Optimal. Leaf size=34 \[ 4+5 \log \left (\frac {x^2 \left (4-\left (-e^{e^x}-\frac {x}{2}\right ) x\right )}{4 \log (x)}\right ) \]
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Rubi [F] time = 2.35, 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 {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{2 e^{e^x} x^2 \log (x)+\left (8 x+x^3\right ) \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 {-40-5 x^2+\left (80+20 x^2\right ) \log (x)+e^{e^x} \left (-10 x+\left (30 x+10 e^x x^2\right ) \log (x)\right )}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)} \, dx\\ &=\int \left (\frac {10 e^{e^x+x} x}{8+2 e^{e^x} x+x^2}+\frac {5 \left (-8-2 e^{e^x} x-x^2+16 \log (x)+6 e^{e^x} x \log (x)+4 x^2 \log (x)\right )}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)}\right ) \, dx\\ &=5 \int \frac {-8-2 e^{e^x} x-x^2+16 \log (x)+6 e^{e^x} x \log (x)+4 x^2 \log (x)}{x \left (8+2 e^{e^x} x+x^2\right ) \log (x)} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \left (\frac {-8+x^2}{x \left (8+2 e^{e^x} x+x^2\right )}+\frac {-1+3 \log (x)}{x \log (x)}\right ) \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \frac {-8+x^2}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx+5 \int \frac {-1+3 \log (x)}{x \log (x)} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \left (-\frac {8}{x \left (8+2 e^{e^x} x+x^2\right )}+\frac {x}{8+2 e^{e^x} x+x^2}\right ) \, dx+5 \operatorname {Subst}\left (\int \frac {-1+3 x}{x} \, dx,x,\log (x)\right )+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx\\ &=5 \int \frac {x}{8+2 e^{e^x} x+x^2} \, dx+5 \operatorname {Subst}\left (\int \left (3-\frac {1}{x}\right ) \, dx,x,\log (x)\right )+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx-40 \int \frac {1}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx\\ &=15 \log (x)-5 \log (\log (x))+5 \int \frac {x}{8+2 e^{e^x} x+x^2} \, dx+10 \int \frac {e^{e^x+x} x}{8+2 e^{e^x} x+x^2} \, dx-40 \int \frac {1}{x \left (8+2 e^{e^x} x+x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.39, size = 26, normalized size = 0.76 \begin {gather*} 5 \left (2 \log (x)+\log \left (8+2 e^{e^x} x+x^2\right )-\log (\log (x))\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 28, normalized size = 0.82 \begin {gather*} 15 \, \log \relax (x) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{x}\right ) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 35, normalized size = 1.03 \begin {gather*} -5 \, x + 5 \, \log \left (x^{2} e^{x} + 2 \, x e^{\left (x + e^{x}\right )} + 8 \, e^{x}\right ) + 10 \, \log \relax (x) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 28, normalized size = 0.82
method | result | size |
risch | \(15 \ln \relax (x )-5 \ln \left (\ln \relax (x )\right )+5 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {x^{2}+8}{2 x}\right )\) | \(28\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.68, size = 29, normalized size = 0.85 \begin {gather*} 15 \, \log \relax (x) + 5 \, \log \left (\frac {x^{2} + 2 \, x e^{\left (e^{x}\right )} + 8}{2 \, x}\right ) - 5 \, \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.09, size = 28, normalized size = 0.82 \begin {gather*} 15\,\ln \relax (x)-5\,\ln \left (\ln \relax (x)\right )+5\,\ln \left (\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+x^2+8}{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.42, size = 27, normalized size = 0.79 \begin {gather*} 15 \log {\relax (x )} + 5 \log {\left (e^{e^{x}} + \frac {x^{2} + 8}{2 x} \right )} - 5 \log {\left (\log {\relax (x )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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