Optimal. Leaf size=29 \[ \frac {10}{-x+2 \left (2+2 (x-5 \log (4))-\log \left (x^2+\log (x)\right )\right )} \]
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Rubi [A] time = 0.36, antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, integrand size = 159, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 6688, 12, 6686} \begin {gather*} \frac {10}{-2 \log \left (x^2+\log (x)\right )+3 x+4 (1-5 \log (4))} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 6686
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20+40 x^2-30 x^3-30 x \log (x)}{24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+x^3 \left (16+400 \log ^2(4)\right )+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx\\ &=\int \frac {10 \left (2+4 x^2-3 x^3-3 x \log (x)\right )}{x \left (x^2+\log (x)\right ) \left (3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )\right )^2} \, dx\\ &=10 \int \frac {2+4 x^2-3 x^3-3 x \log (x)}{x \left (x^2+\log (x)\right ) \left (3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )\right )^2} \, dx\\ &=\frac {10}{3 x+4 (1-5 \log (4))-2 \log \left (x^2+\log (x)\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{4+3 x-20 \log (4)-2 \log \left (x^2+\log (x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 23, normalized size = 0.79
method | result | size |
risch | \(-\frac {10}{2 \ln \left (\ln \relax (x )+x^{2}\right )+40 \ln \relax (2)-3 x -4}\) | \(23\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 22, normalized size = 0.76 \begin {gather*} \frac {10}{3 \, x - 40 \, \log \relax (2) - 2 \, \log \left (x^{2} + \log \relax (x)\right ) + 4} \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 = -0.03 \begin {gather*} -\int \frac {30\,x\,\ln \relax (x)-40\,x^2+30\,x^3-20}{1600\,x^3\,{\ln \relax (2)}^2-\ln \left (\ln \relax (x)+x^2\right )\,\left (16\,x^3-160\,x^3\,\ln \relax (2)+12\,x^4+\ln \relax (x)\,\left (16\,x-160\,x\,\ln \relax (2)+12\,x^2\right )\right )+{\ln \left (\ln \relax (x)+x^2\right )}^2\,\left (4\,x\,\ln \relax (x)+4\,x^3\right )-2\,\ln \relax (2)\,\left (120\,x^4+160\,x^3\right )+\ln \relax (x)\,\left (16\,x-2\,\ln \relax (2)\,\left (120\,x^2+160\,x\right )+1600\,x\,{\ln \relax (2)}^2+24\,x^2+9\,x^3\right )+16\,x^3+24\,x^4+9\,x^5} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.36, size = 22, normalized size = 0.76 \begin {gather*} - \frac {5}{- \frac {3 x}{2} + \log {\left (x^{2} + \log {\relax (x )} \right )} - 2 + 20 \log {\relax (2 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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