Integrand size = 18, antiderivative size = 27 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=-\frac {6 x}{25}-\frac {11}{125 (3+5 x)}+\frac {31}{125} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=-\frac {6 x}{25}-\frac {11}{125 (5 x+3)}+\frac {31}{125} \log (5 x+3) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {6}{25}+\frac {11}{25 (3+5 x)^2}+\frac {31}{25 (3+5 x)}\right ) \, dx \\ & = -\frac {6 x}{25}-\frac {11}{125 (3+5 x)}+\frac {31}{125} \log (3+5 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=\frac {1}{125} \left (-18-30 x-\frac {11}{3+5 x}+31 \log (3+5 x)\right ) \]
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Time = 1.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
risch | \(-\frac {6 x}{25}-\frac {11}{625 \left (x +\frac {3}{5}\right )}+\frac {31 \ln \left (3+5 x \right )}{125}\) | \(20\) |
default | \(-\frac {6 x}{25}-\frac {11}{125 \left (3+5 x \right )}+\frac {31 \ln \left (3+5 x \right )}{125}\) | \(22\) |
norman | \(\frac {-\frac {43}{75} x -\frac {6}{5} x^{2}}{3+5 x}+\frac {31 \ln \left (3+5 x \right )}{125}\) | \(27\) |
parallelrisch | \(\frac {465 \ln \left (x +\frac {3}{5}\right ) x -450 x^{2}+279 \ln \left (x +\frac {3}{5}\right )-215 x}{1125+1875 x}\) | \(32\) |
meijerg | \(\frac {13 x}{45 \left (1+\frac {5 x}{3}\right )}+\frac {31 \ln \left (1+\frac {5 x}{3}\right )}{125}-\frac {2 x \left (5 x +6\right )}{25 \left (1+\frac {5 x}{3}\right )}\) | \(35\) |
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Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=-\frac {150 \, x^{2} - 31 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 90 \, x + 11}{125 \, {\left (5 \, x + 3\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=- \frac {6 x}{25} + \frac {31 \log {\left (5 x + 3 \right )}}{125} - \frac {11}{625 x + 375} \]
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none
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=-\frac {6}{25} \, x - \frac {11}{125 \, {\left (5 \, x + 3\right )}} + \frac {31}{125} \, \log \left (5 \, x + 3\right ) \]
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Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=-\frac {6}{25} \, x - \frac {11}{125 \, {\left (5 \, x + 3\right )}} - \frac {31}{125} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) - \frac {18}{125} \]
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Time = 1.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx=\frac {31\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {6\,x}{25}-\frac {11}{625\,\left (x+\frac {3}{5}\right )} \]
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