Integrand size = 22, antiderivative size = 42 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=\frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (2+3 x)+\frac {25}{121} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {84} \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=\frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (3 x+2)+\frac {25}{121} \log (5 x+3) \]
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Rule 84
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4}{77 (-1+2 x)^2}-\frac {272}{5929 (-1+2 x)}-\frac {27}{49 (2+3 x)}+\frac {125}{121 (3+5 x)}\right ) \, dx \\ & = \frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (2+3 x)+\frac {25}{121} \log (3+5 x) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=\frac {\frac {154}{1-2 x}-136 \log (3-6 x)-1089 \log (2+3 x)+1225 \log (-3 (3+5 x))}{5929} \]
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Time = 2.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {1}{77 \left (x -\frac {1}{2}\right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{121}\) | \(33\) |
default | \(\frac {25 \ln \left (3+5 x \right )}{121}-\frac {2}{77 \left (-1+2 x \right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}\) | \(35\) |
norman | \(-\frac {4 x}{77 \left (-1+2 x \right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{121}\) | \(36\) |
parallelrisch | \(-\frac {2178 \ln \left (\frac {2}{3}+x \right ) x -2450 \ln \left (x +\frac {3}{5}\right ) x +272 \ln \left (x -\frac {1}{2}\right ) x -1089 \ln \left (\frac {2}{3}+x \right )+1225 \ln \left (x +\frac {3}{5}\right )-136 \ln \left (x -\frac {1}{2}\right )+308 x}{5929 \left (-1+2 x \right )}\) | \(53\) |
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Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=\frac {1225 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) - 1089 \, {\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) - 136 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 154}{5929 \, {\left (2 \, x - 1\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=- \frac {136 \log {\left (x - \frac {1}{2} \right )}}{5929} + \frac {25 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{49} - \frac {2}{154 x - 77} \]
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Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=-\frac {2}{77 \, {\left (2 \, x - 1\right )}} + \frac {25}{121} \, \log \left (5 \, x + 3\right ) - \frac {9}{49} \, \log \left (3 \, x + 2\right ) - \frac {136}{5929} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=-\frac {2}{77 \, {\left (2 \, x - 1\right )}} - \frac {9}{49} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) + \frac {25}{121} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \]
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Time = 1.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx=\frac {25\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {9\,\ln \left (x+\frac {2}{3}\right )}{49}-\frac {136\,\ln \left (x-\frac {1}{2}\right )}{5929}-\frac {1}{77\,\left (x-\frac {1}{2}\right )} \]
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