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.
 A: 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.
A: Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:
We will require two basic properties. The first one is the Euler reflection formula:
$$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin z \pi}$$
The second is $\Gamma(z+1)=z\Gamma(z)$. So,
$$\Gamma\left(\frac{5}{3}\right) \Gamma\left(-\frac{2}{3} \right) = \frac{\pi}{\sin \left(-\frac{2}{3} \pi\right)}$$
$$\Gamma\left(\frac{4}{3}\right) \Gamma\left(-\frac{1}{3} \right) = \frac{\pi}{\sin \left(-\frac{1}{3} \pi\right)}$$
We multiply:
$$\Gamma(5/3)\Gamma(-2/3)\Gamma(4/3)\Gamma(-1/3) = \frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)}$$
$$\Gamma(5/3)\Gamma(4/3) = \frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)\Gamma(-2/3)\Gamma(-1/3)}$$
$$\Gamma\left(\frac{1}{3} \right)=\Gamma\left(-\frac{2}{3}+1 \right)=-\frac{2}{3}\Gamma\left(-\frac{2}{3} \right)$$ $$\Gamma\left(\frac{2}{3} \right)=\Gamma\left(-\frac{1}{3}+1 \right)=-\frac{1}{3}\Gamma\left(-\frac{1}{3} \right)$$
Hence
$$\Gamma\left(-\frac{1}{3} \right) = -3\Gamma\left(\frac{2}{3} \right) \ \text{ и} \ \ \Gamma\left(-\frac{2}{3} \right) = -\frac{3}{2}\Gamma\left(\frac{1}{3} \right)$$
Finally,
$$\Gamma\left(\frac{5}{3}\right)\Gamma\left(\frac{4}{3}\right) =\displaystyle{\frac{\pi^2}{\sin \left(-\frac{2}{3} \pi\right) \cdot \sin \left(-\frac{1}{3} \pi\right)\cdot -\frac{3}{2} \Gamma\left(\frac{1}{3}\right) \cdot(-3)\Gamma\left(\frac{2}{3}\right)}}=$$
$$=\frac{2}{9}\cdot\frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)\underbrace{\Gamma\left(1/3\right) \Gamma\left(2/3\right)}_{\displaystyle{\frac{\pi}{\sin(1/3\pi)}}}}=$$
$$\frac{2}{9}\cdot\frac{\pi \cdot \sin (1/3\pi)}{\sin(-2/3\pi)\cdot \sin(-1/3 \pi)}= \frac{4\pi}{9\sqrt{3}}$$
So:
$$\mathrm{B}\left(\frac{5}{3},\frac{4}{3} \right) = \frac{\Gamma(5/3) \Gamma(4/3)}{\Gamma(3)} = \frac{\Gamma(5/3) \Gamma(4/3)}{(3-1)!}=$$
$$= \frac{\Gamma(5/3) \Gamma(4/3)}{2}=\frac{2\pi}{9\sqrt{3}}$$
In conclusion we obtain: 
$$\boxed{\displaystyle{\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} \ dx = (b-a)^2\frac{2\pi}{9\sqrt{3}}}}$$
