Q: Integrate: $$\int \frac{dx}{x\left(x-\sqrt{x^2-1}\right)\left[x^2\left(1-\frac{1}{x^2}+\sqrt{1-\frac{1}{x^2}}\right)-\ln\left(x-\sqrt{x^2-1}\right)\right]}$$
My Attempt:
I used the substitution $x=\sec \theta$ and $dx=\sec \theta \tan \theta \, d\theta$. Then the integral I (say) became:
$$\begin{align} & I=\frac{\tan \theta \, d\theta}{\left(\sec\theta - \tan \theta\right)\left[\tan \theta\left(\sec\theta + \tan \theta\right)-\ln\left(\sec\theta + \tan \theta\right)\right]} \\ & =\frac{\tan \theta \, d\theta}{\tan \theta-\left(\sec\theta - \tan \theta\right)\ln\left(\sec\theta + \tan \theta\right)} \\ & =\frac{\frac{2\tan \frac{\theta}{2}}{1-\tan ^2\frac{\theta}{2}} \, d\theta}{\frac{2\tan \frac{\theta}{2}}{1-\tan ^2\frac{\theta}{2}}-\tan\left(\frac{\pi}{4}- \frac{\theta}{2}\right)\ln\left[\tan\left(\frac{\pi}{4}+ \frac{\theta}{2}\right)\right]} \\ & =\frac{d\theta}{1-\frac{(1-\tan \frac{\theta}{2})^2}{2\tan \frac{\theta}{2}}\ln\left[\tan\left(\frac{\pi}{4}+ \frac{\theta}{2}\right)\right]} \end{align}$$
The I used the substitution $z=\frac{\pi}{4}+ \frac{\theta}{2}$ and $2\, dz=d\theta$. So, I got after some re-arrangement,
$$\begin{align} & I=\frac{2\, dz}{1-\frac{2 \ln \tan z}{\tan^2z-1}} \\ & =\frac{2\, dz}{1-\frac{ \ln \tan^2 z}{\tan^2z-1}}\end{align}$$ Then I changed as $\tan z =p$ and $dz=\frac{dp}{1+p^2}$ and I finally got
$$I=\frac{2(p^2-1)dp}{(1+p^2)(p^2-1-\ln p^2)}$$
But, then I thought of substitutions like $m=p^2$ but I could not progress any further.
Can you tell me how to progress or, if the result is some special function?