Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$ 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?

 A: (This is more a comment than answer, but I couldn't get MathJax to properly show it in comments)
Here is a nice identity (equation (21) of this paper with $x=-1/7$): $$_2F_1 \left(a,a+\frac{1}{2};\frac{4a+5}{6};-\frac{1}{7}\right)=\left(\frac{7}{4}\right)^a {_2}F_1 \left(\frac{a}{3},\frac{a+1}{3};\frac{4a+5}{6};-27\right)$$
It's an example of a cubic transformation. Possibly, one can at this point use contiguous relations to make some progress.
A: 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.
A: The conjecture is true, as are the other cases reported in the comments
where $f(z) := {}_2F_1 \left( \frac13, \frac13; \frac56; z \right)$
takes algebraic values for special rational values of $z$.
There are a few others obtained from the symmetry $z \leftrightarrow 1-z$
(these ${}_2F_1$ parameters correspond to a hyperbolic triangle group
with index $6,6,\infty$ at $c=0,1,\infty$, so the $z=0$ and $z=1$
indices coincide); e.g. $f(-1/3) = 2 / 3^{2/3}$ pairs with
$f(4/3) = 3^{-2/3} (5-\sqrt{-3})/2$.   ($z=1/2$ pairs with itself,
and the pair $f(-4)$ and $f(5)$ has been noted already;
the OP's $f(-27) = -4/7$ pairs with $f(28) = \frac12 - \frac3{14} \sqrt{-3}$.)
Somewhat more exotic are
$$
f\big({-}4\sqrt{13}\,(4+\sqrt{13})^3\big) = \frac7{13\,U_{13}}\\
f\big({-}\sqrt{11}\,(U_{33})^{3/2}\big) = \frac{6}{11^{11/12}\, U_{33}^{1/4}},
$$
with fundamental units $U_{13}=\frac{3+\sqrt{13}}2,\;U_{33}=23+4\sqrt{33}$ and further values at algebraic conjugates and images under
$z \leftrightarrow 1-z$.
In general, for $z<1$ the integral formula for $f(z)$ relates it with
$$
\int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} }
$$
which is half of a "complete real period" for the holomorphic differential
$dx/y$ on the curve $C_z : y^6 = (1-x)^3 x^4 (1-zx)^2$.  This curve
has genus $2$, but is in the special family of genus-$2$ curves
with an automorphism of order $3$ (multiply $y$ by a cube root of unity),
for which both real periods are multiples of the real period of a
single elliptic curve $E_z$ (a.k.a. a complete elliptic integral).
In general the resulting formula doesn't simplify further, but when
$E_z$ has CM (complex multiplication) its periods can be expressed
in terms of gamma functions.  For $z = -27$ and the other special values
listed above, not only does $E_z$ have CM but the CM ring is contained in
${\bf Z}[\rho]$ where $\rho = e^{2\pi i/3} = (-1+\sqrt{-3})/2$.
Then the $\Gamma$ and $\pi$ factors of the period of $E_z$ exactly
match those in the integral formula, leaving us with
an algebraic value of $f(z)$.  It turns out that the choice $z = -27$
makes $E_z$ a curve with complex multiplication by ${\bf Z}[7\rho]$.
The others from the comments lead to ${\bf Z}[m\rho]$ with $m=1,2,3,5$,
and the examples where $z$ is a quadratic irrationality come from
${\bf Z}[13\rho]$ and ${\bf Z}[11\rho]$.
One way to get from $C_z$ to $E_z$ is to start from the change of variable
$u^3 = (1+cx)/x$, which gives
$$
f(z) = \int_{\root 3 \of {1-z}}^\infty \frac{3u \, du}{\sqrt{(u^3+z)(u^3+z-1)}}.
$$
and identifies $C_z$ with the hyperelliptic curve $v^2 = (u^3+z)(u^3+z-1)$.
Now in general a curve $v^2 = u^6+Au^3+B^6$ has an involution $\iota$ taking
$u$ to $B^2/u$, and the quotient by $\iota$ is an elliptic curve;
we compute that this curve has $j$-invariant
$$
j = 6912 \frac{(5+2r)^3}{(2-r)^3(2+r)}
$$
where $A = rB^3$.  (There are two choices of $\iota$, related by
$v \leftrightarrow -v$, and thus two choices of $j$, related by
$r \leftrightarrow -r$; but the corresponding elliptic curves
are $3$-isogenous, so their periods are proportional.)
In our case $r = A/B^3 = -(2z+1)/\sqrt{z^2+z}$ (in which the
$z \leftrightarrow 1-z$ symmetry takes $r$ to $-r$).  Taking $z=-27$
yields $j = -2^{15} 3^4 5^3 (52518123 \pm 11460394\sqrt{21})$,
which are the $j$-invariants of the ${\bf Z}[7\rho]$ curves;
working backwards from the $j$-invariants of the other
${\bf Z}[m\rho]$ curves we find the additional values of $z$
noted in the comments and earlier in this answer.
