Make the substitution $r=ts$ (I'm assuming $t>0$), so that the integral becomes
\begin{align}
\int_0^t(r-t)^{1/3}r^{5/3}\,dr&=t^3\int_0^1s^{5/3}(1-s)^{1/3}\,ds\\
&=t^3B\left(\frac{5}{3}+1,\frac{1}{3}+1\right)\\
&=t^3B\left(\frac{8}{3},\frac{4}{3}\right),
\end{align}
where $B(x,y):=\int_0^1s^{x-1}(1-s)^{y-1}\,ds$ is the Beta function (defined for $x,y\in\Bbb{C}$ such that $\text{Re}(x),\text{Re}(y)>0$). Now, it is a "standard" fact that the Beta function is related to the Gamma function via the identity $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. So, in our case, we have
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{\Gamma\left(\frac{8}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{8}{3}+\frac{4}{3}\right)}=
\frac{\Gamma\left(\frac{8}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(4\right)}
\end{align}
Now, a fundamental property of the Gamma function is that $\Gamma(1+z)=z\Gamma(z)$ (a factorial-like property). Therefore, the above reduces to
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{\left[\frac{5}{3}\cdot\frac{2}{3}\cdot \Gamma\left(\frac{2}{3}\right)\right]\cdot
\left[\frac{1}{3}\cdot \Gamma\left(\frac{1}{3}\right)\right]}{3\cdot 2\cdot \Gamma(1)}=
\frac{5}{81}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma(1)}
\end{align}
Now, another fact about the Gamma function is that $\Gamma(1)=1$ (trivial from the definitions), and the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$. Using this with $z=\frac{1}{3}$, we get
\begin{align}
B\left(\frac{8}{3},\frac{4}{3}\right)&=\frac{5}{81}\cdot\frac{\frac{\pi}{\sin\left(\frac{\pi}{3}\right)}}{1}=\frac{10\pi}{81\sqrt{3}}.
\end{align}
Therefore, our original integral is $t^3\cdot \frac{10\pi}{81\sqrt{3}}$.
Now, in general, integrals with fractional/irrational exponents are not easy at all. We have special functions like the Beta and Gamma and Zeta functions to deal with them, and we have exploit nice identities involving them, and one of the many ways of proving these identities uses several ideas from complex analysis (eg infinite products, contour integration etc). As always with integration, there is no cookie-cutter recipe; you just have to deal with each case/class of integrals separately.
Edit: Alternative Approach.
Rewrite the integral as
\begin{align}
\int_0^1s^{5/3}(1-s)^{1/3}\,ds&=\int_0^1\left(\frac{s}{1-s}\right)^{5/3}(1-s)^2\,ds
=\int_0^{\infty}\frac{x^{5/3}}{(1+x)^4}\,dx\\
&=\int_0^{\infty}x^{2/3}\frac{x}{(1+x)^4}\,dx
\end{align}
where we made the substitution $x=\frac{s}{1-s}$. Now, this last integral is again of a "standard form" found in complex analysis texts, which means you can integrate along a pacman contour/keyhole contour, as I explain in detail in this post (with $\alpha=\frac{2}{3}, R(x)=\frac{x}{(1+x)^4}$). Now, you can have fun calculating residues (this is essentially what I meant in my previous comment regarding proving identities using complex analysis and contour integrals).
