<s>[INST] Evaluating $ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $ I would like to evaluate: $$ \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx $$
$$ \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)} $$
The substitution $ x \rightarrow \sin(x) $ or $ \cos(x) $ can only simplify the denominator, and $ x \rightarrow \sqrt{1+x}$ or $ \sqrt{1-x} $ is also useless...
Can you help me find a useful substitution?
$$ x=\cos(2t) $$
$$  \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt $$
$$ u=\tan(t/2) $$
$$ -4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du $$
But now it looks even more complicated... ?
 [/INST] Would it help you greatly if you transform the integrand to
$$\frac{2-(x+2) \sqrt{1-x}+(x-2) \sqrt{1+x}+2 \sqrt{1-x^2}}{2 x^2}?$$
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