Difficult integral involving trigonometric and hypertrigonometric functions This is definitely the most difficult integral that I've ever seen.
Of course, I'm not able to solve this. 
Could you help me?
$$\int { \sin { x\cos { x } \cosh { \left( \ln { \sqrt { \frac { 1 }{ 1-\sin { x }  }  } +\tanh ^{ -1 }{ \left( \sin x \right) +\tanh ^{ -1 }{ \left( \cos { x }  \right)  }  }  }  \right)  }  } dx } $$
 A: Using the fact that $\cosh t = \frac{1}{\sqrt{1-\tanh^2 t}}$ and   $\sinh t = \frac{\tanh t}{\sqrt{1-\tanh^2 t}} $ $$\cosh(a+b+c) = \cosh a \cosh b \cosh c + \sinh a \sinh b \cosh c + \sinh a \cosh b \sinh c + \cosh a \sinh b \sinh c$$
the integrand simplifies to
$$\int dx \left[\cosh\left(\log\left(\frac{1}{\sqrt{1-\sin x}}\right)\right)\left(1+\sin x \cos x\right) + \sinh\left(\log\left(\frac{1}{\sqrt{1-\sin x}}\right)\right)\left(\sin x + \cos x \right) \right]$$
Next we'll substitute $x = 2z+\frac{\pi}{2}$:
$$\int 2dz \left[\cosh\left(\log\left(\frac{1}{\sqrt{1-\cos 2z}}\right)\right)\left(1-\sin 2z \cos 2z\right) + \sinh\left(\log\left(\frac{1}{\sqrt{1-\cos 2z}}\right)\right)\left(\cos 2z - \sin 2z \right) \right]$$
$$ = \frac{1}{\sqrt2}\int dz (2\sin z + \csc z )(1-\sin 2z \cos 2z)-(2\sin z - \csc z )(\cos 2z - \sin 2z)$$
$$ = \frac{1}{\sqrt2}\int dz\left[4\sin^2 z \cos z(1-\cos 2z) - 2\cos z(1+2\cos 2z)+2\sin z(1-\cos 2z)+ \csc z(1+\cos 2z) \right]$$
$$ =\frac{1}{\sqrt{2}} \int(8\sin^4 z + 4 \sin^ z - 4)\cos z - (4\cos^2 z - 2)\sin z + 2 \csc z dz$$
$$= \sqrt{2}\left(\frac{4}{5}\sin^5 z + \frac{2}{3}\sin^3 z - 2 \sin z + \frac{2}{3}\cos^3 z - 2\cos z - \log|\csc z + \cot z|\right)$$
Therefore our final answer is 
$$\sqrt{2}\left(\frac{4}{5}\sin^5 \left(\frac{x}{2}-\frac{\pi}{4}\right) + \frac{2}{3}\sin^3 \left(\frac{x}{2}-\frac{\pi}{4}\right) - 2 \sin \left(\frac{x}{2}-\frac{\pi}{4}\right) + \frac{2}{3}\cos^3 \left(\frac{x}{2}-\frac{\pi}{4}\right) - 2\cos \left(\frac{x}{2}-\frac{\pi}{4}\right) - \log{\left|\csc \left(\frac{x}{2}-\frac{\pi}{4}\right) + \cot \left(\frac{x}{2}-\frac{\pi}{4}\right)\right|}\right) + C$$
Edit: We can simplify this a bit further with some trig shenanigans. Applying the angle subtraction formulas, we get:
$$\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{1}{3}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^3 - 2 \left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right) + \frac{1}{3}\left(\sin \left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right)^3 - 2\left(\sin \left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)\right) - \sqrt{2}\log{\left|\frac{\sqrt{2}+\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)}\right|} + C$$
$$=\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{2}{3}\sin \left(\frac{x}{2}\right)\left(\sin^2 \left(\frac{x}{2}\right)+3\cos^2\left(\frac{x}{2}\right)\right) - 4 \sin \left(\frac{x}{2}\right) - \sqrt{2}\log{\left|\frac{\sqrt{2}+\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)}\right|} + C$$
$$=\frac{1}{5}\left(\sin \left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^5 + \frac{2}{3}\sin \left(\frac{x}{2}\right)\cos x - \frac{8}{3} \sin \left(\frac{x}{2}\right) - \frac{1}{\sqrt{2}}\log{\left(\frac{3+2\sqrt{2}(\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right))+\sin x}{1-\sin x}\right)} + C$$
And I think I will stop there.
A: Mathematica gives:
$$-\frac{\sqrt{\frac{1}{1-\sin (x)}} \sqrt{\sin ^2(2 x)} \csc^2(x) \\ \left(-90 \sin \left(\frac{x}{2}\right)+35 \sin \left(\frac{3 x}{2}\right)-3 \sin \left(\frac{5 x}{2}\right)+15 \cos \left(\frac{3 x}{2}\right)+3 \cos \left(\frac{5 x}{2}\right)+30 \cos \left(\frac{x}{2}\right) \left(4 \sqrt{\frac{1}{\cos (x)+1}} \log \left(\tan
\left(\frac{x}{2}\right)-1\right)-4 \sqrt{\frac{1}{\cos (x)+1}} \log \left(2 \sqrt{\frac{1}{\cos (x)+1}}+\tan \left(\frac{x}{2}\right)+1\right)+1\right)\right)}{60 \left(\csc \left(\frac{x}{2}\right)+\sec \left(\frac{x}{2}\right)\right)}$$
so I doubt you'll want to work through this by hand.
