Optimal. Leaf size=25 \[ \log \left (\frac {3}{x+e^{-2 x} \left (5+\log \left (-3+e^{x (9+x)}\right )\right )}\right ) \]
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Rubi [F] time = 3.98, 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 {-30+3 e^{2 x}+e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )+\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{-15-3 e^{2 x} x+e^{9 x+x^2} \left (5+e^{2 x} x\right )+\left (-3+e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\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 {30-3 e^{2 x}-e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )-\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx\\ &=\int \left (\frac {e^{11 x+x^2}}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}-\frac {30}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {3 e^{2 x}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {e^{9 x+x^2}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {2 e^{9 x+x^2} x}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}-\frac {6 \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {2 e^{9 x+x^2} \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}\right ) \, dx\\ &=2 \int \frac {e^{9 x+x^2} x}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+2 \int \frac {e^{9 x+x^2} \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+3 \int \frac {e^{2 x}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx-6 \int \frac {\log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx-30 \int \frac {1}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+\int \frac {e^{11 x+x^2}}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+\int \frac {e^{9 x+x^2}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-30+3 e^{2 x}+e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )+\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{-15-3 e^{2 x} x+e^{9 x+x^2} \left (5+e^{2 x} x\right )+\left (-3+e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.52, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 25, normalized size = 1.00
method | result | size |
risch | \(2 x -\ln \left (x \,{\mathrm e}^{2 x}+\ln \left ({\mathrm e}^{\left (x +9\right ) x}-3\right )+5\right )\) | \(25\) |
norman | \(2 x -\ln \left (x \,{\mathrm e}^{2 x}+\ln \left ({\mathrm e}^{x^{2}+9 x}-3\right )+5\right )\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.69, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.57, size = 24, normalized size = 0.96 \begin {gather*} 2\,x-\ln \left (\ln \left ({\mathrm {e}}^{x\,\left (x+9\right )}-3\right )+x\,{\mathrm {e}}^{2\,x}+5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.49, size = 24, normalized size = 0.96 \begin {gather*} 2 x - \log {\left (x e^{2 x} + \log {\left (e^{x^{2} + 9 x} - 3 \right )} + 5 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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