Closed form double integral $ \int_{a}^{c}dr \int_{b}^{d} dr' \, \frac{r r'}{\sqrt{(r - a)(r' - b)(r-c)(r'-d)}} \frac{r_<^{\ell}}{r_>^{\ell+1}}$ Is there a closed form expression for
$$
S_\ell =
\int\limits_{a}^{c}dr
\int\limits_{b}^{d} dr' \,
\frac{r r'}{\sqrt{(r - a)(r' - b)(r-c)(r'-d)}}
\frac{[\min( r , r')]^{\ell}}{[\max(r,r')]^{\ell+1}}
$$
for $0<a<b<c<d$? Here $\max(r,r') = r$ if $r\geq r'$ and $\max(r,r') = r'$ otherwise, $\min(r,r')$ is defined similarly.
For large $\ell$, it may be shown that the integral is dominated by $r\approx r'$ and asymptotically
$$
S_{\ell} \rightarrow \frac{2}{\ell}\int\limits_{b}^{c} dr \,
\frac{r^2}{\sqrt{(r - a)(r - b)(r-c)(r-d)}}
$$
This can be integrated in a closed form using elliptic integrals as shown here.
That solution seemed so general, I wonder if the double integral can also be integrated similarly in a closed form. 
 A: (IN PROGRESS)


For each non-negative integer $\ell\in\mathbb{N}\cup\{0\}$ and $4$-tuple of real parameters $\left(a,b,c,d\right)\in\mathbb{R}^{4}$ with the conditions $0<a<b<c<d$, define the term $\mathcal{S}_{\ell}{\left(a,b,c,d\right)}$ via the integral representation
  $$\mathcal{S}_{\ell}{\left(a,b,c,d\right)}:=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{\ell}\left[\max{\left(x,y\right)}\right]^{-\ell-1}}{\sqrt{\left(x-a\right)\left(y-b\right)\left(x-c\right)\left(y-d\right)}}.\tag{1}$$

The maximum and minimum functions employed in the expression of $(1)$ above may be defined via algebraic functions, respectively, as follows (see [2a] and [2b]):
$$\begin{cases}
&\max{\left(x,y\right)}:=\frac{x+y}{2}+\frac12\sqrt{\left(x-y\right)^{2}};~~~\small{x,y\in\mathbb{R}},\\
&\min{\left(x,y\right)}:=\frac{x+y}{2}-\frac12\sqrt{\left(x-y\right)^{2}};~~~\small{x,y\in\mathbb{R}}.\tag{2}\\
\end{cases}$$
Note that the max/min functions are related by
$$\max{\left(x,y\right)}\cdot\min{\left(x,y\right)}=xy,\tag{3}$$
as may be readily verified using definitions $(2)$ directly.
Another consequence of definition $(2)$ is the usual definition of the minimum function,
$$\min{\left(x,y\right)}=
\begin{cases}
&x;~~~\small{x\le y},\\
&y;~~~\small{x\ge y}.\tag{4}\\
\end{cases}$$
Using the above information, we can reduce the integral representation $(1)$ for $\mathcal{S}_{\ell}{\left(a,b,c,d\right)}$ to integrals whose integrands are purely algebraic, i.e., do not explicitly involve max/min type functions.

For convenience, we introduce as an auxiliary function the quadratic function of two parameters $Q_{\alpha,\beta}:\mathbb{R}\rightarrow\mathbb{R}$, defined by
$$Q_{\alpha,\beta}{\left(x\right)}:=\left(x-\alpha\right)\left(\beta-x\right);~~~\small{\alpha,\beta,x\in\mathbb{R}}.\tag{5}$$
Now, let $\ell\in\mathbb{N}\cup\{0\}\land\left(a,b,c,d\right)\in\mathbb{R}^{4}\land0<a<b<c<d$. We then find
$$\begin{align}
\mathcal{S}_{\ell}{\left(a,b,c,d\right)}
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{\ell}\left[\max{\left(x,y\right)}\right]^{-\ell-1}}{\sqrt{\left(x-a\right)\left(y-b\right)\left(x-c\right)\left(y-d\right)}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{\ell}}{\left[\max{\left(x,y\right)}\right]^{\ell+1}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{\ell}\left[\min{\left(x,y\right)}\right]^{\ell+1}}{\left[\max{\left(x,y\right)}\right]^{\ell+1}\left[\min{\left(x,y\right)}\right]^{\ell+1}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{\left[\max{\left(x,y\right)}\cdot\min{\left(x,y\right)}\right]^{\ell+1}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{xy\,\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{\left(xy\right)^{\ell+1}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{b}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{x}^{c}\mathrm{d}y\,\frac{\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{c}^{d}\mathrm{d}y\,\frac{\left[\min{\left(x,y\right)}\right]^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{b}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{x}^{c}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{c}^{d}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{b}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{b}^{c}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~-\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~+\int_{b}^{c}\mathrm{d}x\int_{c}^{d}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{x^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&~~~~~-\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{x^{2\ell+1}-y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{Q_{a,c}{\left(x\right)}\,Q_{b,d}{\left(y\right)}}}\\
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{x^{\ell+1}}{y^{\ell}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}\\
&~~~~~-\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{x^{2\ell+1}-y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}.\tag{6}\\
\end{align}$$


