Antiderivative of a rational trigonometric function I'm looking for some help in solving the following integral:
$$\int{\frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)+A}}}d\theta$$
I've seen some similar cases with elliptical integrals, but even in this case, I couldn't solve the antiderivative above. It would be great to solve it analytically, but any help is appreciated.
 A: A bit of softening up first:
$$\frac{\sin^2x}{\sqrt{\cos^2x+A}}=\frac1{\sqrt{1+A}}\frac{\sin^2x}{\sqrt{1-\frac1{1+A}\sin^2x}}$$
Then, by Byrd and Friemdan 281.01,
$$\int_0^\varphi\frac{\sin^2x}{\sqrt{\cos^2x+A}}\,dx=\frac1{\sqrt{1+A}}\int_0^\varphi\frac{\sin^2x}{\sqrt{1-\frac1{1+A}\sin^2x}}\,dx$$
$$=\frac1{\sqrt{1+A}}\int_0^{F(\varphi,m=1/(1+A))}\operatorname{sn}^2u\,du$$
By B&F 310.02 this in turn evaluates as
$$\frac1{\sqrt{1+A}}\cdot\frac1{1/(1+A)}(F(\varphi,m)-E(\varphi,m))=\sqrt{1+A}(F(\varphi,m)-E(\varphi,m))$$

In some references like the DLMF, the difference $F(\varphi,m)-E(\varphi,m)$ is denoted in terms of a new function $D(\varphi,m)$:
$$D(\varphi,m)\equiv\int_0^\varphi\frac{\sin^2x}{\sqrt{1-m\sin^2x}}\,dx=\frac1m(F(\varphi,m)-E(\varphi,m))$$
This is because the difference can be evaluated in one step (avoiding a subtraction) using the Carlson symmetric form as
$$D(\varphi,m)=\frac{\sin^3\varphi}3R_D(\cos^2\varphi,1-m\sin^2\varphi,1)$$
The original integral expressed in this symmetric form is
$$\int_0^\varphi\frac{\sin^2x}{\sqrt{\cos^2x+A}}\,dx=\frac1{\sqrt{1+A}}D\left(\varphi,\frac1{1+A}\right)$$
$$=\frac{\sin^3\varphi}{3\sqrt{1+A}}R_D\left(\cos^2\varphi,1-\frac1{1+A}\sin^2\varphi,1\right)\qquad|\varphi|\le\frac\pi2$$
A: $$\frac{\sin^2\theta}{\sqrt{\cos^2\theta+a}}=\frac{a+1}{\sqrt{\cos^2\theta+a}}-{\sqrt{\cos^2\theta+a}}$$ shows the decomposition in elliptic integrals of the first and second kind.
