Elliptic-type integral with nested radical Let:
$$Q(A,Y) = Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1$$
I’m curious as to whether it’s possible to find a closed-form solution for:
$$I(A)=\int_{Y_1(A)}^{Y_2(A)} \frac{1}{\sqrt{\left(1-Y^2\right) Q(A,Y)}}\,dY\\
= \int_{\sin^{-1}(Y_1(A))}^{\sin^{-1}(Y_2(A))}\frac{2 \sqrt{2}}{\sqrt{3-8 A^2+8 A \sin (2 \theta )+4 \cos (2 \theta )+\cos (4 \theta )}}\,d\theta$$
where $0 \le A \le \frac{3 \sqrt{3}}{4}$ and $-1\le Y_1(A), Y_2(A) \le 1$ are the real zeroes of $Q(A,Y)$.
I know a re-parameterisation that makes the zeroes a bit less messy than they would be in terms of $A$.  If we set:
$$A(a) = \frac{\sqrt{a^3 (a+2)^3}}{2 (a+1)}$$
with $0 \le a \le 1$ then we have:
$$Y_{1,2}(a) = \frac{a^2+a\mp (a+2) \sqrt{1-a^2}}{2 (a+1)}$$
I’m not sure if there is any way to explicitly introduce these known zeroes into the integrand, as one could do if $Q(A,Y)$ were a polynomial.  I do have the following relation:
$$Q(A,Y)\,Q(-A,Y)\\
 = Q(A,Y)\,Q(A,-Y)\\
=\left(Y^4-2 Y^2+2 A Y \sqrt{1-Y^2}-A^2+1\right)\left(Y^4-2 Y^2-2 A Y \sqrt{1-Y^2}-A^2+1\right)\\
= P(A,Y)\,P(A,-Y)$$
where:
$$P(A,Y) = Y^4+2Y^3-2Y+A^2-1\\
P(A(a),Y)= (Y-Y_1(a))\,(Y-Y_2(a))\,\left(Y^2+(a+2)Y+\frac{a (a+2)^2+2}{2 (a+1)}\right)$$
$P(A,Y)$ has exactly the same real zeroes as $Q(A,Y)$, but has the opposite sign.
Mathematica is unable to explicitly evaluate the definite integral in either form, but it returns an indefinite integral for the trigonometric form:
$$I(A,\theta) =\\
\frac{4 \sqrt{2} \sqrt{\frac{\left(r_1-r_2\right) \left(r_3-\tan (\theta )\right)}{\left(r_1-r_3\right) \left(r_2-\tan (\theta )\right)}}
   \left(r_1 \cos (\theta )-\sin (\theta )\right) \left(r_4 \cos (\theta )-\sin (\theta )\right) \times \\F\left(\sin
   ^{-1}\left(\sqrt{\frac{\left(r_2-r_4\right) \left(r_1-\tan (\theta )\right)}{\left(r_1-r_4\right) \left(r_2-\tan (\theta
   )\right)}}\right)|\frac{\left(r_2-r_3\right) \left(r_1-r_4\right)}{\left(r_1-r_3\right) \left(r_2-r_4\right)}\right)}{\left(r_1-r_4\right)
   \sqrt{\frac{\left(r_1-r_2\right) \left(r_2-r_4\right) \left(r_1-\tan (\theta )\right) \left(r_4-\tan (\theta
   )\right)}{\left(r_1-r_4\right){}^2 \left(r_2-\tan (\theta )\right){}^2}} \sqrt{3-8 A^2+8 A \sin (2 \theta )+4 \cos (2 \theta )+\cos (4 \theta )}}$$
where the $r_i$ are the roots of:
$$R(A,Y) = A^2 Y^4 - 2 A Y^3 + 2 A^2 Y^2 - 2 A Y + A^2 -1$$
which for $A=A(a)$ are:
$$
\begin{array}{rcl}
r_1 &=& \frac{1-(a+1)^{3/2}\sqrt{1-a}}{a^{3/2} \sqrt{a+2}}\\
&=& \tan(\sin^{-1}(Y_1(a)))\\
r_2 &=& \frac{1+(a+1)^{3/2}\sqrt{1-a}}{a^{3/2} \sqrt{a+2}}\\
& =& \tan(\sin^{-1}(Y_2(a)))\\
r_3 &=& \frac{1-i (a+1) \sqrt{a^2+4 a+3}}{\sqrt{a} (a+2)^{3/2}}\\
r_4 &=& \frac{1+i (a+1) \sqrt{a^2+4 a+3}}{\sqrt{a} (a+2)^{3/2}}
\end{array}
$$
This formula is not very helpful in itself without a deeper understanding of how it was obtained; a naive attempt to get the definite integral from it gives imaginary results, probably because of some issue with branch cuts.
