I am trying to solve $$I=\int_{|z|=3} \frac{cos(\pi z)}{(z-2)^2(z+5)(z+1)} \ dz.$$
My attempt:
Residue Theorem:
Let $$f(z)=\frac{cos(\pi z)}{(z-2)^2(z+5)(z+1)}.$$ Now, $$\text{Res}(f,2)=\lim_{z\to 2} \frac{d}{dz} \left(\frac{cos\pi z}{(z+5)(z+1)}\right)=-\frac{10}{441}.$$ $$\text{Res}(f,-1)=\lim_{z\to -1} \left(\frac{cos\pi z}{(z-2)^2(z+5)}\right)=-\frac{1}{36}.$$ Hence, by the Residue Theorem $$I=2\pi i\left(-\frac{10}{441}-\frac{1}{36}\right)=\pi i\left(-\frac{89}{882}\right).$$
Cauchy's Integral Formula:
Let $\Gamma_1$ and $\Gamma_2$ be a small non-intersecting circles around $z=-1$ and $z=2$ respectively. \begin{align} I&=\int_{\Gamma_1} \frac{\frac{\cos(\pi z)}{(z-2)^2(z+5)}}{(z+1)} \ dz +\int_{\Gamma_2} \frac{\frac{\cos(\pi z)}{(z+1)(z+5)}}{(z-2)^2} \ dz \\ &=2\pi i \left(\frac{\cos(\pi z)}{(z-2)^2(z+5)}\right)_{z=-1}+2\pi i \frac{d}{dz}\left(\frac{\cos(\pi z)}{(z+1)(z+5)}\right)_{z=-2} \\ &=\pi i\left(-\frac{89}{882}\right) \end{align}
Yet the answer I have is $-\frac{131}{198}\pi$. I do not see how I can be incorrect.