resolve differential equation without using hypergeometric function? I can't resolve this differential equation
$$dx=\left(\frac{a +be^{\frac{y}{c}}}{a+b}\right)^{cd} dy,$$
where $a, b, c, d \in \mathbb{R_{\geq 0}}.$
I tried with separable variables, i.e
$$x=\int_{0}^{y}\left(\frac{a +be^{\frac{y}{c}}}{a+b}\right)^{cd} dy \\= \frac{1}{(a+b)^{cd}}\int_{0}^{y}\left(a +be^{\frac{y}{c}}\right)^{cd} dy,$$
but i can't find a solution without calling on Hypergeometric function.
Appreciate your help!!!
 A: Let’s find your integral of:

$$x=  \frac{1}{(a+b)^{cd}}\int_{0}^{y}\left(a +be^{\frac{y}{c}}\right)^{cd} dy =  \frac{a^{cd}}{(a+b)^{cd}}\int_{0}^{y}\left(1 +\frac ba e^{\frac{y}{c}}\right)^{cd} dy $$

where the Incomplete Beta function is equal to:
$$\text B_z(a,b)=\int_0^z t^{a-1}(1-t)^{b-1}dt,\text{Re}(a)>0$$
when we have:
$$\frac{a^{cd}}{(a+b)^{cd}}\int_{0}^{y}\left(1 +\frac ba e^{\frac{y}{c}}\right)^{cd} dy$$
let $$1 +\frac ba e^{\frac{y}{c}}=t$$ or a similar substitution and you will get that:
$$\frac{a^{cd}}{(a+b)^{cd}}\int_{0}^{y}\left(1 +\frac ba e^{\frac{y}{c}}\right)^{cd} dy= -\frac{ca^{cd}}{(a+b)^{cd}}\text B_{\left(\frac{be^{\frac zc}}{a}, \frac{b}{a}\right)}(cd+1,0) $$
where the Generalized Incomplete Beta function equals:
$$\text B_{(\alpha,\beta)}(a,b)= \int_\alpha^\beta t^{a-1}(1-t)^{b-1}dt=\text B_\beta(a,b)-\text B_\alpha(a,b)$$
which is partly simplified as you can also use this formula of you do use a hypergeometric function:

conversion formula 

Please correct me and give me feedback!
