WolframAlpha claims it is
$$\frac{(-1)^{k + 1} (k + 1) (-a - b + k + 1) \binom{b}{k + 1} \binom{a + b - k - 2}{a - k - 1}}{a b},$$
and you can absorb the $k+1$ and $b$, yielding
$$\frac{(-1)^{k + 1} (-a - b + k + 1) \binom{b-1}{k} \binom{a + b - k - 2}{a - k - 1}}{a}.$$