Optimal. Leaf size=24 \[ x-x^2 \log \left (\frac {x^2}{-5 e^{-4 x}+\log (5)}\right ) \]
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Rubi [A] time = 0.82, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 16, number of rules used = 8, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.110, Rules used = {6742, 2184, 2190, 2531, 2282, 6589, 2551, 12} \begin {gather*} x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 2551
Rule 6589
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-2 x+\frac {20 x^2}{-5+e^{4 x} \log (5)}-2 x \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right )\right ) \, dx\\ &=x-x^2-2 \int x \log \left (\frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)}\right ) \, dx+20 \int \frac {x^2}{-5+e^{4 x} \log (5)} \, dx\\ &=x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+(4 \log (5)) \int \frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)} \, dx+\int \frac {2 x \left (5+10 x-e^{4 x} \log (5)\right )}{5-e^{4 x} \log (5)} \, dx\\ &=x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+2 \int \frac {x \left (5+10 x-e^{4 x} \log (5)\right )}{5-e^{4 x} \log (5)} \, dx-2 \int x \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right ) \, dx\\ &=x-x^2-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{2} \int \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right ) \, dx+2 \int \left (x-\frac {10 x^2}{-5+e^{4 x} \log (5)}\right ) \, dx\\ &=x-\frac {4 x^3}{3}-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {1}{5} x \log (5)\right )}{x} \, dx,x,e^{4 x}\right )-20 \int \frac {x^2}{-5+e^{4 x} \log (5)} \, dx\\ &=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+x^2 \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} x \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )-(4 \log (5)) \int \frac {e^{4 x} x^2}{-5+e^{4 x} \log (5)} \, dx\\ &=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )+\frac {1}{2} x \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right )-\frac {1}{8} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )+2 \int x \log \left (1-\frac {1}{5} e^{4 x} \log (5)\right ) \, dx\\ &=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )-\frac {1}{8} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{2} \int \text {Li}_2\left (\frac {1}{5} e^{4 x} \log (5)\right ) \, dx\\ &=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )-\frac {1}{8} \text {Li}_3\left (\frac {1}{5} e^{4 x} \log (5)\right )+\frac {1}{8} \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {1}{5} x \log (5)\right )}{x} \, dx,x,e^{4 x}\right )\\ &=x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.20, size = 31, normalized size = 1.29 \begin {gather*} x-x^2 \log \left (-\frac {e^{4 x} x^2}{5-e^{4 x} \log (5)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.92, size = 27, normalized size = 1.12 \begin {gather*} -x^{2} \log \left (\frac {x^{2} e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} \log \relax (5) - 5}\right ) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 30, normalized size = 1.25 \begin {gather*} -4 \, x^{3} - x^{2} \log \left (x^{2}\right ) + x^{2} \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\right ) + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.29, size = 28, normalized size = 1.17
method | result | size |
norman | \(x -x^{2} \ln \left (\frac {x^{2} {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )\) | \(28\) |
risch | \(-x^{2} \ln \left ({\mathrm e}^{4 x}\right )+x^{2} \ln \left (\ln \relax (5) {\mathrm e}^{4 x}-5\right )-2 x^{2} \ln \relax (x )+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}-i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right ) \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{4 x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{3}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right ) \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (\frac {i x^{2} {\mathrm e}^{4 x}}{\ln \relax (5) {\mathrm e}^{4 x}-5}\right )^{3}}{2}+x\) | \(422\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.72, size = 45, normalized size = 1.88 \begin {gather*} -4 \, x^{3} - 2 \, x^{2} \log \relax (x) + \frac {1}{4} \, {\left (4 \, x^{2} + 1\right )} \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\right ) + x - \frac {1}{4} \, \log \left (e^{\left (4 \, x\right )} \log \relax (5) - 5\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 {10\,x+\ln \left (\frac {x^2\,{\mathrm {e}}^{4\,x}}{{\mathrm {e}}^{4\,x}\,\ln \relax (5)-5}\right )\,\left (10\,x-2\,x\,{\mathrm {e}}^{4\,x}\,\ln \relax (5)\right )+20\,x^2-{\mathrm {e}}^{4\,x}\,\ln \relax (5)\,\left (2\,x-1\right )-5}{{\mathrm {e}}^{4\,x}\,\ln \relax (5)-5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 24, normalized size = 1.00 \begin {gather*} - x^{2} \log {\left (\frac {x^{2} e^{4 x}}{e^{4 x} \log {\relax (5 )} - 5} \right )} + x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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