Integrand size = 20, antiderivative size = 113 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e} \]
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Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2512, 266, 2463, 2441, 2352, 2440, 2438} \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e}-\frac {p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{e}+\frac {p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e} \]
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Rule 266
Rule 2352
Rule 2438
Rule 2440
Rule 2441
Rule 2463
Rule 2512
Rubi steps \begin{align*} \text {integral}& = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {(b p) \int \frac {\log (d+e x)}{\left (a+\frac {b}{x}\right ) x^2} \, dx}{e} \\ & = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {(b p) \int \left (\frac {\log (d+e x)}{b x}-\frac {a \log (d+e x)}{b (b+a x)}\right ) \, dx}{e} \\ & = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \int \frac {\log (d+e x)}{x} \, dx}{e}-\frac {(a p) \int \frac {\log (d+e x)}{b+a x} \, dx}{e} \\ & = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}-p \int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx+p \int \frac {\log \left (\frac {e (b+a x)}{-a d+b e}\right )}{d+e x} \, dx \\ & = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}+\frac {p \text {Li}_2\left (1+\frac {e x}{d}\right )}{e}+\frac {p \text {Subst}\left (\int \frac {\log \left (1+\frac {a x}{-a d+b e}\right )}{x} \, dx,x,d+e x\right )}{e} \\ & = \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}-\frac {p \text {Li}_2\left (\frac {a (d+e x)}{a d-b e}\right )}{e}+\frac {p \text {Li}_2\left (1+\frac {e x}{d}\right )}{e} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e}+\frac {p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e}+\frac {p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e} \]
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Time = 1.51 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.21
method | result | size |
parts | \(\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e}+p b \left (\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b e}-\frac {a \left (\frac {\operatorname {dilog}\left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{a}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {-a d +a \left (e x +d \right )+b e}{-a d +b e}\right )}{a}\right )}{b e}\right )\) | \(137\) |
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\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{e x + d} \,d x } \]
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\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int \frac {\log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{d + e x}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.41 \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b p {\left (\frac {\log \left (e x + d\right ) \log \left (a + \frac {b}{x}\right )}{b} - \frac {\log \left (e x + d\right ) \log \left (-\frac {a e x + a d}{a d - b e} + 1\right ) + {\rm Li}_2\left (\frac {a e x + a d}{a d - b e}\right )}{b} + \frac {\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )}{b}\right )}}{e} - \frac {p \log \left (e x + d\right ) \log \left (a + \frac {b}{x}\right )}{e} + \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \log \left (e x + d\right )}{e} \]
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\[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right )}{e x + d} \,d x } \]
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Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{d+e\,x} \,d x \]
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