Integrand size = 22, antiderivative size = 37 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=-\frac {8 x}{45}+\frac {343}{27 (2+3 x)}-\frac {1421}{27} \log (2+3 x)+\frac {1331}{25} \log (3+5 x) \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=-\frac {8 x}{45}+\frac {343}{27 (3 x+2)}-\frac {1421}{27} \log (3 x+2)+\frac {1331}{25} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {8}{45}-\frac {343}{9 (2+3 x)^2}-\frac {1421}{9 (2+3 x)}+\frac {1331}{5 (3+5 x)}\right ) \, dx \\ & = -\frac {8 x}{45}+\frac {343}{27 (2+3 x)}-\frac {1421}{27} \log (2+3 x)+\frac {1331}{25} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1}{675} \left (-72-120 x+\frac {8575}{2+3 x}-35525 \log (5 (2+3 x))+35937 \log (3+5 x)\right ) \]
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Time = 2.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-\frac {8 x}{45}+\frac {343}{81 \left (\frac {2}{3}+x \right )}-\frac {1421 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{25}\) | \(28\) |
default | \(-\frac {8 x}{45}+\frac {343}{27 \left (2+3 x \right )}-\frac {1421 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{25}\) | \(30\) |
norman | \(\frac {-\frac {1747}{90} x -\frac {8}{15} x^{2}}{2+3 x}-\frac {1421 \ln \left (2+3 x \right )}{27}+\frac {1331 \ln \left (3+5 x \right )}{25}\) | \(35\) |
parallelrisch | \(-\frac {213150 \ln \left (\frac {2}{3}+x \right ) x -215622 \ln \left (x +\frac {3}{5}\right ) x +720 x^{2}+142100 \ln \left (\frac {2}{3}+x \right )-143748 \ln \left (x +\frac {3}{5}\right )+26205 x}{1350 \left (2+3 x \right )}\) | \(45\) |
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Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=-\frac {360 \, x^{2} - 35937 \, {\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) + 35525 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 240 \, x - 8575}{675 \, {\left (3 \, x + 2\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=- \frac {8 x}{45} + \frac {1331 \log {\left (x + \frac {3}{5} \right )}}{25} - \frac {1421 \log {\left (x + \frac {2}{3} \right )}}{27} + \frac {343}{81 x + 54} \]
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Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=-\frac {8}{45} \, x + \frac {343}{27 \, {\left (3 \, x + 2\right )}} + \frac {1331}{25} \, \log \left (5 \, x + 3\right ) - \frac {1421}{27} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=-\frac {8}{45} \, x + \frac {343}{27 \, {\left (3 \, x + 2\right )}} - \frac {412}{675} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) + \frac {1331}{25} \, \log \left ({\left | -\frac {1}{3 \, x + 2} + 5 \right |}\right ) - \frac {16}{135} \]
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Time = 1.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)} \, dx=\frac {1331\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {1421\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {8\,x}{45}+\frac {343}{81\,\left (x+\frac {2}{3}\right )} \]
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