Optimal. Leaf size=30 \[ \frac {\left (-10+\frac {x}{2}+\log (x)\right ) \left (e^{e^{4 x^2}}-x+3 \log (x)\right )}{x} \]
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Rubi [F] time = 0.60, 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 {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{x^2} \, dx\\ &=\frac {1}{2} \int \left (8 e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x))-\frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x)) \, dx\\ &=-\left (\frac {1}{2} \int \left (\frac {2 e^{e^{4 x^2}} (-11+\log (x))}{x^2}+\frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx\right )+4 \int \left (-20 e^{e^{4 x^2}+4 x^2}+e^{e^{4 x^2}+4 x^2} x+2 e^{e^{4 x^2}+4 x^2} \log (x)\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} x \, dx+8 \int e^{e^{4 x^2}+4 x^2} \log (x) \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} (-11+\log (x))}{x^2} \, dx\\ &=-\left (\frac {1}{2} \int \left (\frac {60-x+x^2}{x^2}-\frac {72 \log (x)}{x^2}+\frac {6 \log ^2(x)}{x^2}\right ) \, dx\right )+2 \operatorname {Subst}\left (\int e^{e^{4 x}+4 x} \, dx,x,x^2\right )-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \left (-\frac {11 e^{e^{4 x^2}}}{x^2}+\frac {e^{e^{4 x^2}} \log (x)}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {60-x+x^2}{x^2} \, dx\right )+\frac {1}{2} \operatorname {Subst}\left (\int e^x \, dx,x,e^{4 x^2}\right )-3 \int \frac {\log ^2(x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+36 \int \frac {\log (x)}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} \log (x)}{x^2} \, dx\\ &=\frac {1}{2} e^{e^{4 x^2}}-\frac {36}{x}-\frac {36 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-\frac {1}{2} \int \left (1+\frac {60}{x^2}-\frac {1}{x}\right ) \, dx-6 \int \frac {\log (x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx\\ &=\frac {1}{2} e^{e^{4 x^2}}-\frac {x}{2}+\frac {\log (x)}{2}-\frac {30 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.25, size = 49, normalized size = 1.63 \begin {gather*} \frac {e^{e^{4 x^2}} (-20+x)-x^2+\left (-60+2 e^{e^{4 x^2}}+x\right ) \log (x)+6 \log ^2(x)}{2 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 38, normalized size = 1.27 \begin {gather*} -\frac {x^{2} - {\left (x + 2 \, \log \relax (x) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} - {\left (x - 60\right )} \log \relax (x) - 6 \, \log \relax (x)^{2}}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} - 2 \, {\left (4 \, {\left (x^{3} - 20 \, x^{2}\right )} e^{\left (4 \, x^{2}\right )} + {\left (8 \, x^{2} e^{\left (4 \, x^{2}\right )} - 1\right )} \log \relax (x) + 11\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \relax (x)^{2} - x - 72 \, \log \relax (x) + 60}{2 \, x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 44, normalized size = 1.47
method | result | size |
risch | \(\frac {3 \ln \relax (x )^{2}}{x}-\frac {30 \ln \relax (x )}{x}+\frac {\ln \relax (x )}{2}-\frac {x}{2}+\frac {\left (x -20+2 \ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.39, size = 45, normalized size = 1.50 \begin {gather*} -\frac {1}{2} \, x + \frac {{\left (x + 2 \, \log \relax (x) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \relax (x)^{2} - 60 \, \log \relax (x) - 60}{2 \, x} + \frac {30}{x} + \frac {1}{2} \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.55, size = 42, normalized size = 1.40 \begin {gather*} \frac {\ln \relax (x)}{2}-\frac {x}{2}-\frac {30\,\ln \relax (x)}{x}+\frac {3\,{\ln \relax (x)}^2}{x}+\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x^2}}\,\left (\frac {x}{2}+\ln \relax (x)-10\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 42, normalized size = 1.40 \begin {gather*} - \frac {x}{2} + \frac {\log {\relax (x )}}{2} + \frac {\left (x + 2 \log {\relax (x )} - 20\right ) e^{e^{4 x^{2}}}}{2 x} + \frac {3 \log {\relax (x )}^{2}}{x} - \frac {30 \log {\relax (x )}}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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