Conjecture $\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi$ $$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
The equality numerically holds up to at least $10^4$ decimal digits.

Can we prove that the equality is exact?

An equivalent form of this conjecture is
$$I\left(\frac{\sqrt3}2;\ \frac14,\frac14\right)\stackrel?=\frac23,\tag2$$
where $I\left(z;\ a,b\right)$ is the regularized beta function.

Even simpler case:
$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[6]{9-x}\ \sqrt[3]x}\stackrel?=\frac\pi{\sqrt3},\tag3$$
which is equivalent to
$$I\left(\frac19;\ \frac16,\frac13\right)\stackrel?=\frac12.\tag4$$

A related question.

Here is another approach for evaluating the integral (3).
Using @achille hui's transformation, we can write
$$
I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx 
$$
Using the substitution $x=\sqrt{t}$, we get
$$
I=\frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{2}\frac{1}{\sqrt{t+t^4}}dt
$$
We can now transform this integral into a beta integral using the substitution $t=\frac{1-y}{2+y}$. This gives us
\begin{align*}
I 
&= \frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3} \int_0^1\frac{1}{\sqrt{1-y^3}}dy \\
&= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}B\left(\frac{1}{3},\frac{1}{2} \right) \\
&= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}\left( \frac{\sqrt{3}\Gamma\left(\frac{1}{3}\right)^3}{\pi 2^{\frac{4}{3}}}\right) \\
&= \frac{\pi}{\sqrt{3}}
\end{align*}