But it does suggest that a human who understood the technique that produced it could perform a similar process to obtain an expression for the definite integral in terms of an elliptic integral.
The motivation here is to find a closed-form expression for the probability density function for the area of a triangle whose vertices are chosen uniformly at random from the unit circle, as discussed in this question:
Moments of area of random triangle inscribed in a circle
I believe $Prob(A) = \frac{2}{\pi^2} I(A)$, where $A$ is the area of such a triangle. Numeric integrals for this quantity give the plot below; this goes to infinity at $A=0$, and is finite and non-zero for the maximum value, $A=\frac{3 \sqrt{3}}{4}$.

 A: If we define:
$$\begin{array}{rcl}
g(a)&=&a^2(a+2)^2-3\\
k(a)&=&(a+1)^3\sqrt{(1-a)(a+3)}\\
\end{array}$$
then, by taking limits of the antiderivative provided by Mathematica at the endpoints of the range of integration, we obtain:
$$I_0(a) = \frac{2\sqrt{2}\,(a+1)\,K\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\sqrt{a(a+2)}\,\sqrt{g(a) + k(a) i}}$$
Here $K$ is a complete elliptic integral of the first kind, and the convention used is that followed by Mathematica, where the argument of $K$ appears unsquared in the defining integral.  The parameter $a$ is related to the original parameter $A$ by the equation stated in the question.
When $g(a)\ge 0$, which holds for $a \gt \sqrt{1+\sqrt{3}}-1 \approx 0.652892$, $I_0(a)$ is real-valued and agrees with numerical evaluations of the integral $I(a)$.
However, when $g(a)$ crosses zero, the function $K$ has a branch cut, and $I_0(a)$ jumps discontinuously to an imaginary-valued function.
This problem can be remedied across the full range for the parameter $a$ by using the analytic continuation of $K$ across the branch cut, written as $K'$, which is discussed in detail in the answer to this question on Math StackExchange:
https://math.stackexchange.com/questions/2008090/analytical-continuation-of-complete-elliptic-integral-of-the-first-kind
$$K'(m) = \frac{1}{\sqrt{m}}\left(K\left(\frac{1}{m}\right)+i K\left(1-\frac{1}{m}\right)\right)$$
$K'$ has its own, different branch cut located in a different part of the complex plane, and the argument in this application does not cross it.  So if we define:
$$I_1(a) = \frac{2\sqrt{2}\,(a+1)\,K'\left(\frac{2 k(a) i}{g(a) + k(a) i}\right)}{\sqrt{a(a+2)}\,\sqrt{g(a) + k(a) i}}$$
then $I_1(a)$ is a real-valued function for $0\lt a \le 1$, and it agrees precisely with numerical evaluations of the original integral $I(a)$.
A: If the purpose of this calculation is to test the geometry conjecture, comparing expansions in powers of $A$ or $a$ should be effective. The indefinite integral can readily be evaluated to any order in $A$, but for the definite integral I run into a difficulty. Take the zeroth order term $A=0=a$, when $Y_{1,2}=\pm 1$, $Q=Y^4-2Y^2+1$ and the integral over $Y$ is
$$I(0)=\int_{-1}^1\frac{1}{\sqrt{(1-Y^2)(Y^4-2Y^2+1)}}\,dY,$$
which has a nonintegrable singularity at the end points (the integrand diverges as $(1\pm y)^{-3/2}$).
I have checked that the small-$a$ asymptotic of $I(a)$ is indeed a $1/\sqrt{a}$ divergence. 
