I want the Laurent expansion \begin{equation} f(z) = \sum_{n=-\infty}^{\infty}c_n (z-a)^n \end{equation}
of $f(z) = \frac{1}{(z+1)(z+3)}$ around the isolated pole $a =-1$ (for $|z+1|<2$). From a simple rewriting of the Taylor series, or from using Cauchy integral formula of order $n+1$, I find the correct answer, which is $c_n = 2^{-2-n}(-1)^{n+1}$ for $n\geq -1$ and $0$ otherwise. However, for fun I wanted to ``manually'' integrate the expression \begin{equation} c_n =\frac1{2\pi i} \oint \frac{f(\xi)}{(\xi+1)^{n+1}}d\xi \label{eq:} \end{equation} Selecting the circle with center $\xi =1$ and radius $r<2$, parameterized as $\xi+1 = r e^{i\theta}$ I get \begin{equation} c_n =\frac{r^{-n-1}}{2\pi} \int_0^{2\pi}\frac{e^{-i\theta(n+1)}}{re^{i\theta}+2}d\theta. \end{equation}
However, according to Mathematica, with
FullSimplify[
Integrate[Exp[-I x (n + 1)]/(r Exp[I x] + 2), {x, 0, 2 Pi}],
Assumptions -> Element[n,Integers] && 0 < r < 2],
this integral is $0$. What is wrong?