Optimal. Leaf size=25 \[ 4+\frac {2 x^2}{e^{1+\frac {e^x}{2}}+\log (9 x)} \]
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Rubi [F] time = 1.40, 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^{\frac {1}{2} \left (2+e^x\right )} \left (4 x-e^x x^2\right )+4 x \log (9 x)}{e^{2+e^x}+2 e^{\frac {1}{2} \left (2+e^x\right )} \log (9 x)+\log ^2(9 x)} \, 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 x+e^{\frac {1}{2} \left (2+e^x\right )} \left (4 x-e^x x^2\right )+4 x \log (9 x)}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=\int \left (-\frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}+\frac {2 x \left (-1+2 e^{1+\frac {e^x}{2}}+2 \log (9 x)\right )}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}\right ) \, dx\\ &=2 \int \frac {x \left (-1+2 e^{1+\frac {e^x}{2}}+2 \log (9 x)\right )}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=2 \int \left (-\frac {x}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2}+\frac {2 x}{e^{1+\frac {e^x}{2}}+\log (9 x)}\right ) \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ &=-\left (2 \int \frac {x}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\right )+4 \int \frac {x}{e^{1+\frac {e^x}{2}}+\log (9 x)} \, dx-\int \frac {e^{\frac {1}{2} \left (2+e^x+2 x\right )} x^2}{\left (e^{1+\frac {e^x}{2}}+\log (9 x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.36, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 x^2}{e^{1+\frac {e^x}{2}}+\log (9 x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.89, size = 19, normalized size = 0.76 \begin {gather*} \frac {2 \, x^{2}}{e^{\left (\frac {1}{2} \, e^{x} + 1\right )} + \log \left (9 \, x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.62, size = 139, normalized size = 5.56 \begin {gather*} \frac {2 \, {\left (x^{3} e^{\left (2 \, x\right )} \log \left (9 \, x\right )^{2} + x^{3} e^{\left (2 \, x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right ) - 2 \, x^{2} e^{x} \log \left (9 \, x\right ) - 2 \, x^{2} e^{\left (x + \frac {1}{2} \, e^{x} + 1\right )}\right )}}{x e^{\left (2 \, x\right )} \log \left (9 \, x\right )^{3} + 2 \, x e^{\left (2 \, x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right )^{2} + x e^{\left (2 \, x + e^{x} + 2\right )} \log \left (9 \, x\right ) - 2 \, e^{x} \log \left (9 \, x\right )^{2} - 4 \, e^{\left (x + \frac {1}{2} \, e^{x} + 1\right )} \log \left (9 \, x\right ) - 2 \, e^{\left (x + e^{x} + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 20, normalized size = 0.80
method | result | size |
risch | \(\frac {2 x^{2}}{{\mathrm e}^{\frac {{\mathrm e}^{x}}{2}+1}+\ln \left (9 x \right )}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.95, size = 21, normalized size = 0.84 \begin {gather*} \frac {2 \, x^{2}}{e^{\left (\frac {1}{2} \, e^{x} + 1\right )} + 2 \, \log \relax (3) + \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.60, size = 19, normalized size = 0.76 \begin {gather*} \frac {2\,x^2}{{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}+1}+\ln \left (9\,x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.31, size = 17, normalized size = 0.68 \begin {gather*} \frac {2 x^{2}}{e^{\frac {e^{x}}{2} + 1} + \log {\left (9 x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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