Integrand size = 11, antiderivative size = 46 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=-\frac {1}{64 x}-\frac {3}{64 (2+3 x)^2}-\frac {3}{32 (2+3 x)}-\frac {9 \log (x)}{128}+\frac {9}{128} \log (2+3 x) \]
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Time = 0.01 (sec) , antiderivative size = 46, 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} \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=-\frac {1}{64 x}-\frac {3}{32 (3 x+2)}-\frac {3}{64 (3 x+2)^2}-\frac {9 \log (x)}{128}+\frac {9}{128} \log (3 x+2) \]
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Rule 46
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{64 x^2}-\frac {9}{128 x}+\frac {9}{32 (2+3 x)^3}+\frac {9}{32 (2+3 x)^2}+\frac {27}{128 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{64 x}-\frac {3}{64 (2+3 x)^2}-\frac {3}{32 (2+3 x)}-\frac {9 \log (x)}{128}+\frac {9}{128} \log (2+3 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=\frac {1}{128} \left (-\frac {2 \left (4+27 x+27 x^2\right )}{x (2+3 x)^2}-9 \log (x)+9 \log (2+3 x)\right ) \]
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Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {-\frac {27}{64} x^{2}-\frac {27}{64} x -\frac {1}{16}}{x \left (2+3 x \right )^{2}}-\frac {9 \ln \left (x \right )}{128}+\frac {9 \ln \left (2+3 x \right )}{128}\) | \(36\) |
default | \(-\frac {1}{64 x}-\frac {3}{64 \left (2+3 x \right )^{2}}-\frac {3}{32 \left (2+3 x \right )}-\frac {9 \ln \left (x \right )}{128}+\frac {9 \ln \left (2+3 x \right )}{128}\) | \(37\) |
norman | \(\frac {-\frac {1}{16}+\frac {27}{32} x^{2}+\frac {243}{256} x^{3}}{x \left (2+3 x \right )^{2}}-\frac {9 \ln \left (x \right )}{128}+\frac {9 \ln \left (2+3 x \right )}{128}\) | \(37\) |
meijerg | \(-\frac {1}{64 x}-\frac {15}{256}-\frac {9 \ln \left (x \right )}{128}-\frac {9 \ln \left (3\right )}{128}+\frac {9 \ln \left (2\right )}{128}+\frac {9 x \left (\frac {15 x}{2}+6\right )}{512 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {9 \ln \left (1+\frac {3 x}{2}\right )}{128}\) | \(43\) |
parallelrisch | \(-\frac {162 \ln \left (x \right ) x^{3}-162 \ln \left (\frac {2}{3}+x \right ) x^{3}+16+216 \ln \left (x \right ) x^{2}-216 \ln \left (\frac {2}{3}+x \right ) x^{2}-243 x^{3}+72 \ln \left (x \right ) x -72 \ln \left (\frac {2}{3}+x \right ) x -216 x^{2}}{256 x \left (2+3 x \right )^{2}}\) | \(69\) |
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Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=-\frac {54 \, x^{2} - 9 \, {\left (9 \, x^{3} + 12 \, x^{2} + 4 \, x\right )} \log \left (3 \, x + 2\right ) + 9 \, {\left (9 \, x^{3} + 12 \, x^{2} + 4 \, x\right )} \log \left (x\right ) + 54 \, x + 8}{128 \, {\left (9 \, x^{3} + 12 \, x^{2} + 4 \, x\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=\frac {- 27 x^{2} - 27 x - 4}{576 x^{3} + 768 x^{2} + 256 x} - \frac {9 \log {\left (x \right )}}{128} + \frac {9 \log {\left (x + \frac {2}{3} \right )}}{128} \]
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Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=-\frac {27 \, x^{2} + 27 \, x + 4}{64 \, {\left (9 \, x^{3} + 12 \, x^{2} + 4 \, x\right )}} + \frac {9}{128} \, \log \left (3 \, x + 2\right ) - \frac {9}{128} \, \log \left (x\right ) \]
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Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=-\frac {27 \, x^{2} + 27 \, x + 4}{64 \, {\left (3 \, x + 2\right )}^{2} x} + \frac {9}{128} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {9}{128} \, \log \left ({\left | x \right |}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^2 (4+6 x)^3} \, dx=\frac {9