I was trying to find the normalization constant of the following distribution:$$p(x)=(1-(1-q)x^2)^{\frac{1}{1 - q}}$$ where $1<q<3$, which done by integration over $p(x)$ from $-\infty$ to $\infty$ as $$A=\int_{-\infty }^{\infty}p(x)dx=\int_{-\infty }^{\infty}(1-(1-q)x^2)^{\frac{1}{1 - q}}dx$$ which according to Wolfram Mathematica $A=\frac{\sqrt{\pi} \Gamma(\frac{1}{q-1}-\frac{1}{2})}{\sqrt{q-1} \Gamma(\frac{1}{q-1})}$, I believe such integration can be calculated by hypergeometrical functions, but I don't know how, I would be glad if someone give me a help to find a way to know how to solve such integral?
I also wonder, can such integral be solved by hypergeometric functions $$\int_{-\infty }^{\infty}(1-(1-q)x^2)^{\frac{1}{1 - q}}(1-(1-q)(x+c)^2)^{\frac{1}{1 - q}}dx$$ my question emerges from the fact that the integrand is not an even function, if not, what other methods can be used
Thanks in advance
Effectively, the antiderivative express in terms of the gaussian hypergeometric function $$I=\int\Big[1-(1-q)x^2)\Big]^{\frac{1}{1 - q}}\,dx=x \, _2F_1\left(\frac{1}{2},\frac{1}{q-1};\frac{3}{2};(1-q) x^2\right)$$ and the definite integral $$J=\int_{-\infty}^{+\infty}\Big[1-(1-q)x^2)\Big]^{\frac{1}{1 - q}}\,dx=\sqrt{\frac{\pi}{q-1}}\,\,\frac{\Gamma \left(\frac{3-q}{2 (q-1)}\right)}{\Gamma \left(\frac{1}{q-1}\right)}$$ provided that $$2 \Re\left(\frac{q-2}{q-1}\right)<1\land (\Re(q)\geq 1\lor q\notin \mathbb{R})$$ which is your case with $1<q<3$.
For the second integral, I do not see any possible solution except a series expansion around $c=0$.
For simplicity, let $a=q-1$ and b=$\frac 1 {1-q}$ to make $$K=\int_{-\infty}^{+\infty} \left(1+a x^2\right)^b \left(1+a (x+c)^2\right)^b\,dx$$
using $$\left(1+a (x+c)^2\right)^{b}=\sum_{n=0}^\infty Q_n\, c^n$$
$$K=\sum_{n=0}^\infty c^n\,K_n \quad \text{where}\quad K_n=\int_{-\infty}^{+\infty} \left(1+a x^2\right)^b \,Q_n\,dx$$
We have $K_{2n+1}=0$ and $$K_{2n}=(-1)^{\frac n2} a^{\frac {n-1}2}\sqrt \pi\ \frac{\Gamma \left(\frac{n-1}{2}-2 b\right)}{\Gamma \left(\frac{n}{2}+1\right) \Gamma (n-2 b)}\, \prod_{k=0}^{\frac n2 -1}(b-k)^2$$
This means that $K$ is again an hypergoemetric function. I shall not write it since @Maxim gave the complete result (much more simpler than mine).