Integrand size = 18, antiderivative size = 39 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e} \]
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Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2351, 31} \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=\frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e} \]
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Rule 31
Rule 2351
Rubi steps \begin{align*} \text {integral}& = \frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {(b n) \int \frac {1}{d+e x} \, dx}{d} \\ & = \frac {x \left (a+b \log \left (c x^n\right )\right )}{d (d+e x)}-\frac {b n \log (d+e x)}{d e} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=\frac {-\frac {a+b \log \left (c x^n\right )}{d+e x}+\frac {b n (\log (x)-\log (d+e x))}{d}}{e} \]
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Time = 0.40 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38
method | result | size |
parallelrisch | \(-\frac {\ln \left (e x +d \right ) b e n x +\ln \left (e x +d \right ) b d n -b e x \ln \left (c \,x^{n}\right )-a e x}{d \left (e x +d \right ) e}\) | \(54\) |
risch | \(-\frac {b \ln \left (x^{n}\right )}{e \left (e x +d \right )}-\frac {-i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-2 \ln \left (-x \right ) b e n x +2 \ln \left (e x +d \right ) b e n x -2 \ln \left (-x \right ) b d n +2 \ln \left (e x +d \right ) b d n +2 d b \ln \left (c \right )+2 a d}{2 \left (e x +d \right ) e d}\) | \(173\) |
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Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.31 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=\frac {b e n x \log \left (x\right ) - b d \log \left (c\right ) - a d - {\left (b e n x + b d n\right )} \log \left (e x + d\right )}{d e^{2} x + d^{2} e} \]
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Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (31) = 62\).
Time = 0.62 (sec) , antiderivative size = 153, normalized size of antiderivative = 3.92 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {a}{x} - \frac {b n}{x} - \frac {b \log {\left (c x^{n} \right )}}{x}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {a x - b n x + b x \log {\left (c x^{n} \right )}}{d^{2}} & \text {for}\: e = 0 \\\frac {- \frac {a}{x} - \frac {b n}{x} - \frac {b \log {\left (c x^{n} \right )}}{x}}{e^{2}} & \text {for}\: d = 0 \\- \frac {a d}{d^{2} e + d e^{2} x} - \frac {b d n \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} - \frac {b e n x \log {\left (\frac {d}{e} + x \right )}}{d^{2} e + d e^{2} x} + \frac {b e x \log {\left (c x^{n} \right )}}{d^{2} e + d e^{2} x} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.62 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=-b n {\left (\frac {\log \left (e x + d\right )}{d e} - \frac {\log \left (x\right )}{d e}\right )} - \frac {b \log \left (c x^{n}\right )}{e^{2} x + d e} - \frac {a}{e^{2} x + d e} \]
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Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.64 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=-\frac {b n \log \left (x\right )}{e^{2} x + d e} - \frac {b n \log \left (e x + d\right )}{d e} + \frac {b n \log \left (x\right )}{d e} - \frac {b \log \left (c\right ) + a}{e^{2} x + d e} \]
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Time = 1.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx=-\frac {a}{x\,e^2+d\,e}-\frac {b\,\ln \left (c\,x^n\right )}{e\,\left (d+e\,x\right )}-\frac {2\,b\,n\,\mathrm {atanh}\left (\frac {2\,e\,x}{d}+1\right )}{d\,e} \]
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