Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=\frac {293 x}{81}-\frac {161 x^2}{27}+\frac {188 x^3}{27}-\frac {10 x^4}{3}-\frac {343}{243} \log (2+3 x) \]
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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} \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=-\frac {10 x^4}{3}+\frac {188 x^3}{27}-\frac {161 x^2}{27}+\frac {293 x}{81}-\frac {343}{243} \log (3 x+2) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {293}{81}-\frac {322 x}{27}+\frac {188 x^2}{9}-\frac {40 x^3}{3}-\frac {343}{81 (2+3 x)}\right ) \, dx \\ & = \frac {293 x}{81}-\frac {161 x^2}{27}+\frac {188 x^3}{27}-\frac {10 x^4}{3}-\frac {343}{243} \log (2+3 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=\frac {1}{729} \left (5674+2637 x-4347 x^2+5076 x^3-2430 x^4-1029 \log (2+3 x)\right ) \]
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Time = 1.96 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(-\frac {10 x^{4}}{3}+\frac {188 x^{3}}{27}-\frac {161 x^{2}}{27}+\frac {293 x}{81}-\frac {343 \ln \left (\frac {2}{3}+x \right )}{243}\) | \(26\) |
default | \(\frac {293 x}{81}-\frac {161 x^{2}}{27}+\frac {188 x^{3}}{27}-\frac {10 x^{4}}{3}-\frac {343 \ln \left (2+3 x \right )}{243}\) | \(28\) |
norman | \(\frac {293 x}{81}-\frac {161 x^{2}}{27}+\frac {188 x^{3}}{27}-\frac {10 x^{4}}{3}-\frac {343 \ln \left (2+3 x \right )}{243}\) | \(28\) |
risch | \(\frac {293 x}{81}-\frac {161 x^{2}}{27}+\frac {188 x^{3}}{27}-\frac {10 x^{4}}{3}-\frac {343 \ln \left (2+3 x \right )}{243}\) | \(28\) |
meijerg | \(-\frac {343 \ln \left (1+\frac {3 x}{2}\right )}{243}-\frac {13 x}{3}-\frac {2 x \left (-\frac {9 x}{2}+6\right )}{9}+\frac {4 x \left (9 x^{2}-9 x +12\right )}{9}+\frac {16 x \left (-\frac {405}{8} x^{3}+45 x^{2}-45 x +60\right )}{243}\) | \(52\) |
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Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=-\frac {10}{3} \, x^{4} + \frac {188}{27} \, x^{3} - \frac {161}{27} \, x^{2} + \frac {293}{81} \, x - \frac {343}{243} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=- \frac {10 x^{4}}{3} + \frac {188 x^{3}}{27} - \frac {161 x^{2}}{27} + \frac {293 x}{81} - \frac {343 \log {\left (3 x + 2 \right )}}{243} \]
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Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=-\frac {10}{3} \, x^{4} + \frac {188}{27} \, x^{3} - \frac {161}{27} \, x^{2} + \frac {293}{81} \, x - \frac {343}{243} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=-\frac {10}{3} \, x^{4} + \frac {188}{27} \, x^{3} - \frac {161}{27} \, x^{2} + \frac {293}{81} \, x - \frac {343}{243} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^3 (3+5 x)}{2+3 x} \, dx=\frac {293\,x}{81}-\frac {343\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {161\,x^2}{27}+\frac {188\,x^3}{27}-\frac {10\,x^4}{3} \]
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