# 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*}

• It would be really great if you showed us what you've done even if it hasn't worked :) Apr 2, 2019 at 18:59
• Have edited my original post. Apr 3, 2019 at 9:20
• Thank you! You have earned an upvote from me :) Apr 3, 2019 at 15:17
• Nice work for sure and $\to +1$ Apr 4, 2019 at 8:57
• I wouldn't be optimistic: WolframAlpha seems to fail.
– user65203
Apr 4, 2019 at 9:01

$$\color{brown}{\textbf{Version of 03.05.19}}$$

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 $$(8),(10)$$ $$\color{brown}{\textbf{present the given integral in the closed form}}.$$

• The first identity you state doesn't seem to be correct. If I set $u^3-v^3=1$ then $4(\frac{u-v}{2})^3+3\frac{u-v}{2}=\frac{1}{2}+\frac{3}{2}(u-v)(1-uv)$ which I cannot seem to reduce further unless $uv=1$. Can you provide a proof of the statement? Apr 30, 2019 at 21:40
• Alpha not finding a solution is a good indication that no closed-form exists.
– user65203
May 2, 2019 at 7:34
• In the steps below (3) and above (4) there should be a $$\sqrt[3]{\sinh(2x)}=\sqrt[6]{y^2-1}$$ in the denominator. May 2, 2019 at 7:54
• I agree with the above comment. In fact, according to my working, using the substitution proposed I obtain $$I=\frac{1}{2}\int\left(\frac{\sqrt[3]{\frac{y+1}{2}}+\sqrt[3]{\frac{y-1}{2}}}{\sqrt[3]{\frac{y^2-1}{2}}}\right)^{1/2}\frac{y}{\sqrt{y^{2}-1}}\,\textrm{d}y.$$ Nothing more manageable! May 2, 2019 at 11:17
• @YuriNegometyanov: definitely hard, and definitely interesting ! May 4, 2019 at 13:54

Integration by parts with $$\sf{f(y)=\frac{1-y^6}{(1-y^3)^{17/6}}\implies f'(y)=\frac{y^2(17+5y^3)}{2(1-y^3)^{17/6}}}\\ g'(y)=(1+y)^{1/2}\implies g(y)=\frac23(1+y)^{3/2}$$ gives $$\sf{J=\frac{3}{2}\int\frac{(1-y^6)(1+y)^{1/2}}{(1-y^3)^{17/6}}\,dy=\frac{(1-y^6)(1+y)^{3/2}}{(1-y^3)^{17/6}}-\overbrace{\frac12\int\frac{y^2(1+y)^{3/2}}{(1-y^3)^{17/6}}(17+5y^3)\,dy}^K}$$ and the substitution $$\sf{w=1-y^3}$$ yields $$\small\sf{K=-\frac16\int\frac{(1+\sqrt[3]{1-w})^{3/2}}{w^{17/6}}(22-5w)\,dw=\overbrace{\frac56\int\frac{(1+\sqrt[3]{1-w})^{3/2}}{w^{11/6}}\,dw}^P-\overbrace{\frac{11}3\int\frac{(1+\sqrt[3]{1-w})^{3/2}}{w^{17/6}}\,dw}^Q}.$$ It is possible to simplify further; let us consider $$\sf Q$$ first. Integration by parts again with \begin{align}\sf{f(w)=(1+(1-w)^{1/3})^{3/2}}&\sf{\implies f'(w)=-\frac{(1+(1-w)^{1/3})^{1/2}}{2(1-w)^{2/3}}}\\\sf{g'(w)=\frac1{w^{17/6}}}&\sf{\implies g(w)=-\frac6{11w^{11/6}}}\end{align} gives $$\sf{Q=\frac{11}3\left[-\frac{6(1+\sqrt[3]{1-w})^{3/2}}{11w^{11/6}}-\frac3{11}\int\frac{(1+(1-w)^{1/3})^{1/2}}{w^{11/6}(1-w)^{2/3}}\,dw\right]}$$ and the substitution $$\sf{t=1+(1-w)^{1/3}}$$ yields $$\sf{Q=-\frac{2(1+\sqrt[3]{1-w})^{3/2}}{w^{11/6}}+3\int\frac{\sqrt t}{(1-(t-1)^3)^{11/6}}\,dt}$$ and since $$\sf{1-(t-1)^3=(2-t)(t^2-t+1)}$$, the substitution $$\sf{s=2-t}$$ results in $$\sf{\int\frac{\sqrt t}{(1-(t-1)^3)^{11/6}}\,dt=\int\frac{\sqrt{2-s}}{s^{11/6}(s^2-3s+3)^{11/6}}\,ds.}$$ Using a similar procedure, we find that $$\sf{P=-\frac{(1+\sqrt[3]{1-w})^{3/2}}{w^{5/6}}+\frac32\int\frac{\sqrt{2-s}}{s^{5/6}(s^2-3s+3)^{5/6}}\,ds}$$ and putting back $$\sf{w=1-y^3}$$, we can now derive a simpler expression for $$\sf J$$: $$\sf{J=\frac{(1-y^6)[(1+y)^{3/2}-y^{3/2}]}{(1-y^3)^{17/6}}+3\int\frac{\sqrt{2-s}}{s^{11/6}(s^2-3s+3)^{11/6}}\,ds-\frac32\int\frac{\sqrt{2-s}}{s^{5/6}(s^2-3s+3)^{5/6}}\,ds}.$$

• That font looks sweet, by the way! :P Apr 28, 2019 at 8:30

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.

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.