Complementing the answer by Maximilian Janisch: per Mathematica, for $$J:=\int_0^1 a^\alpha(1-a)^{\beta-1}\cdot\left(1+\frac{a\cdot(b-1)}b\right)^{\delta-1}\,\mathrm da $$ we have $$J=-\frac{\pi \csc (\pi \alpha ) \Gamma (\beta ) \, _2\tilde{F}_1\left(\alpha +1,1-\delta ;\alpha +\beta +1;\frac{1}{b}-1\right)}{\Gamma (-\alpha )} $$ if $0<b\le1/2$ and $$J=b (b-1)^{-\alpha -1} \left(\frac{\Gamma (\alpha +1) b^{\alpha } \Gamma (-\alpha -\delta ) \, _2F_1\left(\alpha +1,1-\beta ;\alpha +\delta +1;\frac{b}{1-b}\right)}{\Gamma (1-\delta )}+\frac{\pi b^{-\delta } \Gamma (\beta ) (\cot (\pi (\alpha +\delta ))+i) (1-b)^{\alpha +\delta } \, _2F_1\left(1-\delta ,-\alpha -\beta -\delta +1;-\alpha -\delta +1;\frac{b}{1-b}\right)}{\Gamma (-\alpha -\delta +1) \Gamma (\alpha +\beta +\delta )}\right) $$ if $1/2<b<1$, where $_2F_1$ is the hypergeometric function and $_2\tilde{F}_1$ is the regularized hypergeometric function.
It appears unlikely that these expressions for $J$ can be significantly simplified.