I think it is worth mentioning the case of the integration of the differential binomials.
The expression of the form
$x^m(a+bx^n)^pdx$ where $m,n,p,a,b$ are constant is called a differential binomial.
THEOREM. (Piskunov)
The integral
$\int x^m(a+bx^n)^pdx$
can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:
$1.$ $p$ is an integer.
$2.$ $\dfrac{m+1}{n}$ is an integer.
$3.$ $\dfrac{m+1}{n}+p$ is an integer.
PROOF
We transform the integral writing $x^n = z$ so $dx = \frac 1 n z^{\frac 1 n -1}$. Then:
$\int {{x^m}} {(a + b{x^n})^p}dx = \int {{z^{{{m + 1} \over n} - 1}}} {(a + bz)^p}dz = \int {{z^q}} {(a + bz)^p}dz$
$1.$ Let $p$ be an integer. Being $q$ a rational number, let it be $\dfrac r s$. This integral then takes the form $\int {R\left( {{z^{q/s}},z} \right)dz} $
which can be reduced by substituting $z=t^s$.
$2.$ If $\dfrac{m+1}{n}$ is an integer. then $q=\dfrac{m+1}{n}-1$ is an integer. $p$ is rational $=\dfrac \lambda \mu$. The integral is reduced to $\int {R\left( {{z^q},{{\left( {a + bz} \right)}^{{\lambda \over \mu }}}} \right)dz} $ which can be reduced substituting $a+bz=t^\mu$
$3.$ If $\dfrac{m+1}{n}+p$ is an integer then $\dfrac{m+1}{n}+p-1=q+p$ is an integer. We tranform the integral into
$\int {{z^{q + p}}{{\left( {{{a + bz} \over z}} \right)}^p}dz} $
where $q+p$ is an integer and $p=\dfrac \lambda \mu$ is rational. The integral is then
$\int {R\left[ {z,{{\left( {{{a + bz} \over z}} \right)}^{{\lambda \over \mu }}}} \right]dz} $
which can be reduced using
${{a + bz} \over z} = {t^\mu }$
Note. P.L. Chebyshev, a russian mathematician, proved the integrals just analysed can't be expressed in terms of elementary functions if it isn't the case $1$ , $2$ or $3$.