Integrand size = 20, antiderivative size = 38 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {18 x}{125}-\frac {11}{1250 (3+5 x)^2}-\frac {64}{625 (3+5 x)}+\frac {87}{625} \log (3+5 x) \]
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Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {18 x}{125}-\frac {64}{625 (5 x+3)}-\frac {11}{1250 (5 x+3)^2}+\frac {87}{625} \log (5 x+3) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {18}{125}+\frac {11}{125 (3+5 x)^3}+\frac {64}{125 (3+5 x)^2}+\frac {87}{125 (3+5 x)}\right ) \, dx \\ & = -\frac {18 x}{125}-\frac {11}{1250 (3+5 x)^2}-\frac {64}{625 (3+5 x)}+\frac {87}{625} \log (3+5 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {295+1172 x+1680 x^2+900 x^3}{250 (3+5 x)^2}+\frac {87}{625} \log (-3 (3+5 x)) \]
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Time = 2.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {18 x}{125}+\frac {-\frac {64 x}{125}-\frac {79}{250}}{\left (3+5 x \right )^{2}}+\frac {87 \ln \left (3+5 x \right )}{625}\) | \(27\) |
default | \(-\frac {18 x}{125}-\frac {11}{1250 \left (3+5 x \right )^{2}}-\frac {64}{625 \left (3+5 x \right )}+\frac {87 \ln \left (3+5 x \right )}{625}\) | \(31\) |
norman | \(\frac {-\frac {283}{375} x -\frac {1549}{450} x^{2}-\frac {18}{5} x^{3}}{\left (3+5 x \right )^{2}}+\frac {87 \ln \left (3+5 x \right )}{625}\) | \(32\) |
parallelrisch | \(\frac {39150 \ln \left (x +\frac {3}{5}\right ) x^{2}-40500 x^{3}+46980 \ln \left (x +\frac {3}{5}\right ) x -38725 x^{2}+14094 \ln \left (x +\frac {3}{5}\right )-8490 x}{11250 \left (3+5 x \right )^{2}}\) | \(46\) |
meijerg | \(\frac {2 x \left (\frac {5 x}{3}+2\right )}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {2 x^{2}}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {x \left (15 x +6\right )}{30 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {87 \ln \left (1+\frac {5 x}{3}\right )}{625}-\frac {9 x \left (\frac {100}{9} x^{2}+30 x +12\right )}{250 \left (1+\frac {5 x}{3}\right )^{2}}\) | \(72\) |
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Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {4500 \, x^{3} + 5400 \, x^{2} - 174 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 2260 \, x + 395}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=- \frac {18 x}{125} - \frac {128 x + 79}{6250 x^{2} + 7500 x + 2250} + \frac {87 \log {\left (5 x + 3 \right )}}{625} \]
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Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {18}{125} \, x - \frac {128 \, x + 79}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {87}{625} \, \log \left (5 \, x + 3\right ) \]
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=-\frac {18}{125} \, x - \frac {128 \, x + 79}{250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {87}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^3} \, dx=\frac {87\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {18\,x}{125}-\frac {\frac {64\,x}{3125}+\frac {79}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]
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