The integrand has an elementary primitive and here is a way to find it:
The curve $y^3=x(1-x)^2$ looks like this:
We note that $(0,0)$ belongs to the curve, and try to parametrize it with $t$ satisfying $y=tx$. This gives us
The parameter $t$, being the direction coefficient from the line $y=tx$, will run from $+\infty$ to $0$ as $x$ runs from $0$ to $1$. A straight-forward calculation (in the last step we set $t=u^2$) gives
The last integral (known by Euler, I've read somewhere) can be taken care of in the "usual" way with partial fraction decomposition and logarithms and inverse tangents will appear (Residue calculus could give a shorter/handier calculation).
I will not write details but just conclude that one could benefit much from knowing how to deal with the Beta function, as JanG demonstrates.