Optimal. Leaf size=158 \[ -\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \text {Li}_2\left (-\frac {e x}{d}\right )}{2 e^2} \]
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Rubi [A]
time = 0.10, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2473, 45, 2393,
2332, 2341, 2354, 2438} \begin {gather*} -\frac {b d^2 m n \text {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{2 e^2}+\frac {b d^2 m n \log (d+e x)}{4 e^2}+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {3 b d m n x}{4 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {1}{4} b m n x^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 2332
Rule 2341
Rule 2354
Rule 2393
Rule 2438
Rule 2473
Rubi steps
\begin {align*} \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b e n) \int \frac {x^2 \log \left (f x^m\right )}{d+e x} \, dx+\frac {1}{4} (b e m n) \int \frac {x^2}{d+e x} \, dx\\ &=-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b e n) \int \left (-\frac {d \log \left (f x^m\right )}{e^2}+\frac {x \log \left (f x^m\right )}{e}+\frac {d^2 \log \left (f x^m\right )}{e^2 (d+e x)}\right ) \, dx+\frac {1}{4} (b e m n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx\\ &=-\frac {b d m n x}{4 e}+\frac {1}{8} b m n x^2+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b n) \int x \log \left (f x^m\right ) \, dx+\frac {(b d n) \int \log \left (f x^m\right ) \, dx}{2 e}-\frac {\left (b d^2 n\right ) \int \frac {\log \left (f x^m\right )}{d+e x} \, dx}{2 e}\\ &=-\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}+\frac {\left (b d^2 m n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{2 e^2}\\ &=-\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \text {Li}_2\left (-\frac {e x}{d}\right )}{2 e^2}\\ \end {align*}
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Mathematica [A]
time = 0.09, size = 164, normalized size = 1.04 \begin {gather*} \frac {\log \left (f x^m\right ) \left (-2 b d^2 n \log (d+e x)+e x \left (2 b d n+2 a e x-b e n x+2 b e x \log \left (c (d+e x)^n\right )\right )\right )+m \left (-3 b d e n x-a e^2 x^2+b e^2 n x^2+b d^2 n (1+2 \log (x)) \log (d+e x)-b e^2 x^2 \log \left (c (d+e x)^n\right )-2 b d^2 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )-2 b d^2 m n \text {Li}_2\left (-\frac {e x}{d}\right )}{4 e^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 1.04, size = 1994, normalized size = 12.62
method | result | size |
risch | \(\text {Expression too large to display}\) | \(1994\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 176, normalized size = 1.11 \begin {gather*} \frac {1}{4} \, {\left (2 \, {\left (\log \left (x e + d\right ) \log \left (-\frac {x e + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {x e + d}{d}\right )\right )} b d^{2} n e^{\left (-2\right )} - {\left (3 \, b d n x e - b d^{2} n \log \left (x e + d\right ) + b x^{2} e^{2} \log \left ({\left (x e + d\right )}^{n}\right ) - {\left (b {\left (n - \log \left (c\right )\right )} - a\right )} x^{2} e^{2}\right )} e^{\left (-2\right )}\right )} m - \frac {1}{4} \, {\left ({\left (2 \, d^{2} e^{\left (-3\right )} \log \left (x e + d\right ) + {\left (x^{2} e - 2 \, d x\right )} e^{\left (-2\right )}\right )} b n e - 2 \, b x^{2} \log \left ({\left (x e + d\right )}^{n} c\right ) - 2 \, a x^{2}\right )} \log \left (f x^{m}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 = -0.01 \begin {gather*} \int x\,\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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