Optimal. Leaf size=21 \[ \log ^4\left (\frac {5}{4} \left (5-5 e^{e^{x^2}} x\right ) \log (x)\right ) \]
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Rubi [F] time = 5.97, 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 {\left (-4+e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (\frac {1}{4} \left (25 \log (x)-25 e^{e^{x^2}} x \log (x)\right )\right )}{-x \log (x)+e^{e^{x^2}} 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 {\left (4-e^{e^{x^2}} \left (4 x+\left (4 x+8 e^{x^2} x^3\right ) \log (x)\right )\right ) \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (1-e^{e^{x^2}} x\right ) \log (x)} \, dx\\ &=\int \left (\frac {8 e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x}+\frac {4 \left (-1+e^{e^{x^2}} x+e^{e^{x^2}} x \log (x)\right ) \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right ) \log (x)}\right ) \, dx\\ &=4 \int \frac {\left (-1+e^{e^{x^2}} x+e^{e^{x^2}} x \log (x)\right ) \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right ) \log (x)} \, dx+8 \int \frac {e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x} \, dx\\ &=4 \int \left (\frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right )}+\frac {(1+\log (x)) \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \log (x)}\right ) \, dx+8 \int \frac {e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x} \, dx\\ &=4 \int \frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right )} \, dx+4 \int \frac {(1+\log (x)) \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \log (x)} \, dx+8 \int \frac {e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x} \, dx\\ &=4 \int \frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right )} \, dx+4 \int \left (\frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x}+\frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \log (x)}\right ) \, dx+8 \int \frac {e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x} \, dx\\ &=4 \int \frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x} \, dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \left (-1+e^{e^{x^2}} x\right )} \, dx+4 \int \frac {\log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{x \log (x)} \, dx+8 \int \frac {e^{e^{x^2}+x^2} x^2 \log ^3\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right )}{-1+e^{e^{x^2}} x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.47, size = 20, normalized size = 0.95 \begin {gather*} \log ^4\left (-\frac {25}{4} \left (-1+e^{e^{x^2}} x\right ) \log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 18, normalized size = 0.86 \begin {gather*} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) + \frac {25}{4} \, \log \relax (x)\right )^{4} \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 {4 \, {\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \relax (x) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) + \frac {25}{4} \, \log \relax (x)\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - x \log \relax (x)}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (8 x^{3} {\mathrm e}^{x^{2}}+4 x \right ) \ln \relax (x )+4 x \right ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-4\right ) \ln \left (-\frac {25 x \ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x^{2}}}}{4}+\frac {25 \ln \relax (x )}{4}\right )^{3}}{x^{2} \ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x^{2}}}-x \ln \relax (x )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, \int \frac {{\left ({\left ({\left (2 \, x^{3} e^{\left (x^{2}\right )} + x\right )} \log \relax (x) + x\right )} e^{\left (e^{\left (x^{2}\right )}\right )} - 1\right )} \log \left (-\frac {25}{4} \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) + \frac {25}{4} \, \log \relax (x)\right )^{3}}{x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - x \log \relax (x)}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.65, size = 19, normalized size = 0.90 \begin {gather*} {\left (\ln \left (\ln \relax (x)-x\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \relax (x)\right )+\ln \left (\frac {25}{4}\right )\right )}^4 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.40, size = 24, normalized size = 1.14 \begin {gather*} \log {\left (- \frac {25 x e^{e^{x^{2}}} \log {\relax (x )}}{4} + \frac {25 \log {\relax (x )}}{4} \right )}^{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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