I am trying to solve
$$ \frac{dy}{dt} = \alpha ((y+1)^2 - \gamma)^n \hspace{2cm} y(0)=0 $$
Here $y$ is a real-valued, monotonically increasing, positive definite function of $t$ in the interval $(0,\infty)$ and is bounded on every closed interval in $(0,\infty)$; $\alpha$ and $\gamma$ are constants.
After a few manipulations, one arrives at this expression:
$$ \int_{x_0}^{x(t)}{\frac{1}{{x^n}\sqrt{x+1}} dx} = \beta t $$
for some $\beta$. For integer $n>1$ the integral can also be performed:
$$ 2^{1-2n} {2(n-1) \choose n-1} \left( \mathrm{tanh^{-1}}(\sqrt{x+1})-\mathrm{tanh}^{-1}(\sqrt{x_0+1}) \right) + \sqrt{x+1}\sum_{k=1}^{n-1}{(-1)^{-k}b_{n,k}\frac{1}{k x^k}} = (-1)^{n}\beta t $$
The coefficients $b_{n,k}$ in the above sum can be represented by
$$ b_{n,k}=\frac{{2(n-1) \choose 2k}{2(n-1-k) \choose n-1-k}}{2^{2(n-1-k)}{n-1 \choose k}^2} = \frac{\Gamma (n - \frac12) \Gamma (k+1)}{\Gamma (n) \Gamma (k + \frac12)} $$
I am not certain how to extend this result to non-integer values of $n$ without the use of special functions, e.g. the Beta function, which is suggested by the integral. In any case, the next step in the calculation would be to take the inverse of the LHS - but I have no idea how to do that. However, the form of the sum seems very suggestive; perhaps at the limit of large $n$, $\sqrt{x+1}\sum_{k=1}^{n-1}{(-1)^{-k}b_{n,k}k^{-1}x^{-k}}$ may be approximated by a constant and for low $n$ by some other function, which will enable further calculations.
The following also seems to hold:
$$ \sqrt{x+1} \sum^{n-1}_{k=1}{(-1)^{-k}\frac{\Gamma(n-\frac12)(k-1)!}{\Gamma(k+\frac12)}x^{-k}} = (-1)^n \frac{2\sqrt{x+1}}{(2n-1)x^{n}} {_{2}F_{1} \left( 1,n;n+\frac12;-\frac1x \right)} -\frac{2 \Gamma(n-\frac12)}{\sqrt{\pi} \Gamma(n)} \mathrm{sinh}^{-1} \left( \frac{1}{\sqrt{x}} \right) $$
although I am uncertain to what degree that may lead to an invertible expression for the integral without the use of more "exotic" special functions; after all the hypergeometric function in this expression appears to be the dominant term. I found that the late-time behaviour may be approximated by the $\int{dx \frac{1}{x^n}}$; however such an approximation misses the early-time behaviour and does not approach the correct asymptote by simply applying the original initial condition.
It would still be very helpful if there exist special exact or approximate formulae for the inverse of the incomplete Beta function, e.g. for special parameter values (the case $n=1$ is already derived here).
A closed form for the integral is the Incomplete Beta function : $$\int{ \frac{1}{\sqrt{x+1}}{x^n}} dx = (-1)^{n+1}B_{-x}(n+1 , 1/2)+constant$$ The inverse function needs numerical calculus to be evaluated : http://www.dtic.mil/dtic/tr/fulltext/u2/a467901.pdf