Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$ I need help with this integral:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
The integrand graph looks like this:
$\hspace{1in}$
The approximate numeric value of the integral:
$$I\approx8.372211626601275661625747121...$$
Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it.
I am also interested in cases when only numerator or only denominator is present under the logarithm.
 A: This is not really an answer, but grossly too long for an comment. I didn't know how to simplify it beyond the final solution.
$$I=\int_{-1}^1 \frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\text{d}{x}$$
Begin with the substitution of $x=-\cos2a$ 
$$I=\int_{-1}^1 \frac{1}{-\cos2a}\sqrt{\frac{1-\cos2a}{1+\cos2a}}\ln\left(\frac{2\cos^2 2a-2\cos 2a+1}{2\cos^2 2a-2\cos2a+1}\right)\text{d}{x}$$
By the tangent and cos double angle properties
$$I=\int_{-1}^1 -\sec2a|\tan a|\ln\left(\frac{-2\cos^22a+\cos 4a+2}{2\cos2a+\cos4a+2}\right)\text{d}{a}$$
Were just getting started. Now replace $a=\frac{1}{2}\text{gd}(b)$ where $\text{gd}$ is the Gudermannian function. 
$$I=\int_{-1}^1 -\sec(\text{gd}(b))|\tan(\text{gd}(\frac{b}{2}))|\ln\left(\frac{-2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}{2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}\right)\text{d}{a}$$
Hehe. Now we get to simplify a bit. This is under the definition of Gudermannian properties.
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{-2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}{2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}\right)$$
Now, use properties of $\tanh$ and $\text{sech} $ to simplify even further
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{(1-\text{sech}^2 b)+2}{(1+\text{sech}^2 b)+2}\right)$$
Our goal is to create an $\text{arctanh}$ function, but that will obviously take some serious effort. Factor out a $3$ to generate that $1$ needed even if it makes an ugly factoring.
$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{3(1-\frac{\text{sech}^2 b}{3})}{3(1+\frac{\text{sech}^2 b}{3})}\right)$$
And now cut out all of the 3's. After this cut, use a property of $\ln$'s to reciprocate the argument of $\ln$. And multiply 2 and 1/2
$$I=\int_{-1}^1 2\text{cosh}\space b|\sinh\frac{b}{2}|\frac{1}{2}\ln\left(\frac{(1+\frac{\text{sech}^2 b}{3})}{(1-\frac{\text{sech}^2 b}{3})}\right)$$
And what do you know! You're there! Use a property of $\ln$ and $\text{arctanh}$ to generate a much CLEANER form (also by throwing the 2 in front).
$$I=2\int_{-1}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$
This function is even, and we can know that because all parts of what is above, $\cosh b,|\sinh b|, $ etc. all even. So we can do the following.
$$I=4\int_{0}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$
This is just an idea, and like I said not a real solution. I have no idea where to continue beyond this, but I thought it may help to come up with a new idea to solve.
A: Eight years later.
Starting from @Ron Gordon's substitution
$$8 \int_0^{\infty}  \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du$$ since the roots of the polynomials in $\color{red}{ u^2}$ are simple, we can use partial fraction decomposition (I shall not type the formulae) and we face four integrals
$$I=\int \frac \alpha {\beta x^2+\gamma}\log(x)\,dx$$ where all coefficients are complex numbers. Then
$$I=\frac{i \alpha  \left(\text{Li}_2\left(\frac{i x \sqrt{\beta }}{\sqrt{\gamma
   }}\right)-\text{Li}_2\left(-\frac{i x \sqrt{\beta }}{\sqrt{\gamma }}\right)+\log (x)
   \left(\log \left(1-\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)-\log
   \left(1+\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)\right)\right)}{2  \sqrt{\beta\gamma }}$$ which make
$$J=\int_0^\infty \frac \alpha {\beta x^2+\gamma}\log(x)\,dx=\frac{i \alpha  \left(\log ^2\left(\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)-\log
   ^2\left(-\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)\right)}{4 
   \sqrt{\beta\gamma }}=-\frac{\pi  \alpha  \log \left(\frac{\beta }{\gamma }\right)}{4 \sqrt{\beta\gamma }}$$ This gives as a  result
$$8 \int_0^{\infty}  \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du=\pi  \left(\pi -\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)\right)$$ I have not been able to simplify further.
Edit
If you look at this question of mine, @Jyrki Lahtonen made the simplification I was not able to do.
