Integrand size = 32, antiderivative size = 27 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (b+a x^{-m}\right )}{b}\right )}{a m} \]
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Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2525, 2459, 2440, 2438} \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,\frac {(1-c) \left (a x^{-m}+b\right )}{b}\right )}{a m} \]
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Rule 2438
Rule 2440
Rule 2459
Rule 2525
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\log \left (c-\frac {a (1-c) x}{b}\right )}{\left (a+\frac {b}{x}\right ) x} \, dx,x,x^{-m}\right )}{m} \\ & = -\frac {\text {Subst}\left (\int \frac {\log \left (c-\frac {a (1-c) x}{b}\right )}{b+a x} \, dx,x,x^{-m}\right )}{m} \\ & = -\frac {\text {Subst}\left (\int \frac {\log \left (1-\frac {(1-c) x}{b}\right )}{x} \, dx,x,b+a x^{-m}\right )}{a m} \\ & = \frac {\text {Li}_2\left (\frac {(1-c) \left (b+a x^{-m}\right )}{b}\right )}{a m} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {(-1+c) x^{-m} \left (a+b x^m\right )}{b}\right )}{a m} \]
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Time = 2.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (\frac {\left (c a -a \right ) x^{-m}}{b}+c \right )}{m a}\) | \(27\) |
default | \(\frac {\operatorname {dilog}\left (\frac {\left (c a -a \right ) x^{-m}}{b}+c \right )}{m a}\) | \(27\) |
risch | \(\text {Expression too large to display}\) | \(1267\) |
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Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\frac {{\rm Li}_2\left (-\frac {b c x^{m} + a c - a}{b x^{m}} + 1\right )}{a m} \]
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Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \]
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\[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int { \frac {\log \left (c + \frac {a {\left (c - 1\right )}}{b x^{m}}\right )}{{\left (b x^{m} + a\right )} x} \,d x } \]
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Timed out. \[ \int \frac {\log \left (c-\frac {a (1-c) x^{-m}}{b}\right )}{x \left (a+b x^m\right )} \, dx=\int \frac {\ln \left (c+\frac {a\,\left (c-1\right )}{b\,x^m}\right )}{x\,\left (a+b\,x^m\right )} \,d x \]
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