A tricky integral involving hyperbolic functions Can anyone suggest a method for solving the integral below? I've tried numerous things but have had no luck yet. To be honest I'm not sure an analytical solution actually exists.
$$I=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x.$$
Thanks.
Here is another attempt I have made:
Let $y=[\tanh(x)]^{2/3}$, and rewrite $I$ such that 
$$I=\int\cosh(2x)[\sinh(x)]^{-1/3}\sqrt{1+[\tanh(x)]^{2/3}}\,\textrm{d}x.$$
Then $\textrm{d}x=(3/2)[\tanh(x)]^{1/3}[\cosh(x)]^{2}\,\textrm{d}y$. Therefore
\begin{align*}
I&=\frac{3}{2}\int\cosh(2x)[\sinh(x)]^{-1/3}[\tanh(x)]^{1/3}[\cosh(x)]^{2}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int\cosh(2x)[\cosh(x)]^{5/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\cosh(x)]^{11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\sinh(x)]^{2}[\cosh(x)]^{5/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\tanh(x)]^{2}[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}\sqrt{1+y}\,\textrm{d}y
+\frac{3}{2}\int[\textrm{sech}(x)]^{-11/3}y^3\sqrt{1+y}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
+\frac{3}{2}\int y^{3}\frac{(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1+y^3)(1+y)^{1/2}}{(1-y^3)^{11/6}}\,\textrm{d}y
\\
&=\frac{3}{2}\int\frac{(1-y^6)(1+y)^{1/2}}{(1-y^3)^{17/6}}\,\textrm{d}y
\end{align*}
 A: $\color{brown}{\textbf{Version of 03.05.19}}$
Are known the identities

$$\begin{cases}
\cosh^2t-\sinh^2t = 1\\
2\cosh^2t = \cosh2t + 1\\
2\sinh^2t = \cosh2t -1\\
2\sinh t\cosh t = \sinh 2t\\
\cosh^{-2}t = 1-\tanh^2x\\
\sinh^{-2}t = \coth^2x-1\\
(\tanh x)' = \cosh^{-2}t\\
(\coth x)' = -\sinh^{-2}t.
\end{cases}\tag1$$

One can get
$$\cosh^{\large^-\frac23}x + \sinh^{\large-^\frac23}x
 = \sqrt[{\large3}]{1-\tanh^2x} + \sqrt[{\large3}]{\coth^2x-1},$$
$$\cosh^{\large^-\frac23}x + \sinh^{\large-^\frac23}x
 = \sqrt[{\large3}]{\coth x - \tanh x}
(\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}),\tag2$$
$$\cosh2x = \dfrac{\cosh^2x+\sinh^2x}{\cosh^2x-\sinh^2x}
=\dfrac{\coth x + \tanh x}{\cot x - \tanh x}.\tag3$$
Therefore,
$$I = \int\cosh2x\sqrt{\cosh^{\large^-\frac23}x + \sinh^{\large^-\frac23}x\,}\,dx,$$
$$I = \int\dfrac{\coth x + \tanh x}{(\coth x - \tanh x)^{\large^\frac56}} \sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}\,dx.\tag4$$
Taking in account that
$$\left(\frac65(\sinh x \cosh x)^{\large^-\frac56}\right)'
= \dfrac{\coth x + \tanh x}{(\coth x - \tanh x)^{\large\frac56}}$$
(see also Wolfram Alpha),
$$\left(\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}\right)'
= \dfrac{\left(\coth^{\large^-\frac23}x (1-\coth^2x)
+ \tanh^{\large^-\frac23}x (1-\tanh^2x)\right)}{6\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}}$$
$$= -\dfrac{(\coth x - \tanh x) 
\left(\sqrt[{\large3}]{\coth x} - \sqrt[{\large3}]{\tanh x}\right)}{6\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}},$$
is possible the integration by parts:
$$I(x) = -\dfrac65 \int\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}\,\mathrm d\left((\sinh x \cosh x)^{\large^-\frac56}\right),$$
$$I(x) = \dfrac35 I_1(x) - \dfrac65\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}\,(\sinh x \cosh x)^{\large^-\frac56},\tag5$$
where
$$I_1(x) = \int \left((\sinh x \cosh x)^{\large^-\frac56}\right) \dfrac{\mathrm d\left(\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}\right)}{\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}}.
