Optimal. Leaf size=34 \[ -x+\log (4)-e^2 \left (-2-x+\frac {x}{-1+\log \left (\frac {3+\frac {2}{x^2}}{x}\right )}\right ) \]
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Rubi [F]
time = 0.54, 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 {-2-3 x^2+e^2 \left (-2+3 x^2\right )+\left (4+6 x^2+e^2 \left (-6-9 x^2\right )\right ) \log \left (\frac {2+3 x^2}{x^3}\right )+\left (-2-3 x^2+e^2 \left (2+3 x^2\right )\right ) \log ^2\left (\frac {2+3 x^2}{x^3}\right )}{2+3 x^2+\left (-4-6 x^2\right ) \log \left (\frac {2+3 x^2}{x^3}\right )+\left (2+3 x^2\right ) \log ^2\left (\frac {2+3 x^2}{x^3}\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 {-2-3 x^2+e^2 \left (-2+3 x^2\right )+\left (4+6 x^2+e^2 \left (-6-9 x^2\right )\right ) \log \left (\frac {2+3 x^2}{x^3}\right )+\left (-2-3 x^2+e^2 \left (2+3 x^2\right )\right ) \log ^2\left (\frac {2+3 x^2}{x^3}\right )}{\left (2+3 x^2\right ) \left (1-\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx\\ &=\int \left (-1+e^2-\frac {3 e^2 \left (2+x^2\right )}{\left (2+3 x^2\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2}-\frac {e^2}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )}\right ) \, dx\\ &=-\left (\left (1-e^2\right ) x\right )-e^2 \int \frac {1}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \, dx-\left (3 e^2\right ) \int \frac {2+x^2}{\left (2+3 x^2\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx\\ &=-\left (\left (1-e^2\right ) x\right )-e^2 \int \frac {1}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \, dx-\left (3 e^2\right ) \int \left (\frac {1}{3 \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2}+\frac {4}{3 \left (2+3 x^2\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2}\right ) \, dx\\ &=-\left (\left (1-e^2\right ) x\right )-e^2 \int \frac {1}{\left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx-e^2 \int \frac {1}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \, dx-\left (4 e^2\right ) \int \frac {1}{\left (2+3 x^2\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx\\ &=-\left (\left (1-e^2\right ) x\right )-e^2 \int \frac {1}{\left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx-e^2 \int \frac {1}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \, dx-\left (4 e^2\right ) \int \left (\frac {i}{2 \sqrt {2} \left (i \sqrt {2}-\sqrt {3} x\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2}+\frac {i}{2 \sqrt {2} \left (i \sqrt {2}+\sqrt {3} x\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2}\right ) \, dx\\ &=-\left (\left (1-e^2\right ) x\right )-e^2 \int \frac {1}{\left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx-e^2 \int \frac {1}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \, dx-\left (i \sqrt {2} e^2\right ) \int \frac {1}{\left (i \sqrt {2}-\sqrt {3} x\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx-\left (i \sqrt {2} e^2\right ) \int \frac {1}{\left (i \sqrt {2}+\sqrt {3} x\right ) \left (-1+\log \left (\frac {2+3 x^2}{x^3}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.05, size = 30, normalized size = 0.88 \begin {gather*} \left (-1+e^2\right ) x-\frac {e^2 x}{-1+\log \left (\frac {2+3 x^2}{x^3}\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 4.72, size = 30, normalized size = 0.88
method | result | size |
risch | \({\mathrm e}^{2} x -x -\frac {{\mathrm e}^{2} x}{\ln \left (\frac {3 x^{2}+2}{x^{3}}\right )-1}\) | \(30\) |
norman | \(\frac {\left (-2 \,{\mathrm e}^{2}+1\right ) x +\left ({\mathrm e}^{2}-1\right ) x \ln \left (\frac {3 x^{2}+2}{x^{3}}\right )}{\ln \left (\frac {3 x^{2}+2}{x^{3}}\right )-1}\) | \(45\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.32, size = 50, normalized size = 1.47 \begin {gather*} \frac {x {\left (e^{2} - 1\right )} \log \left (3 \, x^{2} + 2\right ) - 3 \, x {\left (e^{2} - 1\right )} \log \left (x\right ) - x {\left (2 \, e^{2} - 1\right )}}{\log \left (3 \, x^{2} + 2\right ) - 3 \, \log \left (x\right ) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 49, normalized size = 1.44 \begin {gather*} -\frac {2 \, x e^{2} - {\left (x e^{2} - x\right )} \log \left (\frac {3 \, x^{2} + 2}{x^{3}}\right ) - x}{\log \left (\frac {3 \, x^{2} + 2}{x^{3}}\right ) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.07, size = 24, normalized size = 0.71 \begin {gather*} x \left (-1 + e^{2}\right ) - \frac {x e^{2}}{\log {\left (\frac {3 x^{2} + 2}{x^{3}} \right )} - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.48, size = 55, normalized size = 1.62 \begin {gather*} \frac {x e^{2} \log \left (\frac {3 \, x^{2} + 2}{x^{3}}\right ) - 2 \, x e^{2} - x \log \left (\frac {3 \, x^{2} + 2}{x^{3}}\right ) + x}{\log \left (\frac {3 \, x^{2} + 2}{x^{3}}\right ) - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.32, size = 53, normalized size = 1.56 \begin {gather*} -\frac {x\,\left (\ln \left (\frac {3\,x^2+2}{x^3}\right )+2\,{\mathrm {e}}^2-\ln \left (\frac {3\,x^2+2}{x^3}\right )\,{\mathrm {e}}^2-1\right )}{\ln \left (\frac {3\,x^2+2}{x^3}\right )-1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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