Optimal. Leaf size=26 \[ e^{-2 x} x^{2+2 x} \left (16+\left (-1+\frac {x}{3}\right ) x \log (4)\right ) \]
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Rubi [F] time = 0.74, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{3} e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{-2 x} x^{2 x} \left (96 x+\left (-9 x^2+4 x^3\right ) \log (4)+\left (96 x^2+\left (-6 x^3+2 x^4\right ) \log (4)\right ) \log (x)\right ) \, dx\\ &=\frac {1}{3} \int \left (96 e^{-2 x} x^{1+2 x}+e^{-2 x} x^{2+2 x} (-9+4 x) \log (4)+2 e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \log (x)\right ) \, dx\\ &=\frac {2}{3} \int e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \log (x) \, dx+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} \log (4) \int e^{-2 x} x^{2+2 x} (-9+4 x) \, dx\\ &=-\left (\frac {2}{3} \int \frac {48 \int e^{-2 x} x^{2+2 x} \, dx+\log (4) \left (-3 \int e^{-2 x} x^{3+2 x} \, dx+\int e^{-2 x} x^{4+2 x} \, dx\right )}{x} \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} \log (4) \int \left (-9 e^{-2 x} x^{2+2 x}+4 e^{-2 x} x^{3+2 x}\right ) \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (\frac {2}{3} \int \left (\frac {3 \left (16 \int e^{-2 x} x^{2+2 x} \, dx-\log (4) \int e^{-2 x} x^{3+2 x} \, dx\right )}{x}+\frac {\log (4) \int e^{-2 x} x^{4+2 x} \, dx}{x}\right ) \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (2 \int \frac {16 \int e^{-2 x} x^{2+2 x} \, dx-\log (4) \int e^{-2 x} x^{3+2 x} \, dx}{x} \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=-\left (2 \int \left (\frac {16 \int e^{-2 x} x^{2+2 x} \, dx}{x}-\frac {\log (4) \int e^{-2 x} x^{3+2 x} \, dx}{x}\right ) \, dx\right )+32 \int e^{-2 x} x^{1+2 x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ &=32 \int e^{-2 x} x^{1+2 x} \, dx-32 \int \frac {\int e^{-2 x} x^{2+2 x} \, dx}{x} \, dx-\frac {1}{3} (2 \log (4)) \int \frac {\int e^{-2 x} x^{4+2 x} \, dx}{x} \, dx+\frac {1}{3} (4 \log (4)) \int e^{-2 x} x^{3+2 x} \, dx+(2 \log (4)) \int \frac {\int e^{-2 x} x^{3+2 x} \, dx}{x} \, dx-(3 \log (4)) \int e^{-2 x} x^{2+2 x} \, dx+(32 \log (x)) \int e^{-2 x} x^{2+2 x} \, dx+\frac {1}{3} (2 \log (4) \log (x)) \int e^{-2 x} x^{4+2 x} \, dx-(2 \log (4) \log (x)) \int e^{-2 x} x^{3+2 x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 29, normalized size = 1.12 \begin {gather*} \frac {1}{3} e^{-2 x} x^{2+2 x} \left (48-3 x \log (4)+x^2 \log (4)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 29, normalized size = 1.12 \begin {gather*} \frac {2}{3} \, {\left (24 \, x^{2} + {\left (x^{4} - 3 \, x^{3}\right )} \log \relax (2)\right )} x^{2 \, x} e^{\left (-2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 50, normalized size = 1.92 \begin {gather*} \frac {2}{3} \, x^{4} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \log \relax (2) - 2 \, x^{3} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \log \relax (2) + 16 \, x^{2} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 28, normalized size = 1.08
method | result | size |
risch | \(\frac {2 x^{2} \left (x^{2} \ln \relax (2)-3 x \ln \relax (2)+24\right ) {\mathrm e}^{-2 x} x^{2 x}}{3}\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 31, normalized size = 1.19 \begin {gather*} \frac {2}{3} \, {\left (x^{4} \log \relax (2) - 3 \, x^{3} \log \relax (2) + 24 \, x^{2}\right )} e^{\left (2 \, x \log \relax (x) - 2 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 6.89, size = 27, normalized size = 1.04 \begin {gather*} \frac {2\,x^{2\,x}\,x^2\,{\mathrm {e}}^{-2\,x}\,\left (\ln \relax (2)\,x^2-3\,\ln \relax (2)\,x+24\right )}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 27.55, size = 46, normalized size = 1.77 \begin {gather*} \frac {\left (2 x^{4} e^{- 2 x} \log {\relax (2 )} - 6 x^{3} e^{- 2 x} \log {\relax (2 )} + 48 x^{2} e^{- 2 x}\right ) e^{2 x \log {\relax (x )}}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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