Recognize as a special function?

Question

Is the following function a special function of some kind $$ f(x) = \int_0^x (1+e^{-t})^{b}\,dt, $$ where $b>1$?

Answer 1

If $b$ is integer, then you can use the binomial theorem to get

$$f(x) = \sum_{k=0}^b \binom{b}{k} \int_0^x e^{-kt} dt = \sum_{k=0}^b \binom{b}{k} \frac{1}{k} (1-e^{-kx})$$

which isn't 'special' at all. If $b$ is non-integer then I don't know, but since argument $x$ of $f(x)$ is 'just' the upper integration bound, it doesn't seem like it's in the same league as exciting and well-known special functions like $\Gamma$ or $\zeta$.

You may also write $$f(x) = \int_{e^{-x}}^{1} \frac{(1+u)^b}{u} du$$

which again does not hit me as anything special and probably isn't too hard to solve.