I would like to find an expression for the following integrals, using special functions or otherwise; I suspect they can be written in terms of elliptic functions but I've played with it for a while and haven't arrived at anything useful. The integrals are:
$$ \mathcal{I}_1 = \int \frac{1}{\sqrt{1-\alpha^2(\beta^2-\cos^2\xi)}}\frac{\xi \cos\xi}{\sqrt{\beta^2-\cos^2\xi}} d\xi $$ and $$ \mathcal{I}_2 = \int \frac{1}{\sqrt{1-\frac{1}{4}\sin^2\xi}}\frac{\xi \cos\xi}{\sqrt{\beta^2-\sin^2\xi}} d\xi $$
I would be happy with expressions for the perhaps simpler form:
$$ \mathcal{I}_a = \int \frac{\sin^{-1}\left(B\cos\xi\right)}{\sqrt{A^2-\sin^2\xi)}} d\xi $$ and $$ \mathcal{I}_b = \int \frac{\sin^{-1}\left(B\sin\xi\right)}{\sqrt{A^2-\sin^2\xi)}} d\xi $$
Since $\cos^2(x) = 1 - \sin^2(x)$ both $I_1$ and $I_2$ can be rewritten as some constant multiple of an antiderivative of the form: \begin{align} I(a,b) =\int \frac{x\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x \end{align} for some constants $a,b$. Now, from the definition of the Elliptic integral of the First Kind we see that \begin{align} \frac{1}{b}F\left(\arcsin\left(\frac{x}{a}\right), \frac{a}{b}\right) =\frac{1}{b} \int_0^{\sin\left(\arcsin\left(\frac{x}{a}\right)\right)} \frac{\mathrm{d}t}{\sqrt{1-\frac{a^2}{b^2}t^2}\sqrt{1-t^2}} \overset{u = at}{=} \int_0^{x} \frac{\mathrm{d}u}{\sqrt{a^2-u^2}\sqrt{b^2-u^2}} \end{align} So by taking the substitution $u = \sin(x)$ we get $$ \int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \mathrm{d}x=\frac{1}{b}F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right) $$ And thus \begin{align} I(a,b)\overset{\text{I.B.P.}}{=}& x\int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x - \int\left(\int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x\right)\mathrm{d}x\\ =&\frac{x}{b}F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right) - \frac{1}{b}\int F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right)\mathrm{d}x \end{align} And now, even for the simple case of $a=1$ and $b=\frac{1}{\sqrt{2}}$ the antiderivative in the previous equation already involves hypergeometric functions and generalized hypergeometric functions, so I don't have much hope for a possible closed form for the general case.
Hope this helps!