TITLE: Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$
QUESTION [36 upvotes]: I discovered the following conjecture numerically, but have not been able to prove it yet:
$$_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47.\tag1$$
The equality holds with at least $10000$ decimal digits of precision. It can be written in equivalent forms in terms of definite integrals:
$${\large\int}_0^1\frac{dx}{\sqrt{1-x}\ \sqrt[3]{x^2+(3x)^3}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]4\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right),\tag2$$
or
$${\large\int}_0^\pi\frac{d\phi}{\sqrt[3]{\sin\phi}\,\sqrt[3]{55+12\sqrt{21}\cos\phi}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]4\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right).\tag3$$

Update: A several more equivalent forms:
$$_2F_1\left(\frac13,\frac12;\frac56;\frac{27}{28}\right)\stackrel{\color{#808080}?}=\frac{2^{\small8/3}}{7^{\small2/3}}\tag4$$
$$\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]2\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right)\tag5$$
$$C_{\small-1/3}^{\small(1/3)}(55)\stackrel{\color{#808080}?}=\frac{3}{7\pi^2}\Gamma^3\!\!\left(\tfrac13\right)\tag6$$
$$P_{\small-1/2}^{\small1/6}(55)\stackrel{\color{#808080}?}=\frac{\sqrt2\,\sqrt[4]3\,e^{\small-\pi\,i/12}}{7^{\small13/12}\,\pi^{\small3/2}}\Gamma^2\!\!\left(\tfrac13\right)\tag7$$
where $C_n^{(\lambda)}(x)$ is the Gegenbauer polynomial and $P_l^m(x)$ is the Legendre function of the first kind.


Please suggest ideas how to prove this conjecture.
What are other points where the function $_2F_1\left(\frac13,\frac13;\frac56;z\right)$ takes simple special values?

REPLY [4 votes]: Regarding your secondary question, by appealing to the classical j-function at defined arguments, it seems there are infinitely many algebraic numbers $z$ such that the $_2F_1$ evaluates to an algebraic number. Some examples,
$$_2F_1\left(\frac13,\frac13;\frac56;-z_1\right)= \frac9{17} \big(833+324\cdot17^{1/3}-252\cdot17^{2/3}\big)^{1/6}$$
$$2F_1\left(\frac13,\frac13;\frac56;-z_2\right)= \frac{10}{3\cdot19} \big(2+2\cdot19^{1/3}-19^{2/3}\big)$$
where,
$$z_1 =4\big(19894+7737\cdot17^{1/3}+3009\cdot17^{2/3}\big)$$
$$z_2 =\frac{1}{3}\big(1464289+548752\cdot19^{1/3}+205648\cdot19^{2/3}\big)$$
See also this post.