Optimal. Leaf size=27 \[ \frac {3}{\log \left (4+e^{2 x} x \left (e^{4+x^2}-2 (1+x)\right )\right )} \]
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Rubi [F] time = 5.15, 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 {e^{4+2 x+x^2} \left (-3-6 x-6 x^2\right )+e^{2 x} \left (6+24 x+12 x^2\right )}{\left (4+e^{4+2 x+x^2} x+e^{2 x} \left (-2 x-2 x^2\right )\right ) \log ^2\left (4+e^{4+2 x+x^2} x+e^{2 x} \left (-2 x-2 x^2\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {3 \left (1+2 x+2 x^2\right )}{x \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}+\frac {6 \left (-2-4 x-4 x^2-e^{2 x} x^2+2 e^{2 x} x^3+2 e^{2 x} x^4\right )}{x \left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1+2 x+2 x^2}{x \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx\right )+6 \int \frac {-2-4 x-4 x^2-e^{2 x} x^2+2 e^{2 x} x^3+2 e^{2 x} x^4}{x \left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx\\ &=-\left (3 \int \left (\frac {2}{\log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}+\frac {1}{x \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}+\frac {2 x}{\log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}\right ) \, dx\right )+6 \int \left (-\frac {4}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}-\frac {2}{x \left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}-\frac {4 x}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}-\frac {e^{2 x} x}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}+\frac {2 e^{2 x} x^2}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}+\frac {2 e^{2 x} x^3}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )}\right ) \, dx\\ &=-\left (3 \int \frac {1}{x \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx\right )-6 \int \frac {1}{\log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx-6 \int \frac {x}{\log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx-6 \int \frac {e^{2 x} x}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx-12 \int \frac {1}{x \left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx+12 \int \frac {e^{2 x} x^2}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx+12 \int \frac {e^{2 x} x^3}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx-24 \int \frac {1}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx-24 \int \frac {x}{\left (-4+2 e^{2 x} x-e^{4+2 x+x^2} x+2 e^{2 x} x^2\right ) \log ^2\left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 30, normalized size = 1.11 \begin {gather*} \frac {3}{\log \left (4+e^{4+2 x+x^2} x-2 e^{2 x} x (1+x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.95, size = 29, normalized size = 1.07 \begin {gather*} \frac {3}{\log \left (x e^{\left (x^{2} + 2 \, x + 4\right )} - 2 \, {\left (x^{2} + x\right )} e^{\left (2 \, x\right )} + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.44, size = 34, normalized size = 1.26 \begin {gather*} \frac {3}{\log \left (-2 \, x^{2} e^{\left (2 \, x\right )} + x e^{\left (x^{2} + 2 \, x + 4\right )} - 2 \, x e^{\left (2 \, x\right )} + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 33, normalized size = 1.22
method | result | size |
risch | \(\frac {3}{\ln \left (x \,{\mathrm e}^{x^{2}+2 x +4}+\left (-2 x^{2}-2 x \right ) {\mathrm e}^{2 x}+4\right )}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.89, size = 29, normalized size = 1.07 \begin {gather*} \frac {3}{\log \left (x e^{\left (x^{2} + 2 \, x + 4\right )} - 2 \, {\left (x^{2} + x\right )} e^{\left (2 \, x\right )} + 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.22, size = 34, normalized size = 1.26 \begin {gather*} \frac {3}{\ln \left (x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4-{\mathrm {e}}^{2\,x}\,\left (2\,x^2+2\,x\right )+4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.87, size = 32, normalized size = 1.19 \begin {gather*} \frac {3}{\log {\left (x e^{2 x} e^{x^{2} + 4} + \left (- 2 x^{2} - 2 x\right ) e^{2 x} + 4 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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