complex analysis - Laurent series for $f (z)=frac {sin (2 pi z)}{z (z^2 + 1)}$

How Can find the Laurent series for this function valid for $0 <|z-i|<2$

$$f (z)=\frac {\sin (2 \pi z)}{z (z^2 +1)}$$

Let $g (z) = \sin (\pi z)$

$$\sin (\pi z ) = \sin( 2 \pi (z - i)) \cos (2 \pi i) + \cos (2 \pi (z-i)) \sin (2 \pi i )$$

And Let $h (z)= \frac {1}{z^2 + 1}$

$$\frac {1}{z (z^2 + 1)}= \frac {1}{i (1 -(-(z-i))}[\frac {1/2i}{z-i} +\frac {-1/2i}{2i (1-(-\frac {z-i}{2i}))}]$$

So it's easy to find expansion for $g (z)$ and $h (z)$ and then multiply the two expansions

We notice that $f$ has simple pole at $z = i$ So, we can get the principal part easily Or using this

$$2 \pi i a_1 = \int_{|z-i|=1} f (z) dz$$

Is there a trick to find the Laurent series quickly ?

This question was in my exam .I Calculated the principal part , but I didn't have enough time to calculate the exact form for the analytic part .

Thank you