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?
$$\int { \frac { \sinh { x }  }{ \sqrt { 2\cosh { x+2 }  }  }\frac { 1+\sqrt { \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 }  }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 }  }  }  }{ 1-\sqrt [ 3 ]{ \frac { \sqrt { \cosh { x } +1 } +\sqrt { 2 }  }{ \sqrt { \cosh { x } +1 } +2\sqrt { 2 }  }  }  }  } dx$$

As suggested in comments, let $x=2 \cosh ^{-1}(t)$ to end with
$$I=-2\int\frac{
   \left(\sqrt[3]{\frac{t+1}{t+2}}-\sqrt[6]{\frac{t+1}{t+2}}+1\right)}{\sqrt[6]{\frac{t+1}{t+2}}-1}\,dt$$
Now
$$\sqrt[6]{\frac{t+1}{t+2}}=y \implies t=\frac{1-2 y^6}{y^6-1}$$ makes
$$I=-12 \int\frac{ y^5 (y^2-y+1)}{(y-1) \left(y^6-1\right)^2}\,dy=-12 \int \frac{y^5}{y^5-y^3-y^2+1}\,dy$$
$$y^5-y^3-y^2+1=(y-1)^2 (y+1) \left(y^2+y+1\right)$$
 Partial fraction decomposition of the integrand leads to
$$-\frac{12y^5}{y^5-y^3-y^2+1}=\frac{4 (y+1)}{y^2+y+1}-\frac{7}{y-1}-\frac{2}{(y-1)^2}+\frac{3}{y+1}-12$$ and the final result (hoping no mistake)
$$I=\log \left(\frac{(y+1)^3 \left(y^2+y+1\right)^2}{(y-1)^7}\right)-12
   y+\frac{2}{y-1}+\frac{4 }{\sqrt{3}}\tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)$$
