Optimal. Leaf size=90 \[ \frac {d (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}+\frac {e x^m (f x)^m \left (a+b \log \left (c x^n\right )\right )}{2 f m}-\frac {b d n (f x)^m}{f m^2}-\frac {b e n x^m (f x)^m}{4 f m^2} \]
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Rubi [A] time = 0.12, antiderivative size = 113, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2339, 2338, 266, 43} \[ \frac {x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {b d^2 n x^{1-m} \log (x) (f x)^{m-1}}{2 e m}-\frac {b d n x (f x)^{m-1}}{m^2}-\frac {b e n x^{m+1} (f x)^{m-1}}{4 m^2} \]
Antiderivative was successfully verified.
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Rule 43
Rule 266
Rule 2338
Rule 2339
Rubi steps
\begin {align*} \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx &=\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 ) \operatorname {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 ) \operatorname {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*}
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Mathematica [A] time = 0.07, size = 61, normalized size = 0.68 \[ \frac {(f x)^m \left (2 a m \left (2 d+e x^m\right )+2 b m \log \left (c x^n\right ) \left (2 d+e x^m\right )-b n \left (4 d+e x^m\right )\right )}{4 f m^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 76, normalized size = 0.84 \[ \frac {{\left (2 \, b e m n \log \relax (x) + 2 \, b e m \log \relax (c) + 2 \, a e m - b e n\right )} f^{m - 1} x^{2 \, m} + 4 \, {\left (b d m n \log \relax (x) + b d m \log \relax (c) + a d m - b d n\right )} f^{m - 1} x^{m}}{4 \, m^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.54, size = 166, normalized size = 1.84 \[ \frac {b d \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m} \log \relax (c)}{f m} + \frac {b d f^{m} n x^{m} \log \relax (x)}{f m} + \frac {b f^{m} n x^{2 \, m} e \log \relax (x)}{2 \, f m} + \frac {a d \frac {1}{f}^{m} x^{m} {\left | f \right |}^{2 \, m}}{f m} + \frac {b f^{m} x^{2 \, m} e \log \relax (c)}{2 \, f m} - \frac {b d f^{m} n x^{m}}{f m^{2}} + \frac {a f^{m} x^{2 \, m} e}{2 \, f m} - \frac {b f^{m} n x^{2 \, m} e}{4 \, f m^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.23, size = 426, normalized size = 4.73 \[ \frac {\left (e \,x^{m}+2 d \right ) b x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )+i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i f x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )^{2}-i \pi \mathrm {csgn}\left (i f x \right )^{3}+2 \ln \relax (f )+2 \ln \relax (x )\right )}{2}} \ln \left (x^{n}\right )}{2 m}+\frac {\left (-i \pi b e m \,x^{m} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b e m \,x^{m} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b e m \,x^{m} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b e m \,x^{m} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-2 i \pi b d m \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+2 i \pi b d m \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+2 i \pi b d m \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b d m \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 b e m \,x^{m} \ln \relax (c )+2 a e m \,x^{m}+4 b d m \ln \relax (c )-b e n \,x^{m}+4 a d m -4 b d n \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )+i \pi \,\mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i f x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i f x \right )^{2}-i \pi \mathrm {csgn}\left (i f x \right )^{3}+2 \ln \relax (f )+2 \ln \relax (x )\right )}{2}}}{4 m^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 109, normalized size = 1.21 \[ \frac {b e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )}{2 \, m} + \frac {a e f^{m - 1} x^{2 \, m}}{2 \, m} - \frac {b e f^{m - 1} n x^{2 \, m}}{4 \, m^{2}} - \frac {b d f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b d \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a d}{f m} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (f\,x\right )}^{m-1}\,\left (d+e\,x^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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