Integrand size = 22, antiderivative size = 37 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {1331}{56 (1-2 x)}+\frac {125 x}{12}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (2+3 x) \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125 x}{12}+\frac {1331}{56 (1-2 x)}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (3 x+2) \]
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Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {125}{12}+\frac {1331}{28 (-1+2 x)^2}+\frac {2178}{49 (-1+2 x)}-\frac {1}{147 (2+3 x)}\right ) \, dx \\ & = \frac {1331}{56 (1-2 x)}+\frac {125 x}{12}+\frac {1089}{49} \log (1-2 x)-\frac {1}{441} \log (2+3 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {\frac {83853}{1-2 x}+12250 (2+3 x)+78408 \log (3-6 x)-8 \log (2+3 x)}{3528} \]
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Time = 2.66 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {125 x}{12}-\frac {1331}{112 \left (x -\frac {1}{2}\right )}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\) | \(28\) |
default | \(\frac {125 x}{12}-\frac {1331}{56 \left (-1+2 x \right )}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\) | \(30\) |
norman | \(\frac {-\frac {1217}{21} x +\frac {125}{6} x^{2}}{-1+2 x}+\frac {1089 \ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{441}\) | \(35\) |
parallelrisch | \(-\frac {4 \ln \left (\frac {2}{3}+x \right ) x -39204 \ln \left (x -\frac {1}{2}\right ) x -18375 x^{2}-2 \ln \left (\frac {2}{3}+x \right )+19602 \ln \left (x -\frac {1}{2}\right )+51114 x}{882 \left (-1+2 x \right )}\) | \(45\) |
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Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {73500 \, x^{2} - 8 \, {\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) + 78408 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 36750 \, x - 83853}{3528 \, {\left (2 \, x - 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125 x}{12} + \frac {1089 \log {\left (x - \frac {1}{2} \right )}}{49} - \frac {\log {\left (x + \frac {2}{3} \right )}}{441} - \frac {1331}{112 x - 56} \]
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Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125}{12} \, x - \frac {1331}{56 \, {\left (2 \, x - 1\right )}} - \frac {1}{441} \, \log \left (3 \, x + 2\right ) + \frac {1089}{49} \, \log \left (2 \, x - 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125}{12} \, x - \frac {1331}{56 \, {\left (2 \, x - 1\right )}} - \frac {200}{9} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) - \frac {1}{441} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) - \frac {125}{24} \]
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Time = 1.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2 (2+3 x)} \, dx=\frac {125\,x}{12}+\frac {1089\,\ln \left (x-\frac {1}{2}\right )}{49}-\frac {\ln \left (x+\frac {2}{3}\right )}{441}-\frac {1331}{112\,\left (x-\frac {1}{2}\right )} \]
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