Optimal. Leaf size=25 \[ \frac {5-\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x} \]
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Rubi [F] time = 1.42, 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 {-20-6 x-2 x^2-5 \log \left (\frac {4 e^{x^2}}{x}\right )-5 \log (x)+\left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right ) \log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{4 x^2+x^3+x^2 \log \left (\frac {4 e^{x^2}}{x}\right )+x^2 \log (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 {-20-6 x-2 x^2-5 \log \left (\frac {4 e^{x^2}}{x}\right )-5 \log (x)+\left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right ) \log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx\\ &=\int \left (-\frac {2}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)}-\frac {20}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {6}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {5 \log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {5 \log (x)}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}+\frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx\right )-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \frac {\log (x)}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=-\left (2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx\right )-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \left (\frac {1}{x^2}+\frac {-4-x-\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}\right ) \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx-5 \int \frac {-4-x-\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx-5 \int \frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-5 \int \left (-\frac {4}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}-\frac {\log \left (\frac {4 e^{x^2}}{x}\right )}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}\right ) \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-20 \int \frac {1}{x^2 \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ &=\frac {5}{x}-2 \int \frac {1}{4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)} \, dx+5 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx-6 \int \frac {1}{x \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )} \, dx+\int \frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.06, size = 28, normalized size = 1.12 \begin {gather*} \frac {5}{x}-\frac {\log \left (4+x+\log \left (\frac {4 e^{x^2}}{x}\right )+\log (x)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 18, normalized size = 0.72 \begin {gather*} -\frac {\log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right ) - 5}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 22, normalized size = 0.88 \begin {gather*} -\frac {\log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{x} + \frac {5}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 28, normalized size = 1.12
method | result | size |
default | \(-\frac {\ln \left (\ln \left (\frac {4 \,{\mathrm e}^{x^{2}}}{x}\right )+4+x +\ln \relax (x )\right )}{x}+\frac {5}{x}\) | \(28\) |
risch | \(-\frac {\ln \left (2 \ln \relax (2)+\ln \left ({\mathrm e}^{x^{2}}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x^{2}}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right )\right )}{2}+4+x \right )}{x}+\frac {5}{x}\) | \(83\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.66, size = 311, normalized size = 12.44 \begin {gather*} \frac {{\left (3 \, \log \relax (2) + 1\right )} \log \relax (x)}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} + \frac {{\left (4 \, \log \relax (2)^{2} - 7 \, \log \relax (2) - 25\right )} \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{{\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} \sqrt {8 \, \log \relax (2) + 15}} - \frac {4 \, \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{\sqrt {8 \, \log \relax (2) + 15}} + \frac {5 \, {\left (4 \, \log \relax (2) + 7\right )} \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{{\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} \sqrt {8 \, \log \relax (2) + 15}} - \frac {5 \, \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{2 \, {\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )}} + \frac {3 \, \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right )}{2 \, {\left (\log \relax (2) + 2\right )}} + \frac {5 \, \log \relax (x)}{\log \relax (2)^{2} + 4 \, \log \relax (2) + 4} - \frac {3 \, \log \relax (x)}{\log \relax (2) + 2} + \frac {3 \, \arctan \left (\frac {2 \, x + 1}{\sqrt {8 \, \log \relax (2) + 15}}\right )}{\sqrt {8 \, \log \relax (2) + 15} {\left (\log \relax (2) + 2\right )}} + \frac {10 \, \log \relax (2)^{2} - {\left (x {\left (3 \, \log \relax (2) + 1\right )} + 2 \, \log \relax (2)^{2} + 8 \, \log \relax (2) + 8\right )} \log \left (x^{2} + x + 2 \, \log \relax (2) + 4\right ) + 20 \, \log \relax (2)}{2 \, {\left (\log \relax (2)^{2} + 4 \, \log \relax (2) + 4\right )} x} + \frac {10}{x {\left (\log \relax (2) + 2\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.72, size = 22, normalized size = 0.88 \begin {gather*} -\frac {\ln \left (x+\ln \left (\frac {4}{x}\right )+\ln \relax (x)+x^2+4\right )-5}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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