Optimal. Leaf size=24 \[ \frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \]
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Rubi [F] time = 0.83, 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 {2 x+e^x (-2+4 x)+2 \log \left (-3+e^{2 e^4}\right )}{-15 e^{3 x}-45 e^{2 x} x-45 e^x x^2-15 x^3+\left (45 e^{2 x}+90 e^x x+45 x^2\right ) \log \left (-3+e^{2 e^4}\right )+\left (-45 e^x-45 x\right ) \log ^2\left (-3+e^{2 e^4}\right )+15 \log ^3\left (-3+e^{2 e^4}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-x-e^x (-1+2 x)-\log \left (-3+e^{2 e^4}\right )\right )}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=\frac {2}{15} \int \frac {-x-e^x (-1+2 x)-\log \left (-3+e^{2 e^4}\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=\frac {2}{15} \int \left (\frac {2 x \left (-1+x-\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}-\frac {-1+2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {2}{15} \int \frac {-1+2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx\right )+\frac {4}{15} \int \frac {x \left (-1+x-\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ &=-\left (\frac {2}{15} \int \left (-\frac {1}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}+\frac {2 x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2}\right ) \, dx\right )+\frac {4}{15} \int \left (\frac {x^2}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}-\frac {x \left (1+\log \left (-3+e^{2 e^4}\right )\right )}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3}\right ) \, dx\\ &=\frac {2}{15} \int \frac {1}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx+\frac {4}{15} \int \frac {x^2}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx-\frac {4}{15} \int \frac {x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \, dx-\frac {1}{15} \left (4 \left (1+\log \left (-3+e^{2 e^4}\right )\right )\right ) \int \frac {x}{\left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} \frac {2 x}{15 \left (e^x+x-\log \left (-3+e^{2 e^4}\right )\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 42, normalized size = 1.75 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, {\left (x + e^{x}\right )} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 51, normalized size = 2.12 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, x e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) - 2 \, e^{x} \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 22, normalized size = 0.92
method | result | size |
norman | \(\frac {2 x}{15 \left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )^{2}}\) | \(22\) |
risch | \(\frac {2 x}{15 \left (\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{4}}-3\right )-x -{\mathrm e}^{x}\right )^{2}}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.53, size = 50, normalized size = 2.08 \begin {gather*} \frac {2 \, x}{15 \, {\left (x^{2} + 2 \, {\left (x - \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )\right )} e^{x} - 2 \, x \log \left (e^{\left (2 \, e^{4}\right )} - 3\right ) + \log \left (e^{\left (2 \, e^{4}\right )} - 3\right )^{2} + e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.48, size = 57, normalized size = 2.38 \begin {gather*} \frac {2\,x}{15\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+{\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )}^2+2\,x\,{\mathrm {e}}^x-2\,x\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^4}-3\right )+x^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.19, size = 60, normalized size = 2.50 \begin {gather*} \frac {2 x}{15 x^{2} - 30 x \log {\left (-3 + e^{2 e^{4}} \right )} + \left (30 x - 30 \log {\left (-3 + e^{2 e^{4}} \right )}\right ) e^{x} + 15 e^{2 x} + 15 \log {\left (-3 + e^{2 e^{4}} \right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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