Analytic form of: $ \int \frac{\bigl[\cos^{-1}(x)\sqrt{1-x^2}\bigr]^{-1}}{\ln\bigl( 1+\sin(2x\sqrt{1-x^2})/\pi\bigr)} dx $

Question

Background: On my quest to solve difficult integrals, I chanced upon this site: http://www.durofy.com/5-most-beautiful-questions-from-integral-calculus/

Good problems for me, (novice), although I believe these integrals maybe easy for others.

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I'm having problems with this integral :- $$ \int \frac{\left[\cos^{-1}(x)\left(\sqrt{1-x^2}\right)\right]^{-1}}{\ln\left( 1+\frac{\sin(2x\sqrt{1-x^2})}{\pi}\right)} dx $$

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My effort: I tried to convert $\sqrt{1-x^2}$ into '$ \cos(\arcsin x) $' and then simplify, but I couldn't do it. Then I tried to convert it to '$\sin(\arccos x)$', but it just made it worse.

Edit: As Lucian suggested in the comments, I came this far -

$$ \int \frac{-1}{t\ln\left( 1+\frac{\sin(\sin(2t))}{\pi}\right)} dx $$ I would've thought of doing this, if the 'primary sine function' in the denominator was an 'arcsin function'. Problem is, I'm not able to proceed. What do I do?

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Question: What is this integral's analytic form? What is the underlying trick/substitution/concept needed to solve this integral?

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Note: A non-closed form solution may also exist; I don't know about that. If you do manage to evaluate this integral in terms of even special functions including Bessel, Gamma or Faddeeva, it's okay; you can post it.