Integrand size = 25, antiderivative size = 90 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d n (f x)^m}{f m^2}-\frac {b e n x^m (f x)^m}{4 f m^2}+\frac {d (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}+\frac {e x^m (f x)^m \left (a+b \log \left (c x^n\right )\right )}{2 f m} \]
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Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2377, 2376, 272, 45} \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {b d^2 n x^{1-m} \log (x) (f x)^{m-1}}{2 e m}-\frac {b d n x (f x)^{m-1}}{m^2}-\frac {b e n x^{m+1} (f x)^{m-1}}{4 m^2} \]
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Rule 45
Rule 272
Rule 2376
Rule 2377
Rubi steps \begin{align*} \text {integral}& = \left (x^{1-m} (f x)^{-1+m}\right ) \int x^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx \\ & = \frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac {\left (d+e x^m\right )^2}{x} \, dx}{2 e m} \\ & = \frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \text {Subst}\left (\int \frac {(d+e x)^2}{x} \, dx,x,x^m\right )}{2 e m^2} \\ & = \frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \text {Subst}\left (\int \left (2 d e+\frac {d^2}{x}+e^2 x\right ) \, dx,x,x^m\right )}{2 e m^2} \\ & = -\frac {b d n x (f x)^{-1+m}}{m^2}-\frac {b e n x^{1+m} (f x)^{-1+m}}{4 m^2}-\frac {b d^2 n x^{1-m} (f x)^{-1+m} \log (x)}{2 e m}+\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.68 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(f x)^m \left (2 a m \left (2 d+e x^m\right )-b n \left (4 d+e x^m\right )+2 b m \left (2 d+e x^m\right ) \log \left (c x^n\right )\right )}{4 f m^2} \]
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Time = 1.74 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.17
method | result | size |
parallelrisch | \(-\frac {-2 x \,x^{m} \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b e m -2 x \,x^{m} \left (f x \right )^{m -1} a e m +x \,x^{m} \left (f x \right )^{m -1} b e n -4 x \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b d m -4 x \left (f x \right )^{m -1} a d m +4 x \left (f x \right )^{m -1} b d n}{4 m^{2}}\) | \(105\) |
risch | \(\frac {b \left (e \,x^{m}+2 d \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}} \ln \left (x^{n}\right )}{2 m}-\frac {\left (i \pi b e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{m} m -i \pi b e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m -i \pi b e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m +i \pi b e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{m} m +2 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 ) m -2 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m -2 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m +2 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3} m -2 \ln \left (c \right ) b e \,x^{m} m -4 \ln \left (c \right ) b d m -2 x^{m} a e m +x^{m} b e n -4 a d m +4 b d n \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}}}{4 m^{2}}\) | \(425\) |
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Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left (2 \, b e m n \log \left (x\right ) + 2 \, b e m \log \left (c\right ) + 2 \, a e m - b e n\right )} f^{m - 1} x^{2 \, m} + 4 \, {\left (b d m n \log \left (x\right ) + b d m \log \left (c\right ) + a d m - b d n\right )} f^{m - 1} x^{m}}{4 \, m^{2}} \]
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Time = 3.01 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.73 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \frac {a d x \left (f x\right )^{m - 1}}{m} + \frac {a e x x^{m} \left (f x\right )^{m - 1}}{2 m} + \frac {b d x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {b d n x \left (f x\right )^{m - 1}}{m^{2}} + \frac {b e x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{2 m} - \frac {b e n x x^{m} \left (f x\right )^{m - 1}}{4 m^{2}} & \text {for}\: m \neq 0 \\\frac {\left (d + e\right ) \left (\begin {cases} a \log {\left (x \right )} & \text {for}\: b = 0 \\- \left (- a - b \log {\left (c \right )}\right ) \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}\right )}{f} & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.21 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )}{2 \, m} + \frac {a e f^{m - 1} x^{2 \, m}}{2 \, m} - \frac {b e f^{m - 1} n x^{2 \, m}}{4 \, m^{2}} - \frac {b d f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b d \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a d}{f m} \]
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Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e f^{m} n x^{2 \, m} \log \left (x\right )}{2 \, f m} + \frac {b d f^{m} n x^{m} \log \left (x\right )}{f m} + \frac {b e f^{m} x^{2 \, m} \log \left (c\right )}{2 \, f m} + \frac {b d f^{m} x^{m} \log \left (c\right )}{f m} + \frac {a e f^{m} x^{2 \, m}}{2 \, f m} - \frac {b e f^{m} n x^{2 \, m}}{4 \, f m^{2}} + \frac {a d f^{m} x^{m}}{f m} - \frac {b d f^{m} n x^{m}}{f m^{2}} \]
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Timed out. \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int {\left (f\,x\right )}^{m-1}\,\left (d+e\,x^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]
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