Impossible hyperbolic integral I'm quite good with integral calculation, very rarely happens that I'm not able to solve an indefinite integral. I've tried for 7 days to solve this monster but in the end, I surrendered to this beast.  
I'm sure that it has a solution because it was a challenge from my calculus professor. Unfortunately, he didn't give the solution of this integral. It has been unsolved for years, since 7 days ago when I found it in my old notebook.
Can some expert help me?
$$\int { \frac { \sinh { x }  }{ \sqrt { 2\cosh { x+2 }  }  }\frac { 1+\sqrt { \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 }  }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 }  }  }  }{ 1-\sqrt [ 3 ]{ \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 }  }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 }  }  }  }  } dx$$

First make the substitution $y=\cosh x,$ then simplify the radicals by removing radicals from all denominators, to obtain the simpler $$\int\frac{\sqrt {2y+2}}{2y+2}\frac{\sqrt{(y-7)(y-3-\sqrt{2y+2})}}{\left({(y-7)^2(y-3-\sqrt{2y+2})}\right)^{1/3}}\mathrm d y.$$ Then make the substitution $2y+2=z^2.$ The integral becomes $$\int\left(\frac{z^2-8-2z}{z^2-16}\right)^{1/6}\mathrm d z.$$ Note that $$\frac{z^2-8-2z}{z^2-16}=1-\frac{2}{4+z}.$$ Can you complete it now? OK, seems I need to add that you only need make the substitution $$w^6=\frac{z^2-8-2z}{z^2-16},$$ so that the integral becomes $$12\int\frac{w^6\mathrm d w}{(1-w^6)^2},$$ which is elementary.
