Optimal. Leaf size=22 \[ x \left (5+x+\log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )\right ) \]
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Rubi [A] time = 1.25, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6742, 6688, 2549} \begin {gather*} x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )+5 x \end {gather*}
Antiderivative was successfully verified.
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Rule 2549
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )}+\frac {1+x \log (x)+5 \log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )+2 x \log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )+\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right ) \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \frac {1+x \log (x)+5 \log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )+2 x \log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )+\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right ) \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx\\ &=2 \int \frac {e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \left (5+2 x+\frac {x+\frac {1}{\log (x)}}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )}+\log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )\right ) \, dx\\ &=5 x+x^2+2 \int \frac {e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \frac {x+\frac {1}{\log (x)}}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right ) \, dx\\ &=5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )+2 \int \frac {e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \left (\frac {x}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )}+\frac {1}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )}\right ) \, dx-\int \frac {1+x \left (1+2 e^{x^2} x\right ) \log (x)}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx\\ &=5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )+2 \int \frac {e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx-\int \left (\frac {2 e^{x^2} x^2}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )}+\frac {1+x \log (x)}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )}\right ) \, dx+\int \frac {x}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \frac {1}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx\\ &=5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )+\int \frac {x}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \frac {1}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx-\int \frac {1+x \log (x)}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx\\ &=5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )-\int \left (\frac {x}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )}+\frac {1}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )}\right ) \, dx+\int \frac {x}{\log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx+\int \frac {1}{\log (x) \log \left (4 e^{e^{x^2}+x} \log (x)\right )} \, dx\\ &=5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 26, normalized size = 1.18 \begin {gather*} 5 x+x^2+x \log \left (4 \log \left (4 e^{e^{x^2}+x} \log (x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.20, size = 24, normalized size = 1.09 \begin {gather*} x^{2} + x \log \left (4 \, \log \left (4 \, e^{\left (x + e^{\left (x^{2}\right )}\right )} \log \relax (x)\right )\right ) + 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 28, normalized size = 1.27 \begin {gather*} x^{2} + 2 \, x \log \relax (2) + x \log \left (x + e^{\left (x^{2}\right )} + 2 \, \log \relax (2) + \log \left (\log \relax (x)\right )\right ) + 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.22, size = 97, normalized size = 4.41
method | result | size |
risch | \(x^{2}+x \ln \left (8 \ln \relax (2)+4 \ln \left (\ln \relax (x )\right )+4 \ln \left ({\mathrm e}^{{\mathrm e}^{x^{2}}+x}\right )-2 i \pi \,\mathrm {csgn}\left (i \ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x^{2}}+x}\right ) \left (-\mathrm {csgn}\left (i \ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x^{2}}+x}\right )+\mathrm {csgn}\left (i \ln \relax (x )\right )\right ) \left (-\mathrm {csgn}\left (i \ln \relax (x ) {\mathrm e}^{{\mathrm e}^{x^{2}}+x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{{\mathrm e}^{x^{2}}+x}\right )\right )\right )+5 x\) | \(97\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.87, size = 28, normalized size = 1.27 \begin {gather*} x^{2} + 2 \, x \log \relax (2) + x \log \left (x + e^{\left (x^{2}\right )} + 2 \, \log \relax (2) + \log \left (\log \relax (x)\right )\right ) + 5 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.88, size = 27, normalized size = 1.23 \begin {gather*} 5\,x+x\,\ln \left (4\,x+4\,{\mathrm {e}}^{x^2}+4\,\ln \left (\ln \relax (x)\right )+\ln \left (256\right )\right )+x^2 \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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