$\int (f)^n \, \textrm dx$ 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, $$\int (x^2 +x)^{100} \, \textrm dx?$$
 A: $\int(x^2+x)^{100}~dx$
$=\int x^{100}(x+1)^{100}~dx$
$=\int x^{100}\sum\limits_{n=0}^{100}C_n^{100}x^n~dx$
$=\int\sum\limits_{n=0}^{100}C_n^{100}x^{n+100}~dx$
$=\sum\limits_{n=0}^{100}\dfrac{C_n^{100}x^{n+101}}{n+101}+C$
A: 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
$$\int p(x)^ndx=\text{constant}+\sum_{k_1+...+k_m=n}{n\choose{k_1,...,k_m}}\frac{\prod_{r=1}^{m}a_r^{k_r}}{1+\sum_{r=1}^mrk_r}x^{1+\sum_{r=1}^mrk_r}$$
A: Expanding on clathratus' answer using the multinomial theorem, you have an explicit formula for the integral as (ignoring the constant of integration)
$$\begin{align}
\int_{ }^{ }\left(\sum_{r=0}^{m}a_{r}x^{r}\right)^{n}\,\mathrm{d}x
&=\int_{ }^{ }\sum_{k_{1}+k_{2}+..+k_{m}=n}^{ }{n\choose{k_1,k_2,...,k_m}}\prod_{r=1}^{m}\left(a_{r}x^{r}\right)^{k_{r}}\,\mathrm{d}x
\\
\\
&=\sum_{k_{1}+k_{2}+...+k_{m}=n}^{ }\frac{{n\choose{k_1,k_2,...,k_m}}\prod_{r=1}^{m}\left(a_{r}\right)^{k_{r}}}{1+\sum_{r=1}^{m}rk_{r}}\cdot x^{\left(1+\sum_{r=1}^{m}rk_{r}\right)}
\end{align}$$
where $a_r$ are the polynomial's coefficients, where $\displaystyle {n\choose{k_1,k_2,...,k_m}}$ is a multinomial coefficient, and where the sum is taken over all $m$-tuples of nonnegative integers, $k_i$, that satisfy $k_{1}+k_{2}+...+k_{m}=n$. This provides both an explicit formula and an algorithm but for many cases will be woefully inefficient. More efficient algorithms are presumably possible if you exploit properties of the integrand polynomial such as its factors. This also reflects how you, as a human, should approach the problem. Find the qualities of the polynomial you can use - don't just mechanically substitute it into a formula.

To perform the method, the terms, $k_r$, can be generated via a stars and bars method, e.g., in your case of $n=100$ and $m=2$ (corresponding to $3$ terms, including $x^0$), you would systematically place $(3-1)$ bars between $100$ stars:
$$
\begin{array}{|l|l|}
\hline
\text{Stars and bars}&(k_r)_{1\le r\le 3}\\
\hline
**\ldots *|| & (100,0,0)\\
\hline
**\ldots *|*| & (99,1,0)\\
\hline
**\ldots *|**| & (98,2,0)\\
\hline
\ldots & \ldots\\
\hline
*\ldots *|*\ldots*|*\ldots * & (34,33,33)\\
\hline
\ldots & \ldots\\
\hline
|*|*\ldots * & (0,1,99)\\
\hline
||**\ldots * & (0,0,100)\\
\hline
\end{array}$$
Considering the number of multinomial coefficients, we can see that for an $m$-degree polynomial raised to the power $n$, there are $\displaystyle{\binom{n+(m+1)-1}{(m+1)-1}}$ unsimplified terms as there are $(m+1)$ terms in the polynomial. If we don't improve our algorithm to ignore coefficients of $0$ or to factor $x$, this gives a whopping $5151$ terms for $(m,n)=(2,100)$. Thus, your example would have:
$$\begin{align}
\int_{ }^{ }\left(x^2+x+0\right)^{100}\,\mathrm{d}x
&=\frac{{100\choose{\color{red}{100},\color{blue}{0},\color{green}{0}}}\cdot\left(0^{\color{red}{100}}\cdot1^{\color{blue}{0}}\cdot1^{\color{green}{0}}\right)}{\left(1+1\cdot\color{red}{100}+2\cdot\color{blue}{0}+2\cdot\color{green}{0}\right)}\cdot x^{\left(1+1\cdot\color{red}{100}+2\cdot\color{blue}{0}+2\cdot\color{green}{0}\right)}
\\
&+\frac{{100\choose{\color{red}{99},\color{blue}{1},\color{green}{0}}}\cdot\left(0^{\color{red}{99}}\cdot1^{\color{blue}{1}}\cdot1^{\color{green}{0}}\right)}{\left(1+1\cdot\color{red}{99}+2\cdot\color{blue}{1}+2\cdot\color{green}{0}\right)}\cdot x^{\left(1+1\cdot\color{red}{99}+2\cdot\color{blue}{1}+2\cdot\color{green}{0}\right)}
\\
&+\ldots+\frac{{100\choose{\color{red}{0},\color{blue}{0},\color{green}{100}}}\cdot\left(0^{\color{red}{0}}\cdot1^{\color{blue}{0}}\cdot1^{\color{green}{100}}\right)}{\left(1+1\cdot\color{red}{0}+2\cdot\color{blue}{0}+2\cdot\color{green}{100}\right)}\cdot x^{\left(1+1\cdot\color{red}{0}+2\cdot\color{blue}{0}+2\cdot\color{green}{100}\right)}
\\
&=0+0+\ldots+\frac{x^{3}}{3}+\frac{x^{2}}{2}
\end{align}$$
A: Let’s say that $f(x)$ is a polynomial of degree $m$. Then $(f(x))^{n}$ is a polynomial of degree $mn$. Any polynomial of degree $mn$ can be written as the following:
$$
\begin{aligned}
(f(x))^{n} &= \sum_{i=1}^{mn+1}{a_{i}x^{mn+1-i}}\\
\int{(f(x))^{n}\ dx}&= \sum_{i=1}^{mn+1}{\frac{a_{i}}{mn+2-i}x^{mn+2-1}}\ \ +C
\end{aligned}
$$
We now define:
$$
\begin{aligned}
\{
A
\}\ &,A_{i}=a{i}\\
\{
F
\}\ &,F_{i}=(f(i))^{n}\\
[
K
]\ &,M_{i,j}=i^{mn+1-j}\\
\{
L
\}\ &, L_{i}=\frac{1}{mn+2-i}i^{mn+2-i}\\ 
\end{aligned}
$$
Therefore,
$$
\begin{aligned}
\{A\}&=[K]^{-1}\{F\}\\
\\
\int{(f(x))^{n}\ dx}&=\{A\}\cdot\{L\}+C\\
&=\left([K]^{-1}\{F\}\right)\cdot\{L\}+C
\end{aligned}
$$
Maybe expanding the coefficient is less complicated after all. Or even better if substitution can make the integral simpler.
