Is there a closed-form expression for $\int (a-b\ln(cx))^{-1} \mathrm{d}x$?

Is there a closed form expression for $$\int\frac{1}{a-b\ln(cx)}\,dx\ ?$$

I was wondering how to integrate the above function. I have spent a lot of time on it. First i did an integration by parts until I had to integrate $\ln|a – b \ln(c x)|$ again. Once more, I did an integration by parts and substituted this value back into the original equation. However, now all the terms cancel each other out, effectively giving $0 = 0$. How do I go about integrating $1/(a-b \ln(cx))$? Thanks for the assistance.

Edit: $a$, $b$ and $c$ are constants.

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integrals.wolfram.com suggests that this doesn't have a closed form: it gives an answer in terms of the exponential integral (en.wikipedia.org/wiki/Exponential_integral). –  Qiaochu Yuan May 1 '11 at 17:55
@Arturo: That may be my fault. The original title was: integrate 1/(a-b*ln(cx)) with respect to x which did not seem descriptive of what the OP appeared to be asking in the question. Sorry if my edit resulted in more confusion rather than less. –  cardinal May 1 '11 at 19:07
There is no elementary antiderivative. Note that $a - b \ln(c x) = -b \ln(c e^{-a/b} x)$ (assuming $a$, $b$, $c$, $x$ positive), so (after a change of variables) it all reduces to the case $a=0$, $b=c=1$. You can also prove that the integral is non-elementary using the Rothstein-Trager theorem.
For completeness: the nonelementary integral you'll bump into is $\int\frac{\exp\;u}{u}\mathrm du$, which is the exponential integral $\mathrm{ei}(x)$ that Qiaochu mentions. –  Ｊ. Ｍ. May 1 '11 at 21:05