Tricky integration with log and powers under square root

Question

$$ \int\frac{dr}{\sqrt{r^2 -1-\ln r}} $$ A good substitution for this will be $r=e^x$. $$ \int\frac{e^x dx}{\sqrt{e^{2x}-1-x}} $$ I don't know what to substitute next. Mathematica is unable to solve this.

Answer 1

Let $u=1+\ln r$ ,

Then $r=e^{u-1}$

$dr=e^{u-1}~du$

$\therefore\int\dfrac{dr}{\sqrt{r^2-1-\ln r}}=\int\dfrac{e^{u-1}}{\sqrt{e^{2u-2}-u}}~du$

Case $1$: $ue^{2-2u}\leq1$ , i.e. $\dfrac{1+\ln r}{r^2}\leq1$

Then $\int\dfrac{e^{u-1}}{\sqrt{e^{2u-2}-u}}~du$

$=\int\dfrac{1}{\sqrt{1-ue^{2-2u}}}~du$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(2n)!u^ne^{2n-2nu}}{4^n(n!)^2}\right)~du$

$=u-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(2n)!u^ke^{2n-2nu}}{2^{3n-k+1}n^{n-k+1}n!k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\ln r-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(2n)!(1+\ln r)^k}{2^{3n-k+1}n^{n-k+1}n!k!r^{2n}}+C$