Optimal. Leaf size=24 \[ \frac {\log (2)}{x \log \left (-\left (\left (-1+2 e^{3+x}\right ) \log \left (x^2\right )\right )\right )} \]
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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 {2 \log (2)-4 e^{3+x} \log (2)-2 e^{3+x} x \log (2) \log \left (x^2\right )+\left (\log (2)-2 e^{3+x} \log (2)\right ) \log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{\left (-x^2+2 e^{3+x} x^2\right ) \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {\log (2)}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}-\frac {\log (2) \left (2+x \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )\right )}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}\right ) \, dx\\ &=-\left (\log (2) \int \frac {1}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\right )-\log (2) \int \frac {2+x \log \left (x^2\right )+\log \left (x^2\right ) \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\\ &=-\left (\log (2) \int \left (\frac {2+x \log \left (x^2\right )}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}+\frac {1}{x^2 \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}\right ) \, dx\right )-\log (2) \int \frac {1}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\\ &=-\left (\log (2) \int \frac {1}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\right )-\log (2) \int \frac {2+x \log \left (x^2\right )}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx-\log (2) \int \frac {1}{x^2 \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\\ &=-\left (\log (2) \int \left (\frac {1}{x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}+\frac {2}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )}\right ) \, dx\right )-\log (2) \int \frac {1}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx-\log (2) \int \frac {1}{x^2 \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\\ &=-\left (\log (2) \int \frac {1}{x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\right )-\log (2) \int \frac {1}{\left (-1+2 e^{3+x}\right ) x \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx-\log (2) \int \frac {1}{x^2 \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx-(2 \log (2)) \int \frac {1}{x^2 \log \left (x^2\right ) \log ^2\left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 23, normalized size = 0.96 \begin {gather*} \frac {\log (2)}{x \log \left (\left (1-2 e^{3+x}\right ) \log \left (x^2\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 23, normalized size = 0.96 \begin {gather*} \frac {\log \relax (2)}{x \log \left (-{\left (2 \, e^{\left (x + 3\right )} - 1\right )} \log \left (x^{2}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.60, size = 24, normalized size = 1.00 \begin {gather*} \frac {\log \relax (2)}{x \log \left (-2 \, e^{\left (x + 3\right )} \log \left (x^{2}\right ) + \log \left (x^{2}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.26, size = 831, normalized size = 34.62
method | result | size |
risch | \(\frac {2 i \ln \relax (2)}{x \left (\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right ) \mathrm {csgn}\left (\left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )+\pi \mathrm {csgn}\left (\left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{2}+\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{2}+\pi \mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{3}-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right ) \mathrm {csgn}\left (\left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{2}-\pi \mathrm {csgn}\left (\left ({\mathrm e}^{3+x}-\frac {1}{2}\right ) \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )^{3}-\pi +2 i \ln \left ({\mathrm e}^{3+x}-\frac {1}{2}\right )+2 i \ln \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \ln \relax (x )+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}\right )\right )}\) | \(831\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.51, size = 30, normalized size = 1.25 \begin {gather*} \frac {\log \relax (2)}{{\left (i \, \pi + \log \relax (2)\right )} x + x \log \left (2 \, e^{\left (x + 3\right )} - 1\right ) + x \log \left (\log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,\ln \relax (2)-4\,{\mathrm {e}}^{x+3}\,\ln \relax (2)+\ln \left (-\ln \left (x^2\right )\,\left (2\,{\mathrm {e}}^{x+3}-1\right )\right )\,\ln \left (x^2\right )\,\left (\ln \relax (2)-2\,{\mathrm {e}}^{x+3}\,\ln \relax (2)\right )-2\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{x+3}\,\ln \relax (2)}{{\ln \left (-\ln \left (x^2\right )\,\left (2\,{\mathrm {e}}^{x+3}-1\right )\right )}^2\,\ln \left (x^2\right )\,\left (2\,x^2\,{\mathrm {e}}^{x+3}-x^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.56, size = 19, normalized size = 0.79 \begin {gather*} \frac {\log {\relax (2 )}}{x \log {\left (\left (1 - 2 e^{x + 3}\right ) \log {\left (x^{2} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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