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!
 A: Note
$$ \int_0^1x^{p-1}(1-x)^{q-1}\;dx=B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}. $$
Let
$$ x=2t+3 $$
and then one has
\begin{eqnarray}
I&=&\int_3^5\frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}}\;dx\\
&=&\int_0^1\frac{2^{\frac43}(2t^2+7t+6)}{t^{\frac13}(1-t)^{\frac13}}\;dt\\
&=&2^{\frac43}\bigg[2\int_0^1\frac{t^2}{t^{\frac13}(1-t)^{\frac13}}\;dt+7\int_0^1\frac{t}{t^{\frac13}(1-t)^{\frac13}}\;dt+6\int_0^1\frac{1}{t^{\frac13}(1-t)^{\frac13}}\;dt\bigg]\\
&=&2^{\frac43}\bigg[2\int_0^1t^{\frac53}(1-t)^{-\frac13}\;dt+7\int_0^1t^{\frac23}(1-t)^{-\frac13}\;dt+6\int_0^1t^{-\frac13}(1-t)^{-\frac13}\;dt\bigg]\\
&=&2^{\frac43}\bigg[2B(\frac{8}3,\frac23)+7B(\frac{5}3,\frac23)+6B(\frac{2}3,\frac23)\bigg]\\
&=&2^{\frac43}\bigg[2\frac{\Gamma(\frac{8}3)\Gamma(\frac23)}{\Gamma(\frac{10}3)}+7\frac{\Gamma(\frac{5}3)\Gamma(\frac23)}{\Gamma(\frac73)}+6\frac{\Gamma^2(\frac{2}3)}{\Gamma(\frac43)}\bigg].
\end{eqnarray}
Now
$$ \Gamma(\frac{8}3)\Gamma(\frac23)=\frac{28}{27}\Gamma(\frac23)\Gamma(\frac13)=\frac{28}{27}\frac{2\pi}{\sqrt3}. $$
You can do the rest.
A: Let $x=2t+3$, then $dx=2dt$. The bounds of the integral change like this: $x=3=2t+3\implies t=0$ and $x=2t+3=5\implies t=1$. Hence,
$\begin{align}
\int_3^5\frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}}&=\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{2t(2-2t)}}2dt\\
&=\sqrt[3]{2}\int_0^1\frac{4t^2+14t+12}{\sqrt[3]{t(1-t)}}dt\\
&=\sqrt[3]{2}\int_0^1 (4t^2+14t+12)t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}\int_0^1 t^{\frac{5}{3}}(1-t)^{-\frac{1}{3}}dt+
14\sqrt[3]{2}\int_0^1 t^{\frac{2}{3}}(1-t)^{-\frac{1}{3}}dt+
12\sqrt[3]{2}\int_0^1 t^{-\frac{1}{3}}(1-t)^{-\frac{1}{3}}dt\\
&=4\sqrt[3]{2}B(\frac{8}{3},\frac{2}{3})
+14\sqrt[3]{2}B(\frac{5}{3},\frac{2}{3})
+12\sqrt[3]{2}B(\frac{2}{3},\frac{2}{3})
\end{align}$
A: 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 $\Gamma(\frac 12)=\sqrt{\pi}$.
