Integrand size = 22, antiderivative size = 32 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {121}{25 (3+5 x)}+\frac {49}{3} \log (2+3 x)-\frac {407}{25} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 32, 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} \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {121}{25 (5 x+3)}+\frac {49}{3} \log (3 x+2)-\frac {407}{25} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {49}{2+3 x}+\frac {121}{5 (3+5 x)^2}-\frac {407}{5 (3+5 x)}\right ) \, dx \\ & = -\frac {121}{25 (3+5 x)}+\frac {49}{3} \log (2+3 x)-\frac {407}{25} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {121}{75+125 x}+\frac {49}{3} \log (2+3 x)-\frac {407}{25} \log (-3 (3+5 x)) \]
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Time = 3.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {121}{125 \left (x +\frac {3}{5}\right )}+\frac {49 \ln \left (2+3 x \right )}{3}-\frac {407 \ln \left (3+5 x \right )}{25}\) | \(25\) |
default | \(-\frac {121}{25 \left (3+5 x \right )}+\frac {49 \ln \left (2+3 x \right )}{3}-\frac {407 \ln \left (3+5 x \right )}{25}\) | \(27\) |
norman | \(\frac {121 x}{15 \left (3+5 x \right )}+\frac {49 \ln \left (2+3 x \right )}{3}-\frac {407 \ln \left (3+5 x \right )}{25}\) | \(28\) |
parallelrisch | \(\frac {6125 \ln \left (\frac {2}{3}+x \right ) x -6105 \ln \left (x +\frac {3}{5}\right ) x +3675 \ln \left (\frac {2}{3}+x \right )-3663 \ln \left (x +\frac {3}{5}\right )+605 x}{225+375 x}\) | \(40\) |
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {1221 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1225 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 363}{75 \, {\left (5 \, x + 3\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=- \frac {407 \log {\left (x + \frac {3}{5} \right )}}{25} + \frac {49 \log {\left (x + \frac {2}{3} \right )}}{3} - \frac {121}{125 x + 75} \]
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Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {121}{25 \, {\left (5 \, x + 3\right )}} - \frac {407}{25} \, \log \left (5 \, x + 3\right ) + \frac {49}{3} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=-\frac {121}{25 \, {\left (5 \, x + 3\right )}} - \frac {4}{75} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {49}{3} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]
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Time = 1.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {(1-2 x)^2}{(2+3 x) (3+5 x)^2} \, dx=\frac {49\,\ln \left (x+\frac {2}{3}\right )}{3}-\frac {407\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {121}{125\,\left (x+\frac {3}{5}\right )} \]
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