Integrand size = 17, antiderivative size = 34 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=5-4 (2-x)+2 \left (\log (3)-x \left (-x+\frac {x^2}{x-x \log (3)}\right )\right ) \]
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Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12} \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {4 x}{1-\log (3)}-\frac {2 (x+1)^2 \log (3)}{1-\log (3)} \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \frac {\int (-4+(4+4 x) \log (3)) \, dx}{-1+\log (3)} \\ & = \frac {4 x}{1-\log (3)}-\frac {2 (1+x)^2 \log (3)}{1-\log (3)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {4 \left (-x+x \log (3)+\frac {1}{2} x^2 \log (3)\right )}{-1+\log (3)} \]
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Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53
method | result | size |
norman | \(4 x +\frac {2 \ln \left (3\right ) x^{2}}{\ln \left (3\right )-1}\) | \(18\) |
gosper | \(\frac {2 x \left (x \ln \left (3\right )+2 \ln \left (3\right )-2\right )}{\ln \left (3\right )-1}\) | \(20\) |
default | \(\frac {2 x^{2} \ln \left (3\right )+4 x \ln \left (3\right )-4 x}{\ln \left (3\right )-1}\) | \(24\) |
parallelrisch | \(\frac {\ln \left (3\right ) \left (2 x^{2}+4 x \right )-4 x}{\ln \left (3\right )-1}\) | \(24\) |
risch | \(\frac {2 \ln \left (3\right ) x^{2}}{\ln \left (3\right )-1}+\frac {4 x \ln \left (3\right )}{\ln \left (3\right )-1}-\frac {4 x}{\ln \left (3\right )-1}\) | \(35\) |
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none
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]
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Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 x^{2} \log {\left (3 \right )}}{-1 + \log {\left (3 \right )}} + 4 x \]
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none
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]
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none
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {2 \, {\left ({\left (x^{2} + 2 \, x\right )} \log \left (3\right ) - 2 \, x\right )}}{\log \left (3\right ) - 1} \]
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Time = 0.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-4+(4+4 x) \log (3)}{-1+\log (3)} \, dx=\frac {{\left (\ln \left (3\right )\,\left (4\,x+4\right )-4\right )}^2}{4\,\ln \left (3\right )\,\left (\ln \left (9\right )-2\right )} \]
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