Integrand size = 22, antiderivative size = 37 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=-\frac {8 x}{75}-\frac {1331}{125 (3+5 x)}+\frac {343}{9} \log (2+3 x)-\frac {4719}{125} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 37, 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 = {90} \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=-\frac {8 x}{75}-\frac {1331}{125 (5 x+3)}+\frac {343}{9} \log (3 x+2)-\frac {4719}{125} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{75}+\frac {343}{3 (2+3 x)}+\frac {1331}{25 (3+5 x)^2}-\frac {4719}{25 (3+5 x)}\right ) \, dx \\ & = -\frac {8 x}{75}-\frac {1331}{125 (3+5 x)}+\frac {343}{9} \log (2+3 x)-\frac {4719}{125} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=\frac {-80-120 x-\frac {11979}{3+5 x}+42875 \log (2+3 x)-42471 \log (-3 (3+5 x))}{1125} \]
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Time = 2.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-\frac {8 x}{75}-\frac {1331}{625 \left (x +\frac {3}{5}\right )}+\frac {343 \ln \left (2+3 x \right )}{9}-\frac {4719 \ln \left (3+5 x \right )}{125}\) | \(28\) |
default | \(-\frac {8 x}{75}-\frac {1331}{125 \left (3+5 x \right )}+\frac {343 \ln \left (2+3 x \right )}{9}-\frac {4719 \ln \left (3+5 x \right )}{125}\) | \(30\) |
norman | \(\frac {\frac {1307}{75} x -\frac {8}{15} x^{2}}{3+5 x}+\frac {343 \ln \left (2+3 x \right )}{9}-\frac {4719 \ln \left (3+5 x \right )}{125}\) | \(35\) |
parallelrisch | \(\frac {214375 \ln \left (\frac {2}{3}+x \right ) x -212355 \ln \left (x +\frac {3}{5}\right ) x -600 x^{2}+128625 \ln \left (\frac {2}{3}+x \right )-127413 \ln \left (x +\frac {3}{5}\right )+19605 x}{3375+5625 x}\) | \(45\) |
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Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=-\frac {600 \, x^{2} + 42471 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 42875 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 360 \, x + 11979}{1125 \, {\left (5 \, x + 3\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=- \frac {8 x}{75} - \frac {4719 \log {\left (x + \frac {3}{5} \right )}}{125} + \frac {343 \log {\left (x + \frac {2}{3} \right )}}{9} - \frac {1331}{625 x + 375} \]
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Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=-\frac {8}{75} \, x - \frac {1331}{125 \, {\left (5 \, x + 3\right )}} - \frac {4719}{125} \, \log \left (5 \, x + 3\right ) + \frac {343}{9} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=-\frac {8}{75} \, x - \frac {1331}{125 \, {\left (5 \, x + 3\right )}} - \frac {404}{1125} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {343}{9} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {8}{125} \]
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Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^3}{(2+3 x) (3+5 x)^2} \, dx=\frac {343\,\ln \left (x+\frac {2}{3}\right )}{9}-\frac {8\,x}{75}-\frac {4719\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {1331}{625\,\left (x+\frac {3}{5}\right )} \]
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