Optimal. Leaf size=28 \[ x^2-\left (\frac {1+x}{3}+\log (5)\right )^2+\frac {x^2}{(x+\log (x))^2} \]
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Rubi [F] time = 0.42, 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 {-18 x-2 x^3+16 x^4-6 x^3 \log (5)+\left (18 x-6 x^2+48 x^3-18 x^2 \log (5)\right ) \log (x)+\left (-6 x+48 x^2-18 x \log (5)\right ) \log ^2(x)+(-2+16 x-6 \log (5)) \log ^3(x)}{9 x^3+27 x^2 \log (x)+27 x \log ^2(x)+9 \log ^3(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-18 x+16 x^4+x^3 (-2-6 \log (5))+\left (18 x-6 x^2+48 x^3-18 x^2 \log (5)\right ) \log (x)+\left (-6 x+48 x^2-18 x \log (5)\right ) \log ^2(x)+(-2+16 x-6 \log (5)) \log ^3(x)}{9 x^3+27 x^2 \log (x)+27 x \log ^2(x)+9 \log ^3(x)} \, dx\\ &=\int \frac {2 \left (x \left (-9+8 x^3-x^2 (1+\log (125))\right )+3 x \left (3+8 x^2-x (1+\log (125))\right ) \log (x)+3 x (-1+8 x-3 \log (5)) \log ^2(x)+(-1+8 x-3 \log (5)) \log ^3(x)\right )}{9 (x+\log (x))^3} \, dx\\ &=\frac {2}{9} \int \frac {x \left (-9+8 x^3-x^2 (1+\log (125))\right )+3 x \left (3+8 x^2-x (1+\log (125))\right ) \log (x)+3 x (-1+8 x-3 \log (5)) \log ^2(x)+(-1+8 x-3 \log (5)) \log ^3(x)}{(x+\log (x))^3} \, dx\\ &=\frac {2}{9} \int \left (-1+8 x-\log (125)-\frac {9 x (1+x)}{(x+\log (x))^3}+\frac {9 x}{(x+\log (x))^2}\right ) \, dx\\ &=\frac {8 x^2}{9}-\frac {2}{9} x (1+\log (125))-2 \int \frac {x (1+x)}{(x+\log (x))^3} \, dx+2 \int \frac {x}{(x+\log (x))^2} \, dx\\ &=\frac {8 x^2}{9}-\frac {2}{9} x (1+\log (125))+2 \int \frac {x}{(x+\log (x))^2} \, dx-2 \int \left (\frac {x}{(x+\log (x))^3}+\frac {x^2}{(x+\log (x))^3}\right ) \, dx\\ &=\frac {8 x^2}{9}-\frac {2}{9} x (1+\log (125))-2 \int \frac {x}{(x+\log (x))^3} \, dx-2 \int \frac {x^2}{(x+\log (x))^3} \, dx+2 \int \frac {x}{(x+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 24, normalized size = 0.86 \begin {gather*} \frac {1}{9} x \left (-2 (1+\log (125))+x \left (8+\frac {9}{(x+\log (x))^2}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.43, size = 82, normalized size = 2.93 \begin {gather*} \frac {8 \, x^{4} - 6 \, x^{3} \log \relax (5) - 2 \, x^{3} + 2 \, {\left (4 \, x^{2} - 3 \, x \log \relax (5) - x\right )} \log \relax (x)^{2} + 9 \, x^{2} + 4 \, {\left (4 \, x^{3} - 3 \, x^{2} \log \relax (5) - x^{2}\right )} \log \relax (x)}{9 \, {\left (x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.15, size = 54, normalized size = 1.93 \begin {gather*} \frac {8}{9} \, x^{2} - \frac {2}{9} \, x {\left (3 \, \log \relax (5) + 1\right )} + \frac {x^{3} + x^{2}}{x^{3} + 2 \, x^{2} \log \relax (x) + x \log \relax (x)^{2} + x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 25, normalized size = 0.89
method | result | size |
risch | \(-\frac {2 x \ln \relax (5)}{3}+\frac {8 x^{2}}{9}-\frac {2 x}{9}+\frac {x^{2}}{\left (x +\ln \relax (x )\right )^{2}}\) | \(25\) |
default | \(\frac {-9 \ln \relax (x )^{2}-18 x \ln \relax (x )-2 x^{3}+8 x^{4}-2 x \ln \relax (x )^{2}+16 x^{3} \ln \relax (x )-4 x^{2} \ln \relax (x )+8 x^{2} \ln \relax (x )^{2}}{9 \left (x +\ln \relax (x )\right )^{2}}-\frac {2 x \ln \relax (5)}{3}\) | \(67\) |
norman | \(\frac {-\ln \relax (x )^{2}+\left (-\frac {2}{9}-\frac {2 \ln \relax (5)}{3}\right ) x^{3}-2 x \ln \relax (x )+\left (-\frac {4}{9}-\frac {4 \ln \relax (5)}{3}\right ) x^{2} \ln \relax (x )+\left (-\frac {2}{9}-\frac {2 \ln \relax (5)}{3}\right ) x \ln \relax (x )^{2}+\frac {8 x^{4}}{9}+\frac {8 x^{2} \ln \relax (x )^{2}}{9}+\frac {16 x^{3} \ln \relax (x )}{9}}{\left (x +\ln \relax (x )\right )^{2}}\) | \(75\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 81, normalized size = 2.89 \begin {gather*} \frac {8 \, x^{4} - 2 \, x^{3} {\left (3 \, \log \relax (5) + 1\right )} + 2 \, {\left (4 \, x^{2} - x {\left (3 \, \log \relax (5) + 1\right )}\right )} \log \relax (x)^{2} + 9 \, x^{2} + 4 \, {\left (4 \, x^{3} - x^{2} {\left (3 \, \log \relax (5) + 1\right )}\right )} \log \relax (x)}{9 \, {\left (x^{2} + 2 \, x \log \relax (x) + \log \relax (x)^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.67, size = 89, normalized size = 3.18 \begin {gather*} \frac {8\,x^5+16\,x^4\,\ln \relax (x)+\left (-6\,\ln \relax (5)-2\right )\,x^4+8\,x^3\,{\ln \relax (x)}^2+\left (-12\,\ln \relax (5)-4\right )\,x^3\,\ln \relax (x)+9\,x^3+\left (-6\,\ln \relax (5)-2\right )\,x^2\,{\ln \relax (x)}^2}{9\,x^3+18\,x^2\,\ln \relax (x)+9\,x\,{\ln \relax (x)}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 37, normalized size = 1.32 \begin {gather*} \frac {8 x^{2}}{9} + \frac {x^{2}}{x^{2} + 2 x \log {\relax (x )} + \log {\relax (x )}^{2}} + x \left (- \frac {2 \log {\relax (5 )}}{3} - \frac {2}{9}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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