For each non-negative integer $\ell\in\mathbb{N}\cup\{0\}$ and $4$-tuple of real parameters $\left(a,b,c,d\right)\in\mathbb{R}^{4}$ with the conditions $0<a<b<c<d$, define the term $\mathcal{I}_{\ell}{\left(a,b,c,d\right)}$ via the integral representation
  $$\mathcal{I}_{\ell}{\left(a,b,c,d\right)}:=\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{x^{2\ell+1}-y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}.\tag{7}$$

Then, assuming $\ell\in\mathbb{N}\cup\{0\}\land\left(a,b,c,d\right)\in\mathbb{R}^{4}\land0<a<b<c<d$, we obtain
$$\begin{align}
\mathcal{S}_{\ell}{\left(a,b,c,d\right)}
&=\int_{a}^{c}\mathrm{d}x\int_{b}^{d}\mathrm{d}y\,\frac{x^{\ell+1}}{y^{\ell}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}\\
&~~~~~-\int_{b}^{c}\mathrm{d}x\int_{b}^{x}\mathrm{d}y\,\frac{x^{2\ell+1}-y^{2\ell+1}}{x^{\ell}y^{\ell}\sqrt{\left(x-a\right)\left(c-x\right)\left(y-b\right)\left(d-y\right)}}\\
&=\left(\int_{a}^{c}\frac{x^{\ell+1}}{\sqrt{\left(x-a\right)\left(c-x\right)}}\,\mathrm{d}x\right)\left(\int_{b}^{d}\frac{\mathrm{d}y}{y^{\ell}\sqrt{\left(y-b\right)\left(d-y\right)}}\right)-\mathcal{I}_{\ell}{\left(a,b,c,d\right)}\\
&=\small{\left(\int_{0}^{1}\frac{\left(c-a\right)\left[a+\left(c-a\right)t\right]^{\ell+1}}{\sqrt{\left(c-a\right)^{2}t\left(1-t\right)}}\,\mathrm{d}t\right)\left(\int_{0}^{1}\frac{\left(d-b\right)\left[d-\left(d-b\right)u\right]^{-\ell}}{\sqrt{\left(d-b\right)^{2}u\left(1-u\right)}}\,\mathrm{d}u\right)}\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)};~~~\small{\left[x=a+\left(c-a\right)t\land y=d-\left(d-b\right)u\right]}\\
&=\left(\int_{0}^{1}\frac{\left[a+\left(c-a\right)t\right]^{\ell+1}}{\sqrt{t\left(1-t\right)}}\,\mathrm{d}t\right)\int_{0}^{1}\frac{\mathrm{d}u}{\left[d-\left(d-b\right)u\right]^{\ell}\sqrt{u\left(1-u\right)}}\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)}\\
&=\small{\left(\int_{0}^{1}\frac{\sum_{n=0}^{\ell+1}\binom{\ell+1}{n}a^{\ell-n+1}\left(c-a\right)^{n}t^{n}}{\sqrt{t\left(1-t\right)}}\,\mathrm{d}t\right)\int_{0}^{1}\frac{\mathrm{d}u}{d^{\ell}\left[1-\left(\frac{d-b}{d}\right)u\right]^{\ell}\sqrt{u\left(1-u\right)}}}\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)}\\
&=\small{\left(\sum_{n=0}^{\ell+1}\binom{\ell+1}{n}a^{\ell-n+1}\left(c-a\right)^{n}\int_{0}^{1}\frac{t^{n}}{\sqrt{t\left(1-t\right)}}\,\mathrm{d}t\right)\frac{1}{d^{\ell}}\int_{0}^{1}\frac{\mathrm{d}u}{\left(1-zu\right)^{\ell}\sqrt{u\left(1-u\right)}}}\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)};~~~\small{\left[z:=\frac{d-b}{d}\right]}\\
&=\frac{\pi}{d^{\ell}}\,{_2F_1}{\left(\ell,\frac12;1;z\right)}\sum_{n=0}^{\ell+1}\binom{\ell+1}{n}a^{\ell-n+1}\left(c-a\right)^{n}\int_{0}^{1}t^{n-\frac12}\left(1-t\right)^{-\frac12}\,\mathrm{d}t\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)}\\
&=\frac{\pi}{d^{\ell}}\,{_2F_1}{\left(\ell,\frac12;1;\frac{d-b}{d}\right)}\sum_{n=0}^{\ell+1}\binom{\ell+1}{n}a^{\ell-n+1}\left(c-a\right)^{n}\operatorname{B}{\left(n+\frac12,\frac12\right)}\\
&~~~~~-\mathcal{I}_{\ell}{\left(a,b,c,d\right)}.\tag{8}\\
\end{align}$$
It remains to evaluate the term $\mathcal{I}_{\ell}{\left(a,b,c,d\right)}$.

TBC...

