Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=\frac {1097 x}{81}+\frac {545 x^2}{54}-\frac {475 x^3}{27}-\frac {125 x^4}{6}-\frac {7}{243} \log (2+3 x) \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=-\frac {125 x^4}{6}-\frac {475 x^3}{27}+\frac {545 x^2}{54}+\frac {1097 x}{81}-\frac {7}{243} \log (3 x+2) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1097}{81}+\frac {545 x}{27}-\frac {475 x^2}{9}-\frac {250 x^3}{3}-\frac {7}{81 (2+3 x)}\right ) \, dx \\ & = \frac {1097 x}{81}+\frac {545 x^2}{54}-\frac {475 x^3}{27}-\frac {125 x^4}{6}-\frac {7}{243} \log (2+3 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=\frac {5024+19746 x+14715 x^2-25650 x^3-30375 x^4-42 \log (2+3 x)}{1458} \]
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Time = 0.73 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(-\frac {125 x^{4}}{6}-\frac {475 x^{3}}{27}+\frac {545 x^{2}}{54}+\frac {1097 x}{81}-\frac {7 \ln \left (\frac {2}{3}+x \right )}{243}\) | \(26\) |
default | \(\frac {1097 x}{81}+\frac {545 x^{2}}{54}-\frac {475 x^{3}}{27}-\frac {125 x^{4}}{6}-\frac {7 \ln \left (2+3 x \right )}{243}\) | \(28\) |
norman | \(\frac {1097 x}{81}+\frac {545 x^{2}}{54}-\frac {475 x^{3}}{27}-\frac {125 x^{4}}{6}-\frac {7 \ln \left (2+3 x \right )}{243}\) | \(28\) |
risch | \(\frac {1097 x}{81}+\frac {545 x^{2}}{54}-\frac {475 x^{3}}{27}-\frac {125 x^{4}}{6}-\frac {7 \ln \left (2+3 x \right )}{243}\) | \(28\) |
meijerg | \(-\frac {7 \ln \left (1+\frac {3 x}{2}\right )}{243}+27 x +\frac {5 x \left (-\frac {9 x}{2}+6\right )}{3}-\frac {325 x \left (9 x^{2}-9 x +12\right )}{81}+\frac {100 x \left (-\frac {405}{8} x^{3}+45 x^{2}-45 x +60\right )}{243}\) | \(52\) |
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none
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=-\frac {125}{6} \, x^{4} - \frac {475}{27} \, x^{3} + \frac {545}{54} \, x^{2} + \frac {1097}{81} \, x - \frac {7}{243} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=- \frac {125 x^{4}}{6} - \frac {475 x^{3}}{27} + \frac {545 x^{2}}{54} + \frac {1097 x}{81} - \frac {7 \log {\left (3 x + 2 \right )}}{243} \]
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Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=-\frac {125}{6} \, x^{4} - \frac {475}{27} \, x^{3} + \frac {545}{54} \, x^{2} + \frac {1097}{81} \, x - \frac {7}{243} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=-\frac {125}{6} \, x^{4} - \frac {475}{27} \, x^{3} + \frac {545}{54} \, x^{2} + \frac {1097}{81} \, x - \frac {7}{243} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x) (3+5 x)^3}{2+3 x} \, dx=\frac {1097\,x}{81}-\frac {7\,\ln \left (x+\frac {2}{3}\right )}{243}+\frac {545\,x^2}{54}-\frac {475\,x^3}{27}-\frac {125\,x^4}{6} \]
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