Optimal. Leaf size=37 \[ \log \left (-2+\frac {e^{-e^{x^2} (-4+x)} \log \left (\frac {e^4}{\frac {e^2}{3}-x}\right )}{x}\right ) \]
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Rubi [F]
time = 13.68, 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 {-3 x+\left (e^2-3 x+e^{x^2} \left (-3 x^2+24 x^3-6 x^4+e^2 \left (x-8 x^2+2 x^3\right )\right )\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )}{e^{e^{x^2} (-4+x)} \left (2 e^2 x^2-6 x^3\right )+\left (-e^2 x+3 x^2\right ) \log \left (\frac {3 e^4}{e^2-3 x}\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 {3 x-\left (e^2-3 x\right ) \left (1+e^{x^2} x \left (1-8 x+2 x^2\right )\right ) \left (4+\log (3)+\log \left (\frac {1}{e^2-3 x}\right )\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\int \left (\frac {e^{x^2} \left (1-8 x+2 x^2\right ) \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx\\ &=\int \frac {e^{x^2} \left (1-8 x+2 x^2\right ) \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx+\int \left (\frac {e^{x^2} \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {2 e^{x^2} x^2 \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}+\frac {8 e^{x^2} x \left (4 \left (1+\frac {\log (3)}{4}\right )+\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx\\ &=2 \int \frac {e^{x^2} x^2 \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+8 \int \frac {e^{x^2} x \left (4 \left (1+\frac {\log (3)}{4}\right )+\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \left (-4 \left (1+\frac {\log (3)}{4}\right )-\log \left (\frac {1}{e^2-3 x}\right )\right )}{4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \left (\frac {3 \left (15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )\right )}{e^2 \left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}+\frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{e^2 x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx\\ &=2 \int \left (\frac {e^{x^2} x^2 (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx+8 \int \left (-\frac {e^{x^2} x (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}-\frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx+\frac {\int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\frac {3 \int \frac {15 x \left (1+\frac {\log (3)}{5}\right )-4 e^2 \left (1+\frac {\log (3)}{4}\right )-e^2 \log \left (\frac {1}{e^2-3 x}\right )+3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\int \left (\frac {e^{x^2} (4+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}\right ) \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\frac {\int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{x \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\frac {3 \int \frac {-e^2 (4+\log (3))+3 x (5+\log (3))-\left (e^2-3 x\right ) \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (4-2 e^{e^{x^2} (-4+x)} x+\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\frac {\int \left (\frac {e^2 (4+\log (3))}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 (5+\log (3))}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}-\frac {3 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )}+\frac {e^2 \log \left (\frac {1}{e^2-3 x}\right )}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx}{e^2}+\frac {3 \int \left (\frac {e^2 (4+\log (3))}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 x (5+\log (3))}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}+\frac {e^2 \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}-\frac {3 x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )}\right ) \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx\\ &=2 \int \frac {e^{x^2} x^2 \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+3 \int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx-8 \int \frac {e^{x^2} x \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-\frac {3 \int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx}{e^2}-\frac {9 \int \frac {x \log \left (\frac {1}{e^2-3 x}\right )}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+(4+\log (3)) \int \frac {e^{x^2}}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(4+\log (3)) \int \frac {1}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx+(2 (4+\log (3))) \int \frac {e^{x^2} x^2}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+(3 (4+\log (3))) \int \frac {1}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx-(8 (4+\log (3))) \int \frac {e^{x^2} x}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx-\frac {(3 (5+\log (3))) \int \frac {1}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx}{e^2}-\frac {(9 (5+\log (3))) \int \frac {x}{\left (e^2-3 x\right ) \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx}{e^2}+\int \frac {e^{x^2} \log \left (\frac {1}{e^2-3 x}\right )}{-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )} \, dx+\int \frac {\log \left (\frac {1}{e^2-3 x}\right )}{x \left (-4+2 e^{e^{x^2} (-4+x)} x-\log \left (\frac {3}{e^2-3 x}\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [F]
time = 6.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-3 x+\left (e^2-3 x+e^{x^2} \left (-3 x^2+24 x^3-6 x^4+e^2 \left (x-8 x^2+2 x^3\right )\right )\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )}{e^{e^{x^2} (-4+x)} \left (2 e^2 x^2-6 x^3\right )+\left (-e^2 x+3 x^2\right ) \log \left (\frac {3 e^4}{e^2-3 x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [C] Result contains complex when optimal does not.
time = 5.80, size = 46, normalized size = 1.24
method | result | size |
risch | \(-\left (x -4\right ) {\mathrm e}^{x^{2}}+\ln \left ({\mathrm e}^{\left (x -4\right ) {\mathrm e}^{x^{2}}}+\frac {i \left (2 i \ln \left (3\right )-2 i \ln \left ({\mathrm e}^{2}-3 x \right )+8 i\right )}{4 x}\right )\) | \(46\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 53, normalized size = 1.43 \begin {gather*} -x e^{\left (x^{2}\right )} + \log \left (\frac {2 \, x e^{\left (x e^{\left (x^{2}\right )}\right )} - {\left (\log \left (3\right ) + 4\right )} e^{\left (4 \, e^{\left (x^{2}\right )}\right )} + e^{\left (4 \, e^{\left (x^{2}\right )}\right )} \log \left (-3 \, x + e^{2}\right )}{2 \, x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 45, normalized size = 1.22 \begin {gather*} -{\left (x - 4\right )} e^{\left (x^{2}\right )} + \log \left (\frac {2 \, x e^{\left ({\left (x - 4\right )} e^{\left (x^{2}\right )}\right )} - \log \left (-\frac {3 \, e^{4}}{3 \, x - e^{2}}\right )}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.58, size = 36, normalized size = 0.97 \begin {gather*} \left (4 - x\right ) e^{x^{2}} + \log {\left (e^{\left (x - 4\right ) e^{x^{2}}} - \frac {\log {\left (\frac {3 e^{4}}{- 3 x + e^{2}} \right )}}{2 x} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.49, size = 51, normalized size = 1.38 \begin {gather*} -x e^{\left (x^{2}\right )} + 4 \, e^{\left (x^{2}\right )} + \log \left (-2 \, x e^{\left (x e^{\left (x^{2}\right )} - 4 \, e^{\left (x^{2}\right )}\right )} + \log \left (-\frac {3}{3 \, x - e^{2}}\right ) + 4\right ) - \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.75, size = 48, normalized size = 1.30 \begin {gather*} \ln \left (\ln \left (-\frac {3\,{\mathrm {e}}^4}{3\,x-{\mathrm {e}}^2}\right )-2\,x\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{x^2}}\right )-\ln \left (x\right )-{\mathrm {e}}^{x^2}\,\left (x-4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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