Integrand size = 18, antiderivative size = 33 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=\frac {7}{54 (2+3 x)^2}-\frac {37}{27 (2+3 x)}-\frac {10}{27} \log (2+3 x) \]
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Time = 0.01 (sec) , antiderivative size = 33, 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} \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=-\frac {37}{27 (3 x+2)}+\frac {7}{54 (3 x+2)^2}-\frac {10}{27} \log (3 x+2) \]
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Rule 78
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {7}{9 (2+3 x)^3}+\frac {37}{9 (2+3 x)^2}-\frac {10}{9 (2+3 x)}\right ) \, dx \\ & = \frac {7}{54 (2+3 x)^2}-\frac {37}{27 (2+3 x)}-\frac {10}{27} \log (2+3 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=\frac {1}{54} \left (-\frac {3 (47+74 x)}{(2+3 x)^2}-20 \log (2+3 x)\right ) \]
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Time = 1.73 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {-\frac {37 x}{9}-\frac {47}{18}}{\left (2+3 x \right )^{2}}-\frac {10 \ln \left (2+3 x \right )}{27}\) | \(24\) |
norman | \(\frac {\frac {67}{18} x +\frac {47}{8} x^{2}}{\left (2+3 x \right )^{2}}-\frac {10 \ln \left (2+3 x \right )}{27}\) | \(27\) |
default | \(\frac {7}{54 \left (2+3 x \right )^{2}}-\frac {37}{27 \left (2+3 x \right )}-\frac {10 \ln \left (2+3 x \right )}{27}\) | \(28\) |
parallelrisch | \(-\frac {720 \ln \left (\frac {2}{3}+x \right ) x^{2}+960 \ln \left (\frac {2}{3}+x \right ) x -1269 x^{2}+320 \ln \left (\frac {2}{3}+x \right )-804 x}{216 \left (2+3 x \right )^{2}}\) | \(41\) |
meijerg | \(\frac {3 x \left (\frac {3 x}{2}+2\right )}{16 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {x^{2}}{16 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {5 x \left (\frac {27 x}{2}+6\right )}{54 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {10 \ln \left (1+\frac {3 x}{2}\right )}{27}\) | \(52\) |
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=-\frac {20 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 222 \, x + 141}{54 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=- \frac {74 x + 47}{162 x^{2} + 216 x + 72} - \frac {10 \log {\left (3 x + 2 \right )}}{27} \]
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Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=-\frac {74 \, x + 47}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {10}{27} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=-\frac {74 \, x + 47}{18 \, {\left (3 \, x + 2\right )}^{2}} - \frac {10}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
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Time = 1.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx=-\frac {10\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {\frac {37\,x}{81}+\frac {47}{162}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]
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