Optimal. Leaf size=14 \[ \frac {1}{6} e^{2 x} x^2 \log (5) \]
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Rubi [A] time = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {12, 1593, 2196, 2176, 2194} \begin {gather*} \frac {1}{6} e^{2 x} x^2 \log (5) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 1593
Rule 2176
Rule 2194
Rule 2196
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \log (5) \int e^{2 x} \left (x+x^2\right ) \, dx\\ &=\frac {1}{3} \log (5) \int e^{2 x} x (1+x) \, dx\\ &=\frac {1}{3} \log (5) \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx\\ &=\frac {1}{3} \log (5) \int e^{2 x} x \, dx+\frac {1}{3} \log (5) \int e^{2 x} x^2 \, dx\\ &=\frac {1}{6} e^{2 x} x \log (5)+\frac {1}{6} e^{2 x} x^2 \log (5)-\frac {1}{6} \log (5) \int e^{2 x} \, dx-\frac {1}{3} \log (5) \int e^{2 x} x \, dx\\ &=-\frac {1}{12} e^{2 x} \log (5)+\frac {1}{6} e^{2 x} x^2 \log (5)+\frac {1}{6} \log (5) \int e^{2 x} \, dx\\ &=\frac {1}{6} e^{2 x} x^2 \log (5)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {1}{6} e^{2 x} x^2 \log (5) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, x^{2} e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 11, normalized size = 0.79 \begin {gather*} \frac {1}{6} \, x^{2} e^{\left (2 \, x\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 12, normalized size = 0.86
method | result | size |
gosper | \(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\) | \(12\) |
default | \(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\) | \(12\) |
norman | \(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\) | \(12\) |
risch | \(\frac {x^{2} \ln \relax (5) {\mathrm e}^{2 x}}{6}\) | \(12\) |
meijerg | \(-\frac {\ln \relax (5) \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )}{24}+\frac {\ln \relax (5) \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{12}\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 30, normalized size = 2.14 \begin {gather*} \frac {1}{12} \, {\left ({\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )}\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.36, size = 11, normalized size = 0.79 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (5)}{6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.10, size = 12, normalized size = 0.86 \begin {gather*} \frac {x^{2} e^{2 x} \log {\relax (5 )}}{6} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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