Optimal. Leaf size=27 \[ -x-x^2-\frac {\log \left (\log ^2(x)\right )}{3 \log \left (-4+e^x+x\right )} \]
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Rubi [F] time = 1.74, 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 (8-2 e^x-2 x\right ) \log \left (-4+e^x+x\right )+\left (12 x+21 x^2-6 x^3+e^x \left (-3 x-6 x^2\right )\right ) \log (x) \log ^2\left (-4+e^x+x\right )+\left (x+e^x x\right ) \log (x) \log \left (\log ^2(x)\right )}{\left (-12 x+3 e^x x+3 x^2\right ) \log (x) \log ^2\left (-4+e^x+x\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{3} \left (-3-6 x-\frac {2}{x \log (x) \log \left (-4+e^x+x\right )}+\frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx\\ &=\frac {1}{3} \int \left (-3-6 x-\frac {2}{x \log (x) \log \left (-4+e^x+x\right )}+\frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\left (1+e^x\right ) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \left (\frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )}-\frac {(-5+x) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \frac {(-5+x) \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \left (-\frac {5 \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}+\frac {x \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx\\ &=-x-x^2+\frac {1}{3} \int \frac {\log \left (\log ^2(x)\right )}{\log ^2\left (-4+e^x+x\right )} \, dx-\frac {1}{3} \int \frac {x \log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) \log \left (-4+e^x+x\right )} \, dx+\frac {5}{3} \int \frac {\log \left (\log ^2(x)\right )}{\left (-4+e^x+x\right ) \log ^2\left (-4+e^x+x\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 29, normalized size = 1.07 \begin {gather*} \frac {1}{3} \left (-3 x-3 x^2-\frac {\log \left (\log ^2(x)\right )}{\log \left (-4+e^x+x\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 29, normalized size = 1.07 \begin {gather*} -\frac {3 \, {\left (x^{2} + x\right )} \log \left (x + e^{x} - 4\right ) + \log \left (\log \relax (x)^{2}\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.43, size = 36, normalized size = 1.33 \begin {gather*} -\frac {3 \, x^{2} \log \left (x + e^{x} - 4\right ) + 3 \, x \log \left (x + e^{x} - 4\right ) + \log \left (\log \relax (x)^{2}\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.17, size = 80, normalized size = 2.96
method | result | size |
risch | \(-x^{2}-x -\frac {-i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}+4 \ln \left (\ln \relax (x )\right )}{6 \ln \left ({\mathrm e}^{x}+x -4\right )}\) | \(80\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 29, normalized size = 1.07 \begin {gather*} -\frac {3 \, {\left (x^{2} + x\right )} \log \left (x + e^{x} - 4\right ) + 2 \, \log \left (\log \relax (x)\right )}{3 \, \log \left (x + e^{x} - 4\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.86, size = 171, normalized size = 6.33 \begin {gather*} \frac {2}{3\,x}-\frac {\frac {\ln \left ({\ln \relax (x)}^2\right )}{3}-\frac {2\,\ln \left (x+{\mathrm {e}}^x-4\right )\,\left (x+{\mathrm {e}}^x-4\right )}{3\,x\,\ln \relax (x)\,\left ({\mathrm {e}}^x+1\right )}}{\ln \left (x+{\mathrm {e}}^x-4\right )}-\frac {\frac {2\,\left (x+{\mathrm {e}}^x-4\right )}{3\,x\,\left ({\mathrm {e}}^x+1\right )}-\frac {2\,\ln \relax (x)\,\left (3\,{\mathrm {e}}^x-{\mathrm {e}}^{2\,x}-x^2\,{\mathrm {e}}^x+5\,x\,{\mathrm {e}}^x+4\right )}{3\,x\,{\left ({\mathrm {e}}^x+1\right )}^2}}{\ln \relax (x)}-x-x^2+\frac {2\,\left (5\,x^2-x^3\right )}{3\,x^2\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )}-\frac {2\,\left (-x^3+5\,x^2+5\,x\right )}{3\,x^2\,\left ({\mathrm {e}}^x+1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 2.06, size = 22, normalized size = 0.81 \begin {gather*} - x^{2} - x - \frac {\log {\left (\log {\relax (x )}^{2} \right )}}{3 \log {\left (x + e^{x} - 4 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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