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 $\int_0^1\left|\dfrac{4t^2-1}{3}\right|^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$ :
$\int_0^1\left|\dfrac{4t^2-1}{3}\right|^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$
$=\int_0^\frac{1}{2}\left(\dfrac{1-4t^2}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt+\int_\frac{1}{2}^1\left(\dfrac{4t^2-1}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{3}\right)dt$
$=\int_0^1\left(\dfrac{1-t}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)d\left(\dfrac{\sqrt t}{2}\right)+\int_1^4\left(\dfrac{t-1}{3}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)d\left(\dfrac{\sqrt t}{2}\right)$
$=\dfrac{1}{3^n4}\int_0^1t^{-\frac{1}{2}}(1-t)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)dt+\dfrac{1}{3^n4}\int_1^4t^{-\frac{1}{2}}(t-1)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{3}\right)dt$
$=\dfrac{1}{3^n4}\int_0^1t^n(1-t)^{-\frac{1}{2}}~_2F_1\left(a,1,n+1;-\dfrac{t}{3}\right)dt+\dfrac{3}{4}\int_0^1t^n(3t+1)^{-\frac{1}{2}}~_2F_1(a,1,n+1;t)~dt$
Which may relate to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/21/01/02/02/
If your question is $\int_0^1\left|t^2-\dfrac{1}{4}\right|^n~_2F_1\left(a,1,n+1;t^2-\dfrac{1}{4}\right)dt$ :
$\int_0^1\left|t^2-\dfrac{1}{4}\right|^n~_2F_1\left(a,1,n+1;t^2-\dfrac{1}{4}\right)dt$
$=\int_0^\frac{1}{2}\left(\dfrac{1-4t^2}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{4}\right)dt+\int_\frac{1}{2}^1\left(\dfrac{4t^2-1}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{4t^2-1}{4}\right)dt$
$=\int_0^1\left(\dfrac{1-t}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)d\left(\dfrac{\sqrt t}{2}\right)+\int_1^4\left(\dfrac{t-1}{4}\right)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)d\left(\dfrac{\sqrt t}{2}\right)$
$=\dfrac{1}{4^{n+1}}\int_0^1t^{-\frac{1}{2}}(1-t)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)dt+\dfrac{1}{4^{n+1}}\int_1^4t^{-\frac{1}{2}}(t-1)^n~_2F_1\left(a,1,n+1;\dfrac{t-1}{4}\right)dt$
$=\dfrac{1}{4^{n+1}}\int_0^1t^n(1-t)^{-\frac{1}{2}}~_2F_1\left(a,1,n+1;-\dfrac{t}{4}\right)dt+\int_0^\frac{3}{4}t^n(4t+1)^{-\frac{1}{2}}~_2F_1(a,1,n+1;t)~dt$
Which may relate to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/21/01/02/02/