A: Figured I would contribute and add a self-contained real analytic method:
I will use the following representations that are fairly straight-forward to prove:
$$2\int_{0}^{\infty}\frac{\cos(x)-\cos(t \, x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\ln \left((t^2 -a)^2 + b^2\right)-\ln \left((1-a)^2+b^2\right)$$
$$\int_{0}^{\infty}\frac{\sin(x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\arctan\left(\frac{1}{\sqrt{-a-bi}}\right)+\arctan\left(\frac{1}{\sqrt{-a+bi}}\right)$$
$$\int_{0}^{\infty}\frac{\cos(t)-\cos(t \, x)}{x^2-1} \, dx=\frac{\pi}{2}\sin(t),\> \text{for} \, \> t\geq 0$$
Now we can begin evaluating $I$:
$$I=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right) \, dx$$
Enforce the substitution $\sqrt{\frac{1+x}{1-x}} = u$:
$$\implies I = \int_{0}^{\infty}\frac{4u^2}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right)\, du \stackrel{u \, \mapsto \frac{1}{u}}{=}\int_{0}^{\infty}\frac{4}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$
From adding the two, we can deduce:
$$I = \int_{0}^{\infty}\frac{2}{u^2-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$
Since
$$\ln\left(5 u^4-2 u^2+1\right)-\ln(4)=\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)$$
$$\ln(u^4-2 u^2+5)-\ln(4)=\ln((u^2-1)^2+4)-\ln(4)$$
We can write
$$A=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)\right)\,du\\
B=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln((u^2-1)^2+4)-\ln(4)\right)\, du$$
such that $I = A-B$
Now from the starting representations we deduce:
$$\begin{align} \implies A &= \int_{0}^{\infty}\frac{4}{u^2-1}\int_{0}^{\infty}\frac{\cos(t)-\cos(u \, t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\,dt\,du
\\ &= 2\pi\int_{0}^{\infty}\frac{\sin(t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\, dt 
\\ &= 2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)+2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1+2i}}\right)
\\ &= \Re\left(4\pi \arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)\right)\end{align}$$
Similarly, one finds $$B=\Re\left(4\pi\arctan\left(\frac{1}{\sqrt{-1-2i}}\right)\right)$$
By using the identity:
$$\arctan (x)+\arctan (y)=\arctan \left(\frac{x+y}{1-x y}\right)$$
One deduces
$$\boxed{I=4\pi \operatorname{arccot} \left(\sqrt{\phi}\right)}$$
A: For the purposes of alternative methods, it may be of interest to note that the integrand
$$f(x)=\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)$$ may be rewritten in terms of hyperbolic trigonometric functions. Using 
$$\tanh^{-1}(z) = \frac{1}{2}\log\left(\frac{1+z}{1-z}\right),$$
and we obtain
$$f(x)=\frac{1}{x}e^{\tanh^{-1}x}\log\left(\frac{1+\frac{2x}{1+2x^2}}{1-\frac{2x}{1+2x^2}}\right) = e^{\tanh^{-1} x}\left(\frac{2\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)}{x}\right).$$
The rational function in the bracket, which we will denote $s(x)$, is symmetric about $x=0$.
The desired integral is
$$I=\int_{-1}^1 f(x)dx = \int_{-1}^1e^{\tanh^{-1}x}s(x)dx,$$
which, by adding the indicated useful definite integral to both side, gives
$$I + \int_{-1}^1 e^{-\tanh^{-1}x}s(x)dx = 2\int_{-1}^1 \frac{s(x)dx}{\sqrt{1-x^2}}.$$
Now using the change of variable $x=-y$ we have $$\int_{-1}^1 e^{-\tanh^{-1} x}s(x)dx = -\int_1^{-1} e^{\tanh y}s(-y)dy = \int_{-1}^1 e^{\tanh y}s(y)dy = I,$$ by the symmetry of $s(x)$. Hence, we finally obtain
$$I = \int_{-1}^1\frac{s(x)dx}{\sqrt{1-x^2}} = 2\int_{-1}^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
This integral is symmetric about $x=0$, so we have
$$I=4\int_0^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx,$$ which can be rewritten $$I=-4\int_0^1\left(\frac{d}{dx}\text{sech}^{-1}x\right)\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
Using integration by parts this results in
$$I=8\int_0^1\frac{\text{sech}^{-1}(x)(1-2x^2)}{1+4x^4}dx.$$

We could also make the  change of variable $y=\text{sech}^{-1}x$ to obtain
$$I=8\int_0^\infty\frac{y(\cosh^2(y)-2)\sinh y}{\cosh^4(y)+4}dy= 8\int_0^\infty\frac{y\sinh^3 y}{\cosh^4y+4}dy-8\int_0^\infty\frac{y\sinh y}{\cosh^4 y+4}dy.$$
A: $\large\hspace{3in}I=4\,\pi\operatorname{arccot}$$\sqrt\phi$
A: Noteworthy, RIES (http://mrob.com/pub/ries/index.html) finds closed form from numerical value in the form of an equation:
$$
\cos{\left( \frac{x}{\pi} \right)}+1=\frac{2}{\phi^6}.
$$
Simplifying above, we get another form of the result:
$$
I = \pi \arccos{(17-8\sqrt{5})}.
$$ 
