Conjecture ∫10dx3√x6√1−x√1−x(√6√12+7√3−3√3−6)2=π9(3+√24√27)\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})

Let $t = 1 + y^3$, we can rewrite $B\left(\alpha^2; \frac12, \frac13 \right)$ as

$$\int_0^{\alpha^2} t^{-1/2} (1-t)^{-2/3} dt = \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{3y^2dy}{\sqrt{1+y^3} y^2} = 3 \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{dy}{\sqrt{1+y^3}}$$

Following the setup in my answer to a related question. Let
$\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$ and $\wp(z)$ be the Weierstrass elliptic $\wp$ function with fundamental periods $1$ and $e^{i\pi/3}$. $\wp(z)$ is known to satisfy an ODE of the form
$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$
If one perform variable substitution $\;\displaystyle y = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$, one has

$$\frac{dy}{\sqrt{1+y^3}} = -dz\quad\text{ and }\quad \begin{cases} y\left(\frac{\sqrt{3}\eta}{3}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{3}\right) = 0\\ \\ y\left(\frac{\sqrt{3}\eta}{2}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2}\right) = -1 \end{cases}$$
Using this, we can express conjecture $(8)$ in terms of $y(\cdot)$ and/or $\wp(\cdot)$:
$$\begin{align} & B\left(\alpha^2; \frac12, \frac13 \right) \stackrel{?}{=} \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)} = \frac{\sqrt{3}}{4}\eta\\ \iff & 3\left[y^{-1}(-1) - y^{-1}(-\sqrt[3]{1-\alpha^2})\right] \stackrel{?}{=} \frac{\sqrt{3}}{4}\eta\\ \iff & y^{-1}(-\sqrt[3]{1-\alpha^2}) \stackrel{?}{=} \frac{5\sqrt{3}}{12}\eta\\ \iff & \frac{4}{\eta^2}\wp\left(i\frac{5\sqrt{3}}{12}\right) \stackrel{?}{=} \sqrt[3]{1-\alpha^2} \end{align}$$
Let $u_0 = i\frac{\sqrt{3}}{3}$, $u_{-1} = i\frac{\sqrt{3}}{2}$ and $u = i\frac{5\sqrt{3}}{12} = \frac12(u_0 + u_{-1})$. Using the addition formula for $\wp$ function,
we have

$$\wp(2u) = \wp(u_0 + u_{-1}) = \frac14\left[\frac{\wp'(u_0)-\wp'(u_{-1})}{\wp(u_0)-\wp(u_{-1})}\right]^2 - \wp(u_0) - \wp(u_{-1})\\ =\frac14\left[\frac{-i\frac{\eta^3}{4} - 0}{0 - \frac{\eta^2}{4}}\right]^2 - 0 - \frac{\eta^2}{4} = -\frac{\eta^2}{2}$$
Using the duplication formula of $\wp$ function, we get
$$-\frac{\eta^2}{2} = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u) = \frac{9\wp(u)^4}{4\wp(u)^3-\frac{\eta^4}{16}} - 2\wp(u)$$
Let $Y = \frac{4}{\eta^2}\wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to

$$\begin{align} & Y^4 + 8 Y^3 + 8 Y - 8 = 0\tag{*1a}\\ \iff & Y(8+Y^3) = 8(1-Y^3)\tag{*1b}\\ \implies & (9-A^2)^3(1-A^2) = 512A^6\tag{*1c}\\ \iff & (A^4-24A^3+18A^2-27)(A^4+24A^3+18A^2-27) = 0\tag{*1d} \end{align}$$

• Since $\alpha$ is a root for one of the factors in $(*1d)$, $A = \alpha$ satisfies $(*1d)$ and hence $(*1c)$.
• Since $0 < \alpha < 1$ implies $1 - \alpha^2 > 0$, $(*1c) \implies (*1b)$ in this particular case.
  i.e. $Y = \sqrt[3]{1-\alpha^2}$ satisfies $(*1b)$ and hence $(*1a)$.
• Since $u$ lies between $u_0$ and $u_{-1}$, $\frac{4}{\eta^2}\wp(u) > 0$. Using
  the fact $(*1a)$ has only one positive root, we find $\frac{4}{\eta^2}\wp(u) = \sqrt[3]{1-\alpha^2}$. i.e. conjecture $(8)$ is true.