Integrand size = 22, antiderivative size = 42 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 (3+5 x)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (2+3 x)-\frac {155}{121} \log (3+5 x) \]
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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} \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 (5 x+3)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (3 x+2)-\frac {155}{121} \log (5 x+3) \]
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Rule 84
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{847 (-1+2 x)}+\frac {27}{7 (2+3 x)}+\frac {25}{11 (3+5 x)^2}-\frac {775}{121 (3+5 x)}\right ) \, dx \\ & = -\frac {5}{11 (3+5 x)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (2+3 x)-\frac {155}{121} \log (3+5 x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=\frac {1}{847} \left (-\frac {385}{3+5 x}-4 \log (1-2 x)+1089 \log (4+6 x)-1085 \log (6+10 x)\right ) \]
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Time = 2.58 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {1}{11 \left (x +\frac {3}{5}\right )}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}-\frac {155 \ln \left (3+5 x \right )}{121}\) | \(33\) |
default | \(-\frac {5}{11 \left (3+5 x \right )}-\frac {155 \ln \left (3+5 x \right )}{121}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}\) | \(35\) |
norman | \(\frac {25 x}{33 \left (3+5 x \right )}-\frac {4 \ln \left (-1+2 x \right )}{847}+\frac {9 \ln \left (2+3 x \right )}{7}-\frac {155 \ln \left (3+5 x \right )}{121}\) | \(36\) |
parallelrisch | \(\frac {16335 \ln \left (\frac {2}{3}+x \right ) x -16275 \ln \left (x +\frac {3}{5}\right ) x -60 \ln \left (x -\frac {1}{2}\right ) x +9801 \ln \left (\frac {2}{3}+x \right )-9765 \ln \left (x +\frac {3}{5}\right )-36 \ln \left (x -\frac {1}{2}\right )+1925 x}{7623+12705 x}\) | \(53\) |
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Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {1085 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1089 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 4 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) + 385}{847 \, {\left (5 \, x + 3\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=- \frac {4 \log {\left (x - \frac {1}{2} \right )}}{847} - \frac {155 \log {\left (x + \frac {3}{5} \right )}}{121} + \frac {9 \log {\left (x + \frac {2}{3} \right )}}{7} - \frac {5}{55 x + 33} \]
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Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 \, {\left (5 \, x + 3\right )}} - \frac {155}{121} \, \log \left (5 \, x + 3\right ) + \frac {9}{7} \, \log \left (3 \, x + 2\right ) - \frac {4}{847} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=-\frac {5}{11 \, {\left (5 \, x + 3\right )}} + \frac {9}{7} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {4}{847} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \]
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Time = 1.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx=\frac {9\,\ln \left (x+\frac {2}{3}\right )}{7}-\frac {4\,\ln \left (x-\frac {1}{2}\right )}{847}-\frac {155\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {1}{11\,\left (x+\frac {3}{5}\right )} \]
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