove this conjecture? (And the other two?)

Answer 1

Courtesy of this new question by another OP, we find a simple family with a common form that relates this post with that question using, $$_2F_1\big(a,a;a+\tfrac12;z_1\big)\quad\text{and}\quad_2F_1\big(\tfrac12,a;1;z_2\big)$$ for $a=\frac13,\frac14,\frac16$. (Nemo has tackled the cases $a=\frac13$ and $a=\frac14$, and transformations can connect his results with the family below.)

Note: The relations $(1),(2),(3)$ hold for real $N>1$, but each side is the same algebraic number for integer $N>1$.

I. $a=\frac13$

$$_2F_1\big(\tfrac13,\tfrac13;\tfrac56;-\alpha\big) = \frac{1+N}{4\sqrt[4]{27}\,(1+\alpha)^{1/3}}\frac{_2F_1\big(\tfrac12,\tfrac13;1;\tfrac1{1+\alpha})}{\pi^{-1}\,K(k_3)}\tag1$$ where, $$\alpha =\alpha(\tau)= \frac1{4\sqrt{27}}\big(\lambda^3-\sqrt{27}\lambda^{-3}\big)^2,\quad\lambda =\frac{\eta\big(\frac{\tau+1}3\big)}{\eta(\tau)},\quad \tau=\frac{1+N\sqrt{-3}}2$$

II. $a=\frac14$

$$_2F_1\big(\tfrac14,\tfrac14;\tfrac34;-\beta\big) = \frac{1+N}{4\sqrt[4]{4}\,(1+\beta)^{1/4}}\frac{_2F_1\big(\tfrac12,\tfrac14;1;\tfrac1{1+\beta})}{\pi^{-1}\,K(k_1)}\tag2$$ where, $$\beta=\beta(\tau)= \frac1{4\sqrt{64}}\big(\mu^6-\sqrt{64}\mu^{-6}\big)^2,\quad\mu=\frac{\sqrt2\,\eta\big(2\tau\big)}{e^{2\pi i /48}\,\eta(\tau)},\quad \tau=\frac{1+N\sqrt{-1}}2$$

III. $a=\frac16$

$$_2F_1\big(\tfrac16,\tfrac16;\tfrac23;-\gamma\big) = \frac{1+N}{4\sqrt[4]{3}\,(1+\gamma)^{1/6}}\frac{_2F_1\big(\tfrac12,\tfrac16;1;\tfrac1{1+\gamma})}{\pi^{-1}\,K(k_3)}\tag3$$ where, $$\gamma=\gamma(\tau)=\frac{-72+\sqrt{3}\sqrt{1728-j(\tau)}}{144},\quad \tau=\frac{1+N\sqrt{-3}}2$$ with j-function $j(\tau)$.

IV. Integrals

These $\alpha,\beta,\gamma$ also imply, $$\int_0^{\infty}\frac1{\sqrt[3]{1+2\alpha+\cosh x}}dx=\frac{1+N}{\sqrt3}\,\int_0^{\pi/2}\frac1{\sqrt[3]{2\alpha+2\sin^2 x}}dx$$

$$\int_0^{\infty}\frac1{\sqrt[4]{1+2\beta+\cosh x}}dx=\big(1+N\big)\,\int_0^{\pi/2}\frac1{\sqrt[4]{2\beta+2\sin^2 x}}dx$$

$$\int_0^{\infty}\frac1{\sqrt[6]{1+2\gamma+\cosh x}}dx=1728^{1/6}\,\frac{1+N}{2}\,\int_0^{\pi/2}\frac1{\sqrt[6]{2\gamma+2\sin^2 x}}dx$$

using $\tau$ as defined per section.

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P.S. This family with $a+\tfrac12=b+\tfrac12=c$ joins this second family with $a+b=c=\frac12$ as being two $_2F_1$ functions related in a simple way by eta quotients and $K(k_n)$. Are there any more?