Simplify $\frac{_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2};\frac{3}{4}\right)}{\Pi\left(\frac{1}{4}\big|\frac{1}{\sqrt{3}}\right)}$

Is it possible to simplify the ratio $$\mathcal{E}=\frac{_3F_2\left(\frac{1}{2},\frac{3}{4},\frac{5}{4};\ 1,\frac{3}{2};\ \frac{3}{4}\right)}{\Pi\left(\frac{1}{4}\Big|\frac{1}{\sqrt{3}}\right)},$$ where $\Pi(n|k)$ is the complete elliptic integral of the third kind?

Its numeric value is approximately $\mathcal{E}\approx0.73510519389572273268...$, that looks like $\frac{4}{\pi\sqrt{3}}$.

I will start from @Kirill 's comment

The first thing to do is to rewrite the difference of inverse square roots and remove the $\sqrt t$ term :

$$\left(\frac 1 {\sqrt{1-a}} - \frac 1 {\sqrt {1+a}}\right)^2 = \frac {(\sqrt {1+a} - \sqrt {1-a})^2}{1-a^2} = 2\frac {1 - \sqrt{1-a^2}}{1-a^2}$$

Letting $u = \sqrt {1-t}$ and $ v = \sqrt{4-3t}$ we have $\sqrt{1-a^2} = \sqrt{1-3t/4} = v/2$. Then, taking $w = \sqrt {2-v}$, the difference of inverse square roots is expressed as $ \sqrt 2 \sqrt{1-v/2} / (v/2) = 2\sqrt{2-v}/v = 2w/v$.

After moving the $1/2$ factor inside, the integrand of the left-hand side is expressed as $w dt/tuv$ .

Since $\Bbb C(t,u,v) = \Bbb C((v-1)/u)$, we should simplify things by doing the change of variables $x = (v-1)/u$.

We have $t = (9-x^2)(1-x^2)/(3-x^2)^2 \quad u = 2x/(3-x^2) \quad v = (3+x^2)/(3-x^2)$ and $w^2 = 2-v = (1-x^2)/(1-x^2/3)$

$dt/dx = -2x((1-x^2)(3-x^2)+(9-x^2)(3-x^2)-2(9-x^2)(1-x^2))/(3-x^2)^3 \\ = -2x(3 -4x^2 +x^4 +27 - 12x^2+x^4 -18 + 20x^2-2x^4)/(3-x^2)^3 \\ = -2x(12+4x^2)/(3-x^2)^3 = -8x(3+x^2)/(3-x^2)^2 = -4uv/(3-x^2)$

The integrand becomes $w(dt/dx)dx/tuv = -4wdx/t(3-x^2) = -4w(3-x^2)dx/(9-x^2)(1-x^2)$ or also $ -4 \frac {3-x^2}{9-x^2} \frac {dx} {\sqrt{(1-x^2)(1-x^2/3)}}$.

Also, $t=0$ corresponds to $x=1$ while $t=1$ corresponds to $x=0$. After reversing and dividing both integrals by $4$, the problems becomes proving

$\int_0^1 \left( \frac {3-x^2}{9-x^2} - \frac 1 {4-x^2} \right) \frac {dx} {\sqrt{(1-x^2)(1-x^2/3)}} = 0$

The curve $\Bbb C(x, y=\sqrt{(1-x^2)(1-x^2/3)})$ is an elliptic curve. By parity we can extend the integral to the interval $(-1; 1)$ and then again by symmetry, we can extend it to a loop integral (first our interval with $y = + \sqrt{\ldots}$, then we have $x$ go back with $y = - \sqrt{\ldots}$) on the elliptic curve. Doing so multiplies the value of the integral by $4$. So it is zero on the initial path if and only if it is zero on the loop.

$dx/y$ is an invariant differential, so the divisor of the $1$-form we are integrating is the divisor of $\frac {3-x^2}{9-x^2} - \frac 1 {4-x^2} = \frac{3-6x^2+x^4}{(9-x^2)(4-x^2)}$.

By inspection this has a pole of order $1$ at the eight points given by $x=\pm 2, x= \pm 3$ and no other pole. We can go and compute its residue at each pole. We obtain a residue of $1/4$ at the points $P_1,\ldots P_4$ where $xy >0$ and $-1/4$ at the points $P_5 \ldots P_8$ with $xy < 0$.

A computation shows that $(P_1 + \ldots + P_4) - (P_5 + \ldots + P_8)$ is $0$ in the Picard group of the curve, so our $1$-form is the differential of the logarithm of a rational function, plus an invariant differential (a constant times $dx/y$).

If my computation is not wrong it turns out that indeed, it is the differential of $\frac 14 \log \frac {x^4-13x^2+36}{x^4+12xy+11x^2-36}$.

Finally one has to look at what the argument of $\frac {x^4-13x^2+36}{x^4+12xy+11x^2-36}$ does on the loop of interest. In fact this function stays real and strictly negative on both branches of $(-1;1)$ so the argument is constant, which proves that the integral of this logarithm on that loop is $0$.

All in all the whole computation is filled with miraculous simplifications, so I wouldn't be surprised if there is an more elegant, less brute-force-y method.

Here are the gory details to find the log term.

First, one ugly and easy way is to put the curve in Weierstrass form, and follow the addition procedure to get the needed factor as a big quotient of products of line equations.

Here we were lucky enough to land on the same form for the curve as the one present in the second integral, so I didn't really want to do that if I could help it.

First, looking at the divisor, we must have $f(x,y) = c/f(x,-y)$ for some constant $c$ (actually we have $c=1$ for the function I obtained, which is totally non-obvious when looking at it). So I first tried to find some $g$ such that $f =g(x,y)/g(x,-y)$. This is nice because then to compute its divisor I have to look at $g(x,y)g(x,-y)$ which is a function of $x$ only. The other symmetry also shows that I can search for a $g$ with $g(x,y) = \pm g(-x,-y)$.

After finding out that there is no nontrivial polynomial of degree <= 2 (in $x$) that has $P_1,\ldots,P_4$ as zeros, you eventually stumble on the two degree $3$ polynomials $g_1 = x^3-6y-x$ and $g_2 = xy-2x^2+6$.

Then it was very late and I didn't realize that even though $g_1\bar{g_1}$ and $g_2\bar{g_2}$ were of degree $6$ in $x$ (so had potentially $4$ unwanted zero/pole), we actually have $g_1\bar{g_1} = (x^2-4)(x^2-9)(x^2-1)$ and $g_2\bar{g_2} = (x^2-4)(x^2-9)(1-x^2/3)$ : the extraneous roots are precisely the branch points with $y=0$ so they appear in both $g_i$ and $\bar{g_i}$, and cancel each other when doing the quotient.

Hence both $\log \frac{x^3-6y-x}{x^3+6y-x}$ and $\log \frac{xy-2x^2+6}{-xy-2x^2+6}$ work too (the two quotients are actually opposite of each other)

(instead I found the two linear combinations that made the extra roots go to infinity because why not (I was somehow convinced I couldn't move them to the branch points, lol), but this was all strange because it was over $\Bbb Q(\sqrt 3)$. I proceeded to multiply the two solutions together to cancel the zero/poles at infinity while doubling the other zeros/poles; Then I took the square root with the identity $\sqrt{a/b} = \sqrt{ab}/b$)