how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
 \int_a^bf(x)dx = (b-a)\int_0^1f((b-a)x+a)dx$$
and done substitutions like
$$x=3\cos^2(t)+5\sin^2(t)$$
but it just becomes messy.
I have tried adding the different results hoping that stuff would cancel out or simplify,
but I'm not able to figure it out.
Please help. Thanks!

Noticing that $(x-3)(5-x)=1 -(x-4)^2$, we use the substitution  $x-4=\sin \theta$ to transform the integral into
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(\sin \theta+4)(\sin \theta+5)}{\cos^{\frac 2 3} \theta} \cos \theta d \theta \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^2 \theta+9 \sin \theta+20\right) \cos ^{\frac{1}{3}} \theta d \theta \\
&=2 \int_0^{\frac{\pi}{2}} \sin ^2 \theta \cos ^{\frac{1}{3} }\theta+40 \int_0^{\frac{\pi}{2}} \cos ^{\frac{1}{3} }\theta d \theta \\
&=B\left(\frac{3}{2}, \frac{2}{3}\right)+20 B\left(\frac{1}{2}, \frac{2}{3}\right) \\
&=\frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{13}{6}\right)}+\frac{20\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)}\\&= \frac{143 \sqrt{\pi}}{7} \cdot \frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)},
\end{aligned}
$$
where the last result using the facts $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(\frac 12)=\sqrt{\pi}$.
