Is there a clever way to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$ I'm wondering how to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $\int \frac{w^3}{(w^3+1)^3} dw$ what is quite unpleasant as Wolframalpha says.
Thanks.

It's better to do this without trying to find an antiderivative. Substituting $x=a+(b-a)y$, we find that $x-a = (b-a)y$ and $b-x = (b-a)(1-y)$, so the integral reduces to
$$ (b-a)^2 \int_0^1 y^{2/3} (1-y)^{1/3} \, dy. $$
The remaining integral is $B(\frac{5}{3},\frac{4}{3}) = 2\pi/3^{5/2} $ using properties of the Beta- and Gamma-functions. An elementary way of computing this Beta-function is to set $y=1/(1+u)$, which gives 
$$ \int_0^{\infty} \frac{y^{1/3}}{(1+y)^3} \, dy. $$
This can be calculated by differentiating the integral
$$ \int_0^{\infty} \frac{y^{s-1}}{a+y} \, dy = a^{s-1}\pi\csc{\pi s} $$
a couple of times.
