Integral $\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}$ 
Prove, using elementary methods, that
  $$\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}=\frac{4\pi}{3\sqrt 3}$$

I have seen this integral in the following post, however all answers presented exploits complex analysis or heavy series.
But according to mickep's answer even the indefinite integral possess a primitive in terms of elementary functions. I'm not that insane to try and find that by hand, however it gives me great hope that we can find an elementary approach for the definite integral. 
Although I kept coming back to it for the past months, I still got no success, or relevant progress and I would appreciate some help.
 A: Not an answer (yet), just some thoughts.
$$I=\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}=2 \int_0^{1/2} \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}$$
An obvious substitution:
$$x(1-x)=y$$
$$dx=\frac{dy}{\sqrt{1-4 y}}$$
So we have:
$$I=2 \int_0^{1/4} \frac{y^{-1/3} dy}{(1-y) \sqrt{1-4 y}}$$
Substituting:
$$y=u/4$$
$$I=\frac{4^{1/3}}{2} \int_0^1 \frac{u^{-1/3} du}{(1-\frac14 u) \sqrt{1-u}}$$
This is clearly a hypergeometric function, though it's not considered elementary (which is a pity).
$$I=\frac{4^{1/3}}{2} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right)$$
I'll continue this in a few hours, the integral seems quite interesting.
Wolfram Alpha can't simplify the above expression to its exact value, which is even more interesting.
A more general form (but not really what the OP wants) would be:
$$I(z)=\int_0^1 \frac{dx}{\sqrt[3]{x(1-x)}(1-z x(1-x))}=\frac{4^{1/3}}{2} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{2}{3}; \frac{7}{6}; \frac{z}{4} \right)$$

As for the antiderivative, in terms of Appell function we have:

$$I(a)=\int_0^a \frac{dx}{\sqrt[3]{x(1-x)}(1-x(1-x))}= \\ =\frac32 (a(1-a))^{2/3} F_1 \left(\frac23; \frac12, 1; \frac53;4 a(1-a), a(1-a) \right) \\ 0 < a < \frac12$$

So far no idea about the elementary form.

In addition.
$$\int_0^1 \frac{u^{-1/3} du}{(1-\frac14 u) \sqrt{1-u}}=\frac43 \int_0^1 \frac{v^{-1/2}  dv}{(1+\frac{1}{3} v) (1-v)^{1/3}}$$
Which gives us another hypergeometric form, and another Appell form for the antiderivative.
$$I=I(1)=\frac{2}{3} 4^{1/3} B \left(\frac{1}{2},\frac{2}{3} \right) {_2 F_1} \left(1,\frac{1}{2}; \frac{7}{6}; -\frac{1}{3} \right)$$
Which Wolfram Alpha also can't simplify. I'll see later if Mathematica can do it.
A: Not elemntary at all for the antiderivative.
Considering
$$I=\int \frac{dx}{\sqrt[3]{x(1-x)} (1-x(1-x) )} $$
As  Archis Welankar commented, starting with $x=\sin^2(t)$ leads, after simplifcations, to
$$I=4 \int\frac{ (1-\cos (4 t))^{2/3} \csc (t) \sec (t)}{7+\cos (4 t)}\,dt$$
Now, $t=\frac{1}{4} \cos ^{-1}(u)$ leads to
$$I=-2 \sqrt{2}\int\frac{du}{\sqrt[3]{1-u} \sqrt{u+1} (u+7)}$$
$$I=\frac{12 \sqrt{2}}5 \frac{\sqrt{-u-1}}{(1-u)^{5/6} \sqrt{u+1}}F_1\left(\frac{5}{6};\frac{1}{2},1;\frac{11}{6};-\frac{2}{u-1},-\frac{8}{u-1}\right)$$ where appears  the Appell hypergeometric function of two variables. 
