Integral of (1+bt^z)^(-y)t^x dt = (1/z)*b^(-(x+1)/z) B((x+1)/z,y-(x+1)/z)

The following formula holds for $z>0$, $b>0$, and $0 < \mathrm{Re} \left( \dfrac{x+1}{z} \right) < \mathrm{Re}(y)$: $$\displaystyle\int_0^{\infty} (1+bt^z)^{-y} t^x \mathrm{d}t = \dfrac{1}{z} b^{-\frac{x+1}{z}} B \left( \dfrac{x+1}{z}, y - \dfrac{x+1}{z} \right),$$ where $B$ denotes the beta function.