# Integral involving the Gauss Hypergeometric functions with quadratic argument

I'm trying to calculate this integral

$$\int_{0}^1\left|\frac{4t^2-1}{3}\right|^n\,{}_2F_1\left(a,1,n+1; \frac{4t^2-1}{3}\right)dt$$

where $a\in \mathbb{R}$ and $n\in\mathbb{N}$. We can split

$$\int_{0}^{1/2}\left(\frac{1-4t^2}{3}\right)^n\,{}_2F_1\left(a,1,n+1; \frac{4t^2-1}{3}\right)dt+$$$$\int_{1/2}^{1}\left(\frac{4t^2-1}{3}\right)^n\,{}_2F_1\left(a,1,n+1; \frac{4t^2-1}{3}\right)dt$$

1.-First integral: performing the change $1-4t^2\rightarrow u$ we obtain

$$\frac{n!\sqrt{\pi}}{4\,3^n\,\Gamma(n+3/2)}\,{}_2F_1\left(a,1,n+3/2; \frac{-1}{3}\right)$$

2.-The main problem is second integral. I've tried to use some transformation of variable (https://dlmf.nist.gov/15.8) of Hypergeometric function but none seem useful.

Any help will be wellcome.

If your question is $\int_0^1\left|\dfrac{4t^2-1}{3}\right|^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$ :

$\int_0^1\left|\dfrac{4t^2-1}{3}\right|^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$

$=\int_0^\frac{1}{2}\left(\dfrac{1-4t^2}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt+\int_\frac{1}{2}^1\left(\dfrac{4t^2-1}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$

$=\int_0^1\left(\dfrac{1-t}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)d\left(\dfrac{\sqrt t}{2}\right)+\int_1^4\left(\dfrac{t-1}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)d\left(\dfrac{\sqrt t}{2}\right)$

$=\dfrac{1}{3^n4}\int_0^1t^{-\frac{1}{2}}(1-t)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)dt+\dfrac{1}{3^n4}\int_1^4t^{-\frac{1}{2}}(t-1)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)dt$

$=\dfrac{1}{3^n4}\int_0^1t^n(1-t)^{-\frac{1}{2}}~_2F_1\left(a,1,n+1;-\dfrac{t}{3}\right)dt+\dfrac{3}{4}\int_0^1t^n(3t+1)^{-\frac{1}{2}}~_2F_1(a,1,n+1;t)~dt$

If your question is $\int_0^1\left|t^2-\dfrac{1}{4}\right|^n~_2F_1\left(a,1,n+1;t^2-\dfrac{1}{4}\right)dt$ :

$\int_0^1\left|t^2-\dfrac{1}{4}\right|^n~_2F_1\left(a,1,n+1;t^2-\dfrac{1}{4}\right)dt$

$=\int_0^\frac{1}{2}\left(\dfrac{1-4t^2}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{4}\right)dt+\int_\frac{1}{2}^1\left(\dfrac{4t^2-1}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{4}\right)dt$

$=\int_0^1\left(\dfrac{1-t}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)d\left(\dfrac{\sqrt t}{2}\right)+\int_1^4\left(\dfrac{t-1}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)d\left(\dfrac{\sqrt t}{2}\right)$

$=\dfrac{1}{4^{n+1}}\int_0^1t^{-\frac{1}{2}}(1-t)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)dt+\dfrac{1}{4^{n+1}}\int_1^4t^{-\frac{1}{2}}(t-1)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)dt$

$=\dfrac{1}{4^{n+1}}\int_0^1t^n(1-t)^{-\frac{1}{2}}~_2F_1\left(a,1,n+1;-\dfrac{t}{4}\right)dt+\int_0^\frac{3}{4}t^n(4t+1)^{-\frac{1}{2}}~_2F_1(a,1,n+1;t)~dt$