Optimal. Leaf size=23 \[ \frac {1+2 x+\log (3)}{-x-x^2}+\log (\log (x)) \]
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Rubi [A] time = 0.35, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 7, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {1594, 27, 6688, 14, 893, 2302, 29} \begin {gather*} \log (\log (x))-\frac {1+\log (3)}{x}-\frac {1-\log (3)}{x+1} \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 27
Rule 29
Rule 893
Rule 1594
Rule 2302
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x+2 x^2+x^3+\left (1+2 x+2 x^2+(1+2 x) \log (3)\right ) \log (x)}{x^2 \left (1+2 x+x^2\right ) \log (x)} \, dx\\ &=\int \frac {x+2 x^2+x^3+\left (1+2 x+2 x^2+(1+2 x) \log (3)\right ) \log (x)}{x^2 (1+x)^2 \log (x)} \, dx\\ &=\int \frac {\frac {1+2 x^2+\log (3)+x (2+\log (9))}{(1+x)^2}+\frac {x}{\log (x)}}{x^2} \, dx\\ &=\int \left (\frac {1+2 x^2+\log (3)+x (2+\log (9))}{x^2 (1+x)^2}+\frac {1}{x \log (x)}\right ) \, dx\\ &=\int \frac {1+2 x^2+\log (3)+x (2+\log (9))}{x^2 (1+x)^2} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\int \left (\frac {1-\log (3)}{(1+x)^2}+\frac {1+\log (3)}{x^2}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\frac {1-\log (3)}{1+x}-\frac {1+\log (3)}{x}+\log (\log (x))\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 28, normalized size = 1.22 \begin {gather*} \frac {-1-\log (3)}{x}+\frac {-1-\log (3)+\log (9)}{1+x}+\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.67, size = 26, normalized size = 1.13 \begin {gather*} \frac {{\left (x^{2} + x\right )} \log \left (\log \relax (x)\right ) - 2 \, x - \log \relax (3) - 1}{x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.62, size = 23, normalized size = 1.00 \begin {gather*} -\frac {\log \relax (3) + 1}{x} + \frac {\log \relax (3) - 1}{x + 1} + \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 22, normalized size = 0.96
method | result | size |
risch | \(-\frac {1+2 x +\ln \relax (3)}{\left (x +1\right ) x}+\ln \left (\ln \relax (x )\right )\) | \(22\) |
norman | \(\frac {-2 x -1-\ln \relax (3)}{\left (x +1\right ) x}+\ln \left (\ln \relax (x )\right )\) | \(23\) |
default | \(\ln \left (\ln \relax (x )\right )-\frac {1}{x}-\frac {1}{x +1}-\frac {\ln \relax (3)}{x}+\frac {\ln \relax (3)}{x +1}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.72, size = 20, normalized size = 0.87 \begin {gather*} -\frac {2 \, x + \log \relax (3) + 1}{x^{2} + x} + \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.53, size = 27, normalized size = 1.17 \begin {gather*} \ln \left (\ln \relax (x)\right )-\frac {x\,\left (\ln \relax (3)+1\right )-2\,x^3}{x^3+x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 19, normalized size = 0.83 \begin {gather*} \log {\left (\log {\relax (x )} \right )} + \frac {- 2 x - \log {\relax (3 )} - 1}{x^{2} + x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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