Integrand size = 22, antiderivative size = 43 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {343}{54 (2+3 x)^2}+\frac {1421}{27 (2+3 x)}-\frac {7189}{27} \log (2+3 x)+\frac {1331}{5} \log (3+5 x) \]
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Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {1421}{27 (3 x+2)}+\frac {343}{54 (3 x+2)^2}-\frac {7189}{27} \log (3 x+2)+\frac {1331}{5} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {343}{9 (2+3 x)^3}-\frac {1421}{9 (2+3 x)^2}-\frac {7189}{9 (2+3 x)}+\frac {1331}{3+5 x}\right ) \, dx \\ & = \frac {343}{54 (2+3 x)^2}+\frac {1421}{27 (2+3 x)}-\frac {7189}{27} \log (2+3 x)+\frac {1331}{5} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {49 (41+58 x)}{18 (2+3 x)^2}-\frac {7189}{27} \log (5 (2+3 x))+\frac {1331}{5} \log (3+5 x) \]
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Time = 2.44 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {\frac {1421 x}{9}+\frac {2009}{18}}{\left (2+3 x \right )^{2}}-\frac {7189 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{5}\) | \(32\) |
norman | \(\frac {-\frac {3185}{18} x -\frac {2009}{8} x^{2}}{\left (2+3 x \right )^{2}}-\frac {7189 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{5}\) | \(35\) |
default | \(\frac {343}{54 \left (2+3 x \right )^{2}}+\frac {1421}{27 \left (2+3 x \right )}-\frac {7189 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{5}\) | \(36\) |
parallelrisch | \(-\frac {2588040 \ln \left (\frac {2}{3}+x \right ) x^{2}-2587464 \ln \left (x +\frac {3}{5}\right ) x^{2}+3450720 \ln \left (\frac {2}{3}+x \right ) x -3449952 \ln \left (x +\frac {3}{5}\right ) x +271215 x^{2}+1150240 \ln \left (\frac {2}{3}+x \right )-1149984 \ln \left (x +\frac {3}{5}\right )+191100 x}{1080 \left (2+3 x \right )^{2}}\) | \(63\) |
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none
Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {71874 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x + 3\right ) - 71890 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 42630 \, x + 30135}{270 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=- \frac {- 2842 x - 2009}{162 x^{2} + 216 x + 72} + \frac {1331 \log {\left (x + \frac {3}{5} \right )}}{5} - \frac {7189 \log {\left (x + \frac {2}{3} \right )}}{27} \]
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none
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {49 \, {\left (58 \, x + 41\right )}}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {1331}{5} \, \log \left (5 \, x + 3\right ) - \frac {7189}{27} \, \log \left (3 \, x + 2\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {49 \, {\left (58 \, x + 41\right )}}{18 \, {\left (3 \, x + 2\right )}^{2}} + \frac {1331}{5} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {7189}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^3}{(2+3 x)^3 (3+5 x)} \, dx=\frac {1331\,\ln \left (x+\frac {3}{5}\right )}{5}-\frac {7189\,\ln \left (x+\frac {2}{3}\right )}{27}+\frac {\frac {1421\,x}{81}+\frac {2009}{162}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]
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