TITLE: What is the most efficient method to evaluate this indefinite integral?
QUESTION [20 upvotes]: $$\int x^5 e^x\,\mathrm{d}x$$
Is there another, more efficient way to solve this integral that is not integration by parts?

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
&\color{#66f}{\large\int x^{5}\expo{x}\,\dd x}
=\pars{\totald[5]{}{\mu}\int\expo{\mu x}\,\dd x\,}_{\,\mu\ =\ 1}
=\left.\totald[5]{}{\mu}\pars{\expo{\mu x} \over \mu}\,\right\vert_{\,\mu\ =\ 1}
\\[3mm]&=\left.\frac{x^5 e^{\mu  x}}{\mu }-\frac{5 x^4 e^{\mu  x}}{\mu ^2}
+\frac{20 x^3 e^{\mu  x}}{\mu ^3}-\frac{60 x^2 e^{\mu  x}}{\mu ^4}
-\frac{120 e^{\mu  x}}{\mu ^6}+\frac{120 x e^{\mu  x}}{\mu ^5}
\,\right\vert_{\,\mu\ =\ 1}
\\[3mm]&\color{#66f}{\large%
=\left(x^5-5 x^4+20 x^3-60 x^2+120 x-120\right)\expo{x}}
\end{align}