Optimal. Leaf size=23 \[ x \log \left (-1+e^{\frac {5 x}{-2 x+e^2 x^5}}\right ) \]
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Rubi [A] time = 55.41, antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 84, number of rules used = 5, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 6742, 6725, 2548, 12} \begin {gather*} x \log \left (e^{-\frac {5}{2-e^2 x^4}}-1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2548
Rule 6688
Rule 6725
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {20 e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right )\right ) \, dx\\ &=-\left (20 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\right )+\int \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-20 \int \left (\frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{5 \left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}-\frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{5 \left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}\right ) \, dx-\int -\frac {20 e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+20 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )-4 \int \left (\frac {2 e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )}\right ) \, dx+4 \int \left (\frac {2 e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}+\frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )}\right ) \, dx+20 \int \left (\frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{5 \left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}-\frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{5 \left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2}\right ) \, dx\\ &=x \log \left (-1+e^{-\frac {5}{2-e^2 x^4}}\right )+4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} x^4}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx-4 \int \frac {e^{2+\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right ) x^4}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx-4 \int \frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )} \, dx+4 \int \frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )} \, dx-8 \int \frac {e^{\frac {5}{-2+e^2 x^4}}}{\left (-1+e^{\frac {1}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx+8 \int \frac {e^{\frac {5}{-2+e^2 x^4}} \left (4+3 e^{\frac {1}{-2+e^2 x^4}}+2 e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}\right )}{\left (1+e^{\frac {1}{-2+e^2 x^4}}+e^{\frac {2}{-2+e^2 x^4}}+e^{\frac {3}{-2+e^2 x^4}}+e^{\frac {4}{-2+e^2 x^4}}\right ) \left (-2+e^2 x^4\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 4.26, size = 20, normalized size = 0.87 \begin {gather*} x \log \left (-1+e^{\frac {5}{-2+e^2 x^4}}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 33, normalized size = 1.43 \begin {gather*} x \log \left (-{\left (e^{2} - e^{\left (\frac {2 \, x^{4} e^{2} + 1}{x^{4} e^{2} - 2}\right )}\right )} e^{\left (-2\right )}\right ) \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*} \int \frac {20 \, x^{4} e^{\left (\frac {5}{x^{4} e^{2} - 2} + 2\right )} + {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} - {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} + 4\right )} e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} + 4\right )} \log \left (e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} - 1\right )}{x^{8} e^{4} - 4 \, x^{4} e^{2} - {\left (x^{8} e^{4} - 4 \, x^{4} e^{2} + 4\right )} e^{\left (\frac {5}{x^{4} e^{2} - 2}\right )} + 4}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.21, size = 19, normalized size = 0.83
method | result | size |
risch | \(x \ln \left ({\mathrm e}^{\frac {5}{x^{4} {\mathrm e}^{2}-2}}-1\right )\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.74, size = 72, normalized size = 3.13 \begin {gather*} x \log \left (e^{\left (\frac {4}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {3}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {2}{x^{4} e^{2} - 2}\right )} + e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} + 1\right ) + x \log \left (e^{\left (\frac {1}{x^{4} e^{2} - 2}\right )} - 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.65, size = 18, normalized size = 0.78 \begin {gather*} x\,\ln \left ({\mathrm {e}}^{\frac {5}{x^4\,{\mathrm {e}}^2-2}}-1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.60, size = 15, normalized size = 0.65 \begin {gather*} x \log {\left (e^{\frac {5}{x^{4} e^{2} - 2}} - 1 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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