Optimal. Leaf size=25 \[ -4 x \left (1+\frac {x}{1+x}\right ) \left (4+\frac {2}{\log (2)}\right )+\log \left (x^2\right ) \]
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Rubi [A] time = 0.09, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 5, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 1594, 27, 1620} \begin {gather*} -\frac {16 x (1+\log (4))}{\log (2)}+\frac {\log (4) \log (x)}{\log (2)}-\frac {8 (1+\log (4))}{(x+1) \log (2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 1594
Rule 1620
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x+2 x^2+x^3} \, dx}{\log (2)}\\ &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x \left (1+2 x+x^2\right )} \, dx}{\log (2)}\\ &=\frac {\int \frac {-8 x-32 x^2-16 x^3+\left (2-12 x-62 x^2-32 x^3\right ) \log (2)}{x (1+x)^2} \, dx}{\log (2)}\\ &=\frac {\int \left (\frac {\log (4)}{x}-16 (1+\log (4))+\frac {8 (1+\log (4))}{(1+x)^2}\right ) \, dx}{\log (2)}\\ &=-\frac {16 x (1+\log (4))}{\log (2)}-\frac {8 (1+\log (4))}{(1+x) \log (2)}+\frac {\log (4) \log (x)}{\log (2)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 30, normalized size = 1.20 \begin {gather*} -\frac {2 \left (8 x (1+\log (4))+\frac {4+\log (256)}{1+x}-\log (2) \log (x)\right )}{\log (2)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 43, normalized size = 1.72 \begin {gather*} \frac {2 \, {\left ({\left (x + 1\right )} \log \relax (2) \log \relax (x) - 8 \, x^{2} - 8 \, {\left (2 \, x^{2} + 2 \, x + 1\right )} \log \relax (2) - 8 \, x - 4\right )}}{{\left (x + 1\right )} \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 35, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (16 \, x \log \relax (2) - \log \relax (2) \log \left ({\left | x \right |}\right ) + 8 \, x + \frac {4 \, {\left (2 \, \log \relax (2) + 1\right )}}{x + 1}\right )}}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 34, normalized size = 1.36
method | result | size |
default | \(\frac {-32 x \ln \relax (2)-16 x +2 \ln \relax (2) \ln \relax (x )-\frac {2 \left (8 \ln \relax (2)+4\right )}{x +1}}{\ln \relax (2)}\) | \(34\) |
risch | \(-32 x -\frac {16 x}{\ln \relax (2)}-\frac {16}{x +1}-\frac {8}{\ln \relax (2) \left (x +1\right )}+2 \ln \left (-x \right )\) | \(36\) |
norman | \(\frac {-\frac {8 \left (1+2 \ln \relax (2)\right ) x}{\ln \relax (2)}-\frac {16 \left (1+2 \ln \relax (2)\right ) x^{2}}{\ln \relax (2)}}{x +1}+2 \ln \relax (x )\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 35, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (8 \, x {\left (2 \, \log \relax (2) + 1\right )} - \log \relax (2) \log \relax (x) + \frac {4 \, {\left (2 \, \log \relax (2) + 1\right )}}{x + 1}\right )}}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.11, size = 35, normalized size = 1.40 \begin {gather*} 2\,\ln \relax (x)-\frac {16\,\ln \relax (2)+8}{\ln \relax (2)+x\,\ln \relax (2)}-\frac {x\,\left (32\,\ln \relax (2)+16\right )}{\ln \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.37, size = 27, normalized size = 1.08 \begin {gather*} - x \left (\frac {16}{\log {\relax (2 )}} + 32\right ) + 2 \log {\relax (x )} - \frac {8 + 16 \log {\relax (2 )}}{x \log {\relax (2 )} + \log {\relax (2 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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