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 $$\frac{(-1)^{k + 1} (-a - b + k + 1) \binom{b-1}{k} \binom{a + b - k - 2}{a - k - 1}}{a}.$$$$\frac{(-1)^{k+1} (-a - b + k + 1) \binom{b-1}{k} \binom{a + b - k - 2}{a - k - 1}}{a}.$$ A little simpler: $$\frac{(-1)^k b \binom{b-1}{k} \binom{a + b - k - 1}{b}}{a}.$$
