Integrand size = 22, antiderivative size = 43 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {343}{9 (2+3 x)}-\frac {1331}{25 (3+5 x)}+\frac {3136}{9} \log (2+3 x)-\frac {8712}{25} \log (3+5 x) \]
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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} \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {343}{9 (3 x+2)}-\frac {1331}{25 (5 x+3)}+\frac {3136}{9} \log (3 x+2)-\frac {8712}{25} \log (5 x+3) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {343}{3 (2+3 x)^2}+\frac {3136}{3 (2+3 x)}+\frac {1331}{5 (3+5 x)^2}-\frac {8712}{5 (3+5 x)}\right ) \, dx \\ & = -\frac {343}{9 (2+3 x)}-\frac {1331}{25 (3+5 x)}+\frac {3136}{9} \log (2+3 x)-\frac {8712}{25} \log (3+5 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {49683+78812 x-78400 \left (6+19 x+15 x^2\right ) \log (5 (2+3 x))+78408 \left (6+19 x+15 x^2\right ) \log (3+5 x)}{225 (2+3 x) (3+5 x)} \]
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Time = 0.80 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {343}{9 \left (2+3 x \right )}-\frac {1331}{25 \left (3+5 x \right )}+\frac {3136 \ln \left (2+3 x \right )}{9}-\frac {8712 \ln \left (3+5 x \right )}{25}\) | \(36\) |
risch | \(\frac {-\frac {78812 x}{225}-\frac {16561}{75}}{\left (2+3 x \right ) \left (3+5 x \right )}+\frac {3136 \ln \left (2+3 x \right )}{9}-\frac {8712 \ln \left (3+5 x \right )}{25}\) | \(39\) |
norman | \(\frac {\frac {16561}{30} x^{2}+\frac {10469}{30} x}{\left (2+3 x \right ) \left (3+5 x \right )}+\frac {3136 \ln \left (2+3 x \right )}{9}-\frac {8712 \ln \left (3+5 x \right )}{25}\) | \(42\) |
parallelrisch | \(\frac {2352000 \ln \left (\frac {2}{3}+x \right ) x^{2}-2352240 \ln \left (x +\frac {3}{5}\right ) x^{2}+2979200 \ln \left (\frac {2}{3}+x \right ) x -2979504 \ln \left (x +\frac {3}{5}\right ) x +248415 x^{2}+940800 \ln \left (\frac {2}{3}+x \right )-940896 \ln \left (x +\frac {3}{5}\right )+157035 x}{450 \left (2+3 x \right ) \left (3+5 x \right )}\) | \(70\) |
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Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {78408 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 78400 \, {\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 78812 \, x + 49683}{225 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=- \frac {78812 x + 49683}{3375 x^{2} + 4275 x + 1350} - \frac {8712 \log {\left (x + \frac {3}{5} \right )}}{25} + \frac {3136 \log {\left (x + \frac {2}{3} \right )}}{9} \]
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Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {78812 \, x + 49683}{225 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} - \frac {8712}{25} \, \log \left (5 \, x + 3\right ) + \frac {3136}{9} \, \log \left (3 \, x + 2\right ) \]
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Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.30 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=-\frac {1331}{25 \, {\left (5 \, x + 3\right )}} + \frac {1715}{3 \, {\left (\frac {1}{5 \, x + 3} + 3\right )}} + \frac {8}{225} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) + \frac {3136}{9} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^3}{(2+3 x)^2 (3+5 x)^2} \, dx=\frac {3136\,\ln \left (x+\frac {2}{3}\right )}{9}-\frac {8712\,\ln \left (x+\frac {3}{5}\right )}{25}-\frac {\frac {78812\,x}{3375}+\frac {16561}{1125}}{x^2+\frac {19\,x}{15}+\frac {2}{5}} \]
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