Integrand size = 15, antiderivative size = 43 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {1}{10 (1-4 x)^2}-\frac {3}{25 (1-4 x)}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \]
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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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {46} \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=-\frac {3}{25 (1-4 x)}+\frac {1}{10 (1-4 x)^2}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \]
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Rule 46
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {27}{125 (-2+3 x)}-\frac {4}{5 (-1+4 x)^3}-\frac {12}{25 (-1+4 x)^2}-\frac {36}{125 (-1+4 x)}\right ) \, dx \\ & = \frac {1}{10 (1-4 x)^2}-\frac {3}{25 (1-4 x)}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {-5+120 x+18 (1-4 x)^2 \log (8-12 x)-18 (1-4 x)^2 \log (-1+4 x)}{250 (1-4 x)^2} \]
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Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {\frac {12 x}{25}-\frac {1}{50}}{\left (-1+4 x \right )^{2}}+\frac {9 \ln \left (-2+3 x \right )}{125}-\frac {9 \ln \left (-1+4 x \right )}{125}\) | \(32\) |
norman | \(\frac {\frac {8}{25} x +\frac {8}{25} x^{2}}{\left (-1+4 x \right )^{2}}+\frac {9 \ln \left (-2+3 x \right )}{125}-\frac {9 \ln \left (-1+4 x \right )}{125}\) | \(35\) |
default | \(\frac {1}{10 \left (-1+4 x \right )^{2}}+\frac {3}{25 \left (-1+4 x \right )}-\frac {9 \ln \left (-1+4 x \right )}{125}+\frac {9 \ln \left (-2+3 x \right )}{125}\) | \(36\) |
parallelrisch | \(\frac {144 \ln \left (x -\frac {2}{3}\right ) x^{2}-144 \ln \left (x -\frac {1}{4}\right ) x^{2}-72 \ln \left (x -\frac {2}{3}\right ) x +72 \ln \left (x -\frac {1}{4}\right ) x +40 x^{2}+9 \ln \left (x -\frac {2}{3}\right )-9 \ln \left (x -\frac {1}{4}\right )+40 x}{125 \left (-1+4 x \right )^{2}}\) | \(63\) |
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Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=-\frac {18 \, {\left (16 \, x^{2} - 8 \, x + 1\right )} \log \left (4 \, x - 1\right ) - 18 \, {\left (16 \, x^{2} - 8 \, x + 1\right )} \log \left (3 \, x - 2\right ) - 120 \, x + 5}{250 \, {\left (16 \, x^{2} - 8 \, x + 1\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 x - 1}{800 x^{2} - 400 x + 50} + \frac {9 \log {\left (x - \frac {2}{3} \right )}}{125} - \frac {9 \log {\left (x - \frac {1}{4} \right )}}{125} \]
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Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 \, x - 1}{50 \, {\left (16 \, x^{2} - 8 \, x + 1\right )}} - \frac {9}{125} \, \log \left (4 \, x - 1\right ) + \frac {9}{125} \, \log \left (3 \, x - 2\right ) \]
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Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 \, x - 1}{50 \, {\left (4 \, x - 1\right )}^{2}} - \frac {9}{125} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) + \frac {9}{125} \, \log \left ({\left | 3 \, x - 2 \right |}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {\frac {3\,x}{100}-\frac {1}{800}}{x^2-\frac {x}{2}+\frac {1}{16}}-\frac {18\,\mathrm {atanh}\left (\frac {24\,x}{5}-\frac {11}{5}\right )}{125} \]
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