Impossible hyperbolic integral

Question

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?

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

Answer 1

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) $$I=\log \left(\frac{(y+1)^3 \left(y^2+y+1\right)^2}{(y-1)^7}\right)-12 y+\frac{2}{y-1}+\frac{4 }{\sqrt{3}}\tan ^{-1}\left(\frac{2 y+1}{\sqrt{3}}\right)$$

Answer 2

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.