On Reshetnikov's integral $\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{N}\,\frac{2\pi}{\sqrt{3}\,|\alpha|}$

(Too long for a comment. And courtesy of V. Reshetnikov's result here, though as he points out it is tentative.)

The algebraic number $\alpha$ that solves,

$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}=\frac{1}{11}\,\frac{2\pi}{\sqrt{3}\;\alpha}$$

seems to have a $40$-deg minpoly. However, it turns out we can also reduce its degree and express it using the common form above. Let,

$$\alpha=\frac{3^{1/2}-v^{1/2}}{3^{-1/2}+v^{1/2}}$$

where $v$ is a the second largest positive root ($r_9$ in Mathematica syntax) of,

$$\small P(v)=-3^{10} + 23816430 v^2 - 323903448 v^3 + 2177615583 v^4 - 9297934272 v^5 + 25869358152 v^6 - 37475802144 v^7 - 16459141842 v^8 + 180065426112 v^9 - 338100745356 v^{10} + 329418595440 v^{11} - 211367836746 v^{12} + 102243404736 v^{13} - 8162926200 v^{14} - 9999738144 v^{15} + 1006439643 v^{16} - 134177472 v^{17} - 2246706 v^{18} + 30888 v^{19} + 11 v^{20} = 0$$

Also,

$$I\big(\alpha^2;\tfrac{1}{2},\tfrac{1}{3}\big)=\tfrac{1}{5}\,I\big(\tfrac{1-\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\tfrac{1}{6}\,I\big(\tfrac{1+\alpha}{2};\tfrac{1}{3},\tfrac{1}{3}\big)=\frac{1}{11}$$

with regularized beta function $I(z;a,b)$. Furthermore, if

$$y =\frac{r_1+r_9+r_{13}+r_{14}}{12}$$

then $y$ is a root of the solvable quintic,

$$67 - 1748 y - 7033 y^2 - 1378 y^3 + 234 y^4 + y^5=0$$

with discriminant divisible by $11^4\times23^4$. Using the other quartic symmetric polynomials show that the $20$-deg is just a quartic in disguise, hence is solvable. All these suggest that $P(v)$ is the correct polynomial for $N=11$.

I. Duplication

Following Nemo's lead in this answer, we find the formula, $$\frac{1}{2}I(p^2;\tfrac{1}{2},\tfrac{1}{3})=I(1+q^3;\tfrac{1}{2},\tfrac{1}{3})$$ where $p,q$ are related by the $12$-deg, $$p^2(-2 + 2 q + q^2)^6 = 36(1 + q^3) (4 + 4 q + 6 q^2 - 2 q^3 + q^4)^2$$ This then enables us to find infinitely many $\displaystyle\frac{1}{2^n N}$.

For example, since $I(p^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{3}$ is known, then solving for $I(\alpha^2;\tfrac{1}{2},\tfrac{1}{3})=\frac{1}{6}$ turns out to involve a $36$-deg equation.

II. Triplication

(Courtesy of Nemo.) Starting with, $$B\left(z;\frac{1}{2},\frac{1}{3}\right)=2 \sqrt{z} \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right). $$ The transformation $$ \, _2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};-\frac{3 z \left(1-\frac{z}{9}\right)^2}{(1-z)^2}\right)=\frac{(1-z) \, }{1-\frac{z}{9}}{}_2F_1\left(\frac{1}{2},\frac{2}{3};\frac{3}{2};z\right) $$ applied two times gives $$ \frac{1}{3} B\left({\frac{(9-z)^2 z \left(z^3+225 z^2-405 z+243\right)^2}{729 (1-z)^2 (z+3)^6}};\frac{1}{2},\frac{1}{3}\right)=B\left(z;\frac{1}{2},\frac{1}{3}\right). $$