Optimal. Leaf size=17 \[ 4-x+x (-9+\log (3))+\frac {1}{(2+\log (x))^2} \]
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Rubi [A] time = 0.39, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 6688, 6742, 2302, 30} \begin {gather*} \frac {1}{(\log (x)+2)^2}-x (10-\log (3)) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 30
Rule 2302
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+x (-80+8 \log (3))+(-120 x+12 x \log (3)) \log (x)+(-60 x+6 x \log (3)) \log ^2(x)+(-10 x+x \log (3)) \log ^3(x)}{8 x+12 x \log (x)+6 x \log ^2(x)+x \log ^3(x)} \, dx\\ &=\int \frac {-2+8 x (-10+\log (3))+12 x (-10+\log (3)) \log (x)+6 x (-10+\log (3)) \log ^2(x)+x (-10+\log (3)) \log ^3(x)}{x (2+\log (x))^3} \, dx\\ &=\int \left (-10 \left (1-\frac {\log (3)}{10}\right )-\frac {2}{x (2+\log (x))^3}\right ) \, dx\\ &=-x (10-\log (3))-2 \int \frac {1}{x (2+\log (x))^3} \, dx\\ &=-x (10-\log (3))-2 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,2+\log (x)\right )\\ &=-x (10-\log (3))+\frac {1}{(2+\log (x))^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.16, size = 14, normalized size = 0.82 \begin {gather*} -10 x+x \log (3)+\frac {1}{(2+\log (x))^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.60, size = 48, normalized size = 2.82 \begin {gather*} \frac {{\left (x \log \relax (3) - 10 \, x\right )} \log \relax (x)^{2} + 4 \, x \log \relax (3) + 4 \, {\left (x \log \relax (3) - 10 \, x\right )} \log \relax (x) - 40 \, x + 1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 19, normalized size = 1.12 \begin {gather*} x {\left (\log \relax (3) - 10\right )} + \frac {1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 15, normalized size = 0.88
method | result | size |
risch | \(x \ln \relax (3)-10 x +\frac {1}{\left (\ln \relax (x )+2\right )^{2}}\) | \(15\) |
default | \(\frac {1-40 x -40 x \ln \relax (x )-10 x \ln \relax (x )^{2}}{\left (\ln \relax (x )+2\right )^{2}}+x \ln \relax (3)\) | \(30\) |
norman | \(\frac {\left (-40+4 \ln \relax (3)\right ) x +1+\left (-40+4 \ln \relax (3)\right ) x \ln \relax (x )+\left (\ln \relax (3)-10\right ) x \ln \relax (x )^{2}}{\left (\ln \relax (x )+2\right )^{2}}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.60, size = 19, normalized size = 1.12 \begin {gather*} x {\left (\log \relax (3) - 10\right )} + \frac {1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.28, size = 35, normalized size = 2.06 \begin {gather*} x\,\ln \relax (3)-10\,x+\frac {x^2}{x^2\,{\ln \relax (x)}^2+4\,x^2\,\ln \relax (x)+4\,x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.12, size = 19, normalized size = 1.12 \begin {gather*} x \left (-10 + \log {\relax (3 )}\right ) + \frac {1}{\log {\relax (x )}^{2} + 4 \log {\relax (x )} + 4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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