Optimal. Leaf size=363 \[ \frac {b k n (g x)^{2 m}}{4 g m^2}-\frac {3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \text {Li}_2\left (1+\frac {f x^m}{e}\right )}{2 f^2 g m^2} \]
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Rubi [A]
time = 0.26, antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 12, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2505, 20,
272, 45, 2423, 16, 32, 30, 19, 2504, 2441, 2352} \begin {gather*} \frac {b e^2 k n x^{-2 m} (g x)^{2 m} \text {PolyLog}\left (2,\frac {f x^m}{e}+1\right )}{2 f^2 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^2 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}+\frac {b k n (g x)^{2 m}}{4 g m^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 16
Rule 19
Rule 20
Rule 30
Rule 32
Rule 45
Rule 272
Rule 2352
Rule 2423
Rule 2441
Rule 2504
Rule 2505
Rubi steps
\begin {align*} \int (g x)^{-1+2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right ) \, dx &=-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-(b n) \int \left (-\frac {k (g x)^{2 m}}{4 g m x}+\frac {e k x^{-1-m} (g x)^{2 m}}{2 f g m}-\frac {e^2 k x^{-1-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{2 f^2 g m}+\frac {(g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{2 g m x}\right ) \, dx\\ &=-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {(b n) \int \frac {(g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{x} \, dx}{2 g m}+\frac {(b k n) \int \frac {(g x)^{2 m}}{x} \, dx}{4 g m}+\frac {\left (b e^2 k n\right ) \int x^{-1-2 m} (g x)^{2 m} \log \left (e+f x^m\right ) \, dx}{2 f^2 g m}-\frac {(b e k n) \int x^{-1-m} (g x)^{2 m} \, dx}{2 f g m}\\ &=-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {(b n) \int (g x)^{-1+2 m} \log \left (d \left (e+f x^m\right )^k\right ) \, dx}{2 m}+\frac {(b k n) \int (g x)^{-1+2 m} \, dx}{4 m}+\frac {\left (b e^2 k n x^{-2 m} (g x)^{2 m}\right ) \int \frac {\log \left (e+f x^m\right )}{x} \, dx}{2 f^2 g m}-\frac {\left (b e k n x^{-2 m} (g x)^{2 m}\right ) \int x^{-1+m} \, dx}{2 f g m}\\ &=\frac {b k n (g x)^{2 m}}{8 g m^2}-\frac {b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {(b f k n) \int \frac {x^{-1+m} (g x)^{2 m}}{e+f x^m} \, dx}{4 g m}+\frac {\left (b e^2 k n x^{-2 m} (g x)^{2 m}\right ) \text {Subst}\left (\int \frac {\log (e+f x)}{x} \, dx,x,x^m\right )}{2 f^2 g m^2}\\ &=\frac {b k n (g x)^{2 m}}{8 g m^2}-\frac {b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {\left (b e^2 k n x^{-2 m} (g x)^{2 m}\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {f x}{e}\right )}{e+f x} \, dx,x,x^m\right )}{2 f g m^2}+\frac {\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \int \frac {x^{-1+3 m}}{e+f x^m} \, dx}{4 g m}\\ &=\frac {b k n (g x)^{2 m}}{8 g m^2}-\frac {b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \text {Li}_2\left (1+\frac {f x^m}{e}\right )}{2 f^2 g m^2}+\frac {\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \text {Subst}\left (\int \frac {x^2}{e+f x} \, dx,x,x^m\right )}{4 g m^2}\\ &=\frac {b k n (g x)^{2 m}}{8 g m^2}-\frac {b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \text {Li}_2\left (1+\frac {f x^m}{e}\right )}{2 f^2 g m^2}+\frac {\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \text {Subst}\left (\int \left (-\frac {e}{f^2}+\frac {x}{f}+\frac {e^2}{f^2 (e+f x)}\right ) \, dx,x,x^m\right )}{4 g m^2}\\ &=\frac {b k n (g x)^{2 m}}{4 g m^2}-\frac {3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}-\frac {k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac {e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac {e^2 k x^{-2 m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^m\right )}{2 f^2 g m}-\frac {b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac {(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}+\frac {b e^2 k n x^{-2 m} (g x)^{2 m} \text {Li}_2\left (1+\frac {f x^m}{e}\right )}{2 f^2 g m^2}\\ \end {align*}
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Mathematica [A]
time = 0.23, size = 352, normalized size = 0.97 \begin {gather*} \frac {x^{-2 m} (g x)^{2 m} \left (2 a e f k m x^m-3 b e f k n x^m-a f^2 k m x^{2 m}+b f^2 k n x^{2 m}+2 b e^2 k m^2 n \log ^2(x)+2 b e f k m x^m \log \left (c x^n\right )-b f^2 k m x^{2 m} \log \left (c x^n\right )-2 a e^2 k m \log \left (e-e x^m\right )+b e^2 k n \log \left (e-e x^m\right )-2 b e^2 k m \log \left (c x^n\right ) \log \left (e-e x^m\right )+2 b e^2 k n \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )+e^2 k m \log (x) \left (-2 a m+b n-2 b m \log \left (c x^n\right )+2 b n \log \left (e-e x^m\right )-2 b n \log \left (e+f x^m\right )\right )+2 a f^2 m x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )-b f^2 n x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )+2 b f^2 m x^{2 m} \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )+2 b e^2 k n \text {Li}_2\left (1+\frac {f x^m}{e}\right )\right )}{4 f^2 g m^2} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (g x \right )^{-1+2 m} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \,x^{m}\right )^{k}\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 301, normalized size = 0.83 \begin {gather*} -\frac {2 \, b g^{2 \, m - 1} k m n e^{2} \log \left ({\left (f x^{m} + e\right )} e^{\left (-1\right )}\right ) \log \left (x\right ) + 2 \, b g^{2 \, m - 1} k n {\rm Li}_2\left (-{\left (f x^{m} + e\right )} e^{\left (-1\right )} + 1\right ) e^{2} + {\left (b f^{2} k m \log \left (c\right ) + a f^{2} k m - b f^{2} k n - {\left (2 \, b f^{2} m \log \left (c\right ) + 2 \, a f^{2} m - b f^{2} n\right )} \log \left (d\right ) + {\left (b f^{2} k m n - 2 \, b f^{2} m n \log \left (d\right )\right )} \log \left (x\right )\right )} g^{2 \, m - 1} x^{2 \, m} - {\left (2 \, b f k m n e \log \left (x\right ) + 2 \, b f k m e \log \left (c\right ) + {\left (2 \, a f k m - 3 \, b f k n\right )} e\right )} g^{2 \, m - 1} x^{m} - {\left ({\left (2 \, b f^{2} k m n \log \left (x\right ) + 2 \, b f^{2} k m \log \left (c\right ) + 2 \, a f^{2} k m - b f^{2} k n\right )} g^{2 \, m - 1} x^{2 \, m} - {\left (2 \, b k m e^{2} \log \left (c\right ) + {\left (2 \, a k m - b k n\right )} e^{2}\right )} g^{2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, f^{2} m^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \ln \left (d\,{\left (e+f\,x^m\right )}^k\right )\,{\left (g\,x\right )}^{2\,m-1}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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