Integral of $(4\cot^{3}x+\cot^{2}x+\cot x-2)e^{x}dx$

Question

How to integrate the following expression $$(4\cot^{3}x+\cot^{2}x+\cot x-2)e^{x}dx$$

Answer 1

We are going to integrate it by parts one by one in descending order.

$$ \begin{aligned} 4\int e^x \cot ^3 x & =4 \int e^x\left(\csc^2 x-1\right) \cot x d x \\ & =-2 \int e^x d\left(\cot ^2 x\right)-4 \int e^x \cot x d x \\ & =-2 e^x \cot ^2 x+2 \int e^x \cot ^2 xdx-4 \int e^x \cot xdx \end{aligned} $$

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$$ I=-2 e^x \cot ^2 x+3 \int e^x \cot ^2 x d x-3 \int e^x \cot x d x-2 e^x $$

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$$ \begin{aligned} \int e^x \cot ^2 x d x & =\int e^x\left(\csc ^2 x-1\right) d x \\ & =-\int e^x d(\cot x)-e^x \\ & =-e^x \cot x+\int e^x \cot x d x-e^x \end{aligned} $$

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Plugging back yields $$ \begin{aligned} \boxed{I =-e^x\left(2 \cot ^2 x+3 \cot x+5\right)+C} \end{aligned} $$

Answer 2

I think this integral can't expressed as closed form. If you can use hypergeometric function, you can integrate since the following equality holds.

$$ \int\cot xe^xdx=-ie^x{}_2F_1\left(-\frac i2,1;1-\frac i2;e^{2ix}\right)-\left(\frac25+\frac i5\right)e^{(1+2i)x}{}_2F_1\left(1,1-\frac i2;2-\frac i2;e^{2ix}\right)+\mathit{Const.} $$

This identity can be easily obtained by expressing cot as an exponential function.