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Greg Egan
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I have a partial answer. 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 $g(a) \lt 0$, $I_0(a)$ is imaginary-valued, and the absolute value does not match the integral.

I would hope that there’s some way to remedy this, either by finding a second solution for the portion of the domain where this formula fails, or, better still, by finding a simplification of the formula that applies to the entire domain.

Greg Egan
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