Definite integral of a hypergeometric function

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This question is related to Bounding a polynomial from below but perhaps more straightforward.

Let $\sigma >0$ be fixed. For even $k \in \mathbb{N} \cup \{0\}$, we consider the hypergeometric function \begin{equation} \varphi_k(x) = {}_{2}F_{1}(-k,k+\sigma+\frac12;\frac12;x^2), \quad x \in (-1,1). \end{equation}

My question is the following:

Can we find a "closed form" for the definite integral \begin{equation} \int_{-1}^1 (1-x^2)^\sigma \varphi_k(x)^2 dx \end{equation} for all even $k\in \mathbb{N} \cup \{0\}$?


Some remarks:

Since $k$ is an integer, $\varphi_k$ terminates and is thus a polynomial of the form \begin{equation} \varphi_k(x) = \sum_{j=0}^{k} (-1)^j {k \choose j} b_j \, x^{2j} \quad x \in (-1,1), \end{equation} where $b_j = \frac{\Big(k+\sigma+\frac12\Big)_j}{\Big(\frac12 \Big)_j}$ and for $s \in \mathbb{R}$, $(s)_j$ denotes the Pochhammer symbol \begin{equation} (s)_{j}={\begin{cases}1&j=0\\s(s+1)\cdots (s+j-1)&j>0.\end{cases}} \end{equation} In particular, $\varphi_0(x) =1$.

Writing the square of the sum as the double sum of the product of the coefficients, and using the linearity of the integral, I obtain something of the form \begin{equation} \sum_{j=0}^k \sum_{\ell = 0}^k (-1)^{j+\ell} {k \choose \ell} {k \choose j} \frac{b_j b_\ell}{2(j+\ell)+1} \int_{-1}^1 (1-x^2)^\sigma x^{2(\ell+j)} dx. \end{equation} Moreover, according to Wolfram Alpha, \begin{equation} \int_{-1}^1 (1-x^2)^\sigma x^{2(\ell+j)} dx = \frac{\Gamma(j+\ell+\frac12)\Gamma(\sigma+1)}{\Gamma(\sigma+\ell+j+\frac32)}. \end{equation} I am interested however in knowing if this can be reduced to an expression which does not contain a double sum and is more "compact".

I made this remark in the previous question, and perhaps they might be useful.