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.

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 $I(a)$.