I need to solve the following integral:
$\int \exp \big( -c \ {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) \big) z \ \mathrm{d} z$
with $c>0$, $d>0$, and $\alpha>2$.
My attempt was to make
$\tfrac{\mathrm{d}}{\mathrm{d} z} \Big( -c \ {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) \Big) = \tfrac{2 c}{z} \Big( \tfrac{1}{1+dz^\alpha} - {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) \Big)$
appear outside the exponential and use $\int u \ \mathrm{d} v = u v \int v \ \mathrm{d} u$ with $u = \tfrac{z}{\frac{\mathrm{d}}{\mathrm{d} z} \big( -c {}_{2}F_{1} ( 1, -\frac{2}{\alpha}, 1 -\frac{2}{\alpha}, -d z^{\alpha}) \big)}$ and $v = \exp \big( -c \ {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) \big)$, but apparently it doens't make it any simpler.
If it's of any help, I've obtained the primitive
$\int {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) \ \mathrm{d} z = \tfrac{r}{3} \Big( {}_{2}F_{1} \big( 1, -\tfrac{2}{\alpha}, 1 -\tfrac{2}{\alpha}, -d z^{\alpha} \big) + 2 {}_{2}F_{1} \big( 1, \tfrac{1}{\alpha}, 1 + \tfrac{1}{\alpha}, -d z^{\alpha} \big) \Big).$
Let us start simplifying the notation by noticing that $$ f_\alpha(z)\stackrel{\text{def}}{=}\phantom{}_2 F_1\left(1,-\frac{2}{\alpha};1-\frac{2}{\alpha};z\right) = -\sum_{n\geq 0}\frac{2}{\alpha n-2}z^n=1-\sum_{n\geq 1}\frac{2z^n}{\alpha n\color{red}{-2}} \tag{1}$$ If the red term were absent, we would simply have $f_\alpha(z)=1-\frac{2}{\alpha}\log(1-z)$, and the given integral, up to the substitution $z\mapsto z^{1/\alpha}$, would simply be a value of an incomplete Beta function. Now it would be really useful to know if you are interested in a simple closed form for a primitive (that probably does not exist) or if the integration range is fixed and you are good with accurate approximations or asymptotics for large $\alpha$s. Please let me know and I will improve the current answer accordingly.