\tag6$$
Let
$$\left(\sqrt[{\large3}]{\coth x} - \sqrt[{\large3}]{\tanh x}\right)^2 = y,\tag7$$
then
$$\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x} 
= \sqrt{y+4}\,$$
$$(\sinh x \cosh x)^{-1} = \dfrac{\cosh^2x - \sinh^2x}{\sinh x \cosh x} = \coth x - \tanh x = \sqrt y (y+3),$$
$$I(x) = \dfrac3{10} J\left(\left(\sqrt[{\large3}]{\coth x} - \sqrt[{\large3}]{\tanh x}\right)^2\right) - \dfrac65\sqrt{\sqrt[{\large3}]{\coth x} + \sqrt[{\large3}]{\tanh x}}\,(\sinh x \cosh x)^{\large^-\frac56},\tag8$$
where
$$J(y) = \int (y+3)^{\large^\frac56}(y+4)^{\large^-\frac34}\,y^{\large^\frac5{12}}\,dy\tag9,$$
wherein the last integral can be expressed in the closed form of
$$\begin{align}
&J(y) = \frac2{51} y^{\large^\frac5{12}} \left(-17\cdot 3^{\large^\frac56} F_1\left(\frac5{12}; \frac34, \frac16; \frac{17}{12};-y, -\frac y3\right)\right.\\
& + \left. 5\cdot 3^{\large^\frac56} y F_1\left(\frac{17}{12}; \frac34, \frac16;\frac{29}{12}; -y, -\frac y3\right) + 17 (y+1)^{\large^\frac14} (y+3)^{\large^\frac56}\right) + \mathrm{constant}
\end{align}\tag{10}$$
via the Appell hypergeometric function of two variables.
Formulas $(8),(10)$ $\color{brown}{\textbf{present the given integral in the closed form}}.$
A: Other attempt:
With $t=e^{2x}$, and omitting a constant factor, you obtain the rational form
$$\int (t+t^{-1})\sqrt{(t^{1/2}-t^{-1/2})^{-2/3}+(t^{1/2}+t^{-1/2})^{-2/3}}\frac{dt}t$$
or
$$\int (t^2+1)(t^2-1)^{-1/3}\sqrt{(t-1)^{2/3}+(t+1)^{2/3}}\,t^{-4/3}\,dt.$$
Nothing more appetizing.
A: Thanks to Yuri for his working. Very helpful. I believe I have arrived at the solution now
\begin{align*}
 I&=\int\cosh(2x)\sqrt{[\sinh(x)]^{-2/3}+[\cosh(x)]^{-2/3}}\,\textrm{d}x
\\
&=\int\cosh(2x)\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1/6}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
 \\
 &=\frac{6}{5}\int\frac{\textrm{d}}{\textrm{d}x}\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}
 [\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
 \\
 &=\frac{6}{5}\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}
 \\&\quad-\frac{6}{5}
 \int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x.
\end{align*}
Now
\begin{equation*}
 \frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}
 =\frac{1}{6}\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1}\frac{[\sqrt[3]{\tanh(x)}-\sqrt[3]{\coth(x)}]}{[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}}.
\end{equation*}
Therefore
\begin{align*}
 I_{2}&=\frac{6}{5}
 \int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\frac{\textrm{d}}{\textrm{d}x}[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}\,\textrm{d}x
\\&=
 \frac{1}{5}\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{-1/6}
 \frac{[\sqrt[3]{\tanh(x)}-\sqrt[3]{\coth(x)}]}{[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}}\,\textrm{d}x.
\end{align*}
Let $y=[\sqrt[3]{\tanh(x)}+\sqrt[3]{\coth(x)}]^{1/2}$ then
\begin{equation*}
 I_{2}=
 \frac{6}{5}\int\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}\,\textrm{d}y=\frac{6}{5}\int[\coth(x)-\tanh(x)]^{-5/6}\,\textrm{d}y
.
\end{equation*}
After a bit of work it is possible to show that
\begin{equation*}
\coth(x)-\tanh(x)=(y^{4}-1)\sqrt{y^{4}-4}.
\end{equation*}
Therefore
$$I=\frac{6}{5}\biggl\{\biggl[\frac{\sinh(2x)}{2}\biggr]^{5/6}y-\int(y^{4}-1)^{-5/6}(y^{4}-4)^{-5/12}\,\textrm{d}y\biggr\},$$
where $y$ is as above and the integral can be written in terms of the hypergeometric function.
A very interesting problem!
Thanks to all.
