Integrand size = 22, antiderivative size = 53 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {4}{539 (1-2 x)}+\frac {9}{49 (2+3 x)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (2+3 x)+\frac {125}{121} \log (3+5 x) \]
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Time = 0.02 (sec) , antiderivative size = 53, 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} \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {4}{539 (1-2 x)}+\frac {9}{49 (3 x+2)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (3 x+2)+\frac {125}{121} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {8}{539 (-1+2 x)^2}-\frac {808}{41503 (-1+2 x)}-\frac {27}{49 (2+3 x)^2}-\frac {1053}{343 (2+3 x)}+\frac {625}{121 (3+5 x)}\right ) \, dx \\ & = \frac {4}{539 (1-2 x)}+\frac {9}{49 (2+3 x)}-\frac {404 \log (1-2 x)}{41503}-\frac {351}{343} \log (2+3 x)+\frac {125}{121} \log (3+5 x) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {-\frac {8239}{-2+x+6 x^2}+\frac {14322 x}{-2+x+6 x^2}-404 \log (5-10 x)-42471 \log (5 (2+3 x))+42875 \log (3+5 x)}{41503} \]
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Time = 2.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {125 \ln \left (3+5 x \right )}{121}-\frac {4}{539 \left (-1+2 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}+\frac {9}{49 \left (2+3 x \right )}-\frac {351 \ln \left (2+3 x \right )}{343}\) | \(44\) |
risch | \(\frac {\frac {186 x}{539}-\frac {107}{539}}{\left (-1+2 x \right ) \left (2+3 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}-\frac {351 \ln \left (2+3 x \right )}{343}+\frac {125 \ln \left (3+5 x \right )}{121}\) | \(47\) |
norman | \(\frac {-\frac {1116 x^{2}}{539}+\frac {265}{539}}{\left (-1+2 x \right ) \left (2+3 x \right )}-\frac {404 \ln \left (-1+2 x \right )}{41503}-\frac {351 \ln \left (2+3 x \right )}{343}+\frac {125 \ln \left (3+5 x \right )}{121}\) | \(48\) |
parallelrisch | \(-\frac {254826 \ln \left (\frac {2}{3}+x \right ) x^{2}-257250 \ln \left (x +\frac {3}{5}\right ) x^{2}+2424 \ln \left (x -\frac {1}{2}\right ) x^{2}-20405+42471 \ln \left (\frac {2}{3}+x \right ) x -42875 \ln \left (x +\frac {3}{5}\right ) x +404 \ln \left (x -\frac {1}{2}\right ) x +85932 x^{2}-84942 \ln \left (\frac {2}{3}+x \right )+85750 \ln \left (x +\frac {3}{5}\right )-808 \ln \left (x -\frac {1}{2}\right )}{41503 \left (-1+2 x \right ) \left (2+3 x \right )}\) | \(90\) |
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Time = 0.23 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {42875 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (5 \, x + 3\right ) - 42471 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (3 \, x + 2\right ) - 404 \, {\left (6 \, x^{2} + x - 2\right )} \log \left (2 \, x - 1\right ) + 14322 \, x - 8239}{41503 \, {\left (6 \, x^{2} + x - 2\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {186 x - 107}{3234 x^{2} + 539 x - 1078} - \frac {404 \log {\left (x - \frac {1}{2} \right )}}{41503} + \frac {125 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {351 \log {\left (x + \frac {2}{3} \right )}}{343} \]
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Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {186 \, x - 107}{539 \, {\left (6 \, x^{2} + x - 2\right )}} + \frac {125}{121} \, \log \left (5 \, x + 3\right ) - \frac {351}{343} \, \log \left (3 \, x + 2\right ) - \frac {404}{41503} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {9}{49 \, {\left (3 \, x + 2\right )}} + \frac {24}{3773 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}} + \frac {125}{121} \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) - \frac {404}{41503} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(1-2 x)^2 (2+3 x)^2 (3+5 x)} \, dx=\frac {125\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {351\,\ln \left (x+\frac {2}{3}\right )}{343}-\frac {404\,\ln \left (x-\frac {1}{2}\right )}{41503}+\frac {\frac {31\,x}{539}-\frac {107}{3234}}{x^2+\frac {x}{6}-\frac {1}{3}} \]
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