$\int (f)^n \, \textrm dx$ where $f$ is a polynomial and $n$ is a positive integer Besides expanding the integrand, is there some general method for solving indefinite integrals of the form $\int (f)^n \, \textrm dx$ where $f$ is a polynomial and $n$ is a positive integer? For example, $$\int (x^2 +x)^{100} \, \textrm dx?$$

Most of the time, you can simplify your task by performing substitutions, integrating by parts, or manipulating expressions with elementary algebra.
If worst comes to worst, and things can't be handled with any clever tricks and manipulations, I suggest using the binomial theorem
$$(a+b)^n=\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}a^kb^{n-k},$$
or the more general multinomial theorem
$$(x_1+x_2+...+x_m)^n=\sum_{k_1+k_2+...+k_m=n}{n\choose{k_1,k_2,...,k_m}}\prod_{r=1}^{m}x_r^{k_r},\tag{1}$$
where $${n\choose{k_1,k_2,...,k_m}}=\frac{n!}{k_1!k_2!\cdots k_m!}.$$
Note that the summation in $(1)$ runs over all integer $m$-tuples $(k_1,...,k_m)$, where $0\le k_i\le n$, and $\sum_ik_i=n$. See here for a proof.
Of course, this requires a lot of manual calculation if you want to get rid of some notation, but it does provide you with a viable method in the calculation of the integral of any polynomial raised to a non-negative integer power. Tell me if you were looking for anything other than this.
Beyond this, there isn't really anything else, because what we have is general enough to tackle any such power of a polynomial integral.

Addendum (10/13/2020):
If we write $x_j=a_jx^j$ and $p(x)=\sum_{j=0}^{m}a_jx^j$ we have
$$p(x)^n=\sum_{k_1+...+k_m=n}x^{\sum_{r=1}^{m}rk_r}{n\choose {k_1,...,k_m}}\prod_{r=1}^{m}a_r^{k_r},$$
and hence
$$\int p(x)^ndx=\text{constant}+\sum_{k_1+...+k_m=n}{n\choose{k_1,...,k_m}}\frac{\prod_{r=1}^{m}a_r^{k_r}}{1+\sum_{r=1}^mrk_r}x^{1+\sum_{r=1}^mrk_r}$$
