Optimal. Leaf size=27 \[ \frac {4 e^2 \left (1+\left (3+2 x+x^4\right )^2\right )}{\log \left (\frac {e^{10}}{9}\right )} \]
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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(27)=54\).
time = 0.02, antiderivative size = 79, normalized size of antiderivative = 2.93, number of steps
used = 2, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12}
\begin {gather*} \frac {4 e^2 x^8}{10-\log (9)}+\frac {16 e^2 x^5}{10-\log (9)}+\frac {24 e^2 x^4}{10-\log (9)}+\frac {16 e^2 x^2}{10-\log (9)}+\frac {48 e^2 x}{10-\log (9)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^2 \int \left (48+32 x+96 x^3+80 x^4+32 x^7\right ) \, dx}{10-\log (9)}\\ &=\frac {48 e^2 x}{10-\log (9)}+\frac {16 e^2 x^2}{10-\log (9)}+\frac {24 e^2 x^4}{10-\log (9)}+\frac {16 e^2 x^5}{10-\log (9)}+\frac {4 e^2 x^8}{10-\log (9)}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.00, size = 35, normalized size = 1.30 \begin {gather*} -\frac {16 e^2 \left (3 x+x^2+\frac {3 x^4}{2}+x^5+\frac {x^8}{4}\right )}{-10+\log (9)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.54, size = 24, normalized size = 0.89
method | result | size |
default | \(\frac {4 \,{\mathrm e}^{2} \left (x^{4}+2 x +3\right )^{2}}{\ln \left (\frac {{\mathrm e}^{10}}{9}\right )}\) | \(24\) |
gosper | \(\frac {4 \left (x^{7}+4 x^{4}+6 x^{3}+4 x +12\right ) {\mathrm e}^{2} x}{\ln \left (\frac {{\mathrm e}^{10}}{9}\right )}\) | \(33\) |
norman | \(-\frac {24 \,{\mathrm e}^{2} x}{\ln \left (3\right )-5}-\frac {8 \,{\mathrm e}^{2} x^{2}}{\ln \left (3\right )-5}-\frac {12 \,{\mathrm e}^{2} x^{4}}{\ln \left (3\right )-5}-\frac {8 \,{\mathrm e}^{2} x^{5}}{\ln \left (3\right )-5}-\frac {2 \,{\mathrm e}^{2} x^{8}}{\ln \left (3\right )-5}\) | \(65\) |
risch | \(\frac {4 \,{\mathrm e}^{2} x^{8}}{-2 \ln \left (3\right )+10}+\frac {16 \,{\mathrm e}^{2} x^{5}}{-2 \ln \left (3\right )+10}+\frac {24 \,{\mathrm e}^{2} x^{4}}{-2 \ln \left (3\right )+10}+\frac {16 \,{\mathrm e}^{2} x^{2}}{-2 \ln \left (3\right )+10}+\frac {48 \,{\mathrm e}^{2} x}{-2 \ln \left (3\right )+10}+\frac {36 \,{\mathrm e}^{2}}{-2 \ln \left (3\right )+10}\) | \(87\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \left (\frac {1}{9} \, e^{10}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 32, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \left (3\right ) - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 66 vs.
\(2 (22) = 44\).
time = 0.01, size = 66, normalized size = 2.44 \begin {gather*} - \frac {2 x^{8} e^{2}}{-5 + \log {\left (3 \right )}} - \frac {8 x^{5} e^{2}}{-5 + \log {\left (3 \right )}} - \frac {12 x^{4} e^{2}}{-5 + \log {\left (3 \right )}} - \frac {8 x^{2} e^{2}}{-5 + \log {\left (3 \right )}} - \frac {24 x e^{2}}{-5 + \log {\left (3 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.38, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \left (\frac {1}{9} \, e^{10}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.32, size = 64, normalized size = 2.37 \begin {gather*} -\frac {4\,{\mathrm {e}}^2\,x^8}{\ln \left (9\right )-10}-\frac {16\,{\mathrm {e}}^2\,x^5}{\ln \left (9\right )-10}-\frac {24\,{\mathrm {e}}^2\,x^4}{\ln \left (9\right )-10}-\frac {16\,{\mathrm {e}}^2\,x^2}{\ln \left (9\right )-10}-\frac {48\,{\mathrm {e}}^2\,x}{\ln \left (9\right )-10} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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