Evaluating $\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$ The question is to evaluate $$\int \frac{x \sqrt{2\sin(x^2+1) - \sin2(x^2+1)}}{2 \sin(x^2+1)+\sin2(x^2+1)} dx$$ if $x^2 \neq (n \pi-1) \forall n\in N$
I tried to rewrite the integral as $$\int \frac{x\sqrt{(2\sin(x^2+1))(1-\cos(x^2+1))}}{(2\sin(x^2+1))(1+\cos(x^2+1))}$$
If there would have been a radical sign in denominator too the problem would be quite simple but I couldn't understand how to proceed from here.Any ideas?

Note that the substitution $x^2+1=y$ and $2x\,\mathrm dx=\mathrm dy$ leads to\begin{align}\frac12\int\frac{\sqrt{2\sin(y)-2\sin(y)\cos(y)}}{2\bigl(\sin(y)+2\sin(y)\cos(y)\bigr)}\,\mathrm dy&=\frac{\sqrt2}4\int\frac{\sqrt{\sin(y)-\sin(y)\cos(y)}}{\sin(y)+2\sin(y)\cos(y)}\,\mathrm dy\\&=\frac{\sqrt2}4\int\frac{\sqrt{\csc(y)-\cot(y)}}{1+\cos y}\mathrm dy.\end{align}Usinge the formulas for $\sin(2\theta)$, $\cos(2\theta)$, and $\cot(2\theta)$, it is not hard to see that this is equal to $\frac{\sqrt2}4\cdot\frac23\tan^{3/2}\left(\frac y2\right)$.
