Integrand size = 11, antiderivative size = 31 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=-\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x) \]
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Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^3 (4+6 x)} \, dx=-\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (3 x+2) \]
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Rule 46
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4 x^3}-\frac {3}{8 x^2}+\frac {9}{16 x}-\frac {27}{16 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=-\frac {1}{8 x^2}+\frac {3}{8 x}+\frac {9 \log (x)}{16}-\frac {9}{16} \log (2+3 x) \]
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Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74
method | result | size |
norman | \(\frac {-\frac {1}{8}+\frac {3 x}{8}}{x^{2}}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\) | \(23\) |
risch | \(\frac {-\frac {1}{8}+\frac {3 x}{8}}{x^{2}}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\) | \(23\) |
default | \(-\frac {1}{8 x^{2}}+\frac {3}{8 x}+\frac {9 \ln \left (x \right )}{16}-\frac {9 \ln \left (2+3 x \right )}{16}\) | \(24\) |
parallelrisch | \(\frac {9 \ln \left (x \right ) x^{2}-9 \ln \left (\frac {2}{3}+x \right ) x^{2}-2+6 x}{16 x^{2}}\) | \(27\) |
meijerg | \(-\frac {1}{8 x^{2}}+\frac {3}{8 x}+\frac {9 \ln \left (x \right )}{16}+\frac {9 \ln \left (3\right )}{16}-\frac {9 \ln \left (2\right )}{16}-\frac {9 \ln \left (1+\frac {3 x}{2}\right )}{16}\) | \(32\) |
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none
Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=-\frac {9 \, x^{2} \log \left (3 \, x + 2\right ) - 9 \, x^{2} \log \left (x\right ) - 6 \, x + 2}{16 \, x^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=\frac {9 \log {\left (x \right )}}{16} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{16} + \frac {3 x - 1}{8 x^{2}} \]
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none
Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=\frac {3 \, x - 1}{8 \, x^{2}} - \frac {9}{16} \, \log \left (3 \, x + 2\right ) + \frac {9}{16} \, \log \left (x\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=\frac {3 \, x - 1}{8 \, x^{2}} - \frac {9}{16} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + \frac {9}{16} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^3 (4+6 x)} \, dx=\frac {\frac {3\,x}{8}-\frac {1}{8}}{x^2}-\frac {9\,\mathrm {atanh}\left (3\,x+1\right )}{8} \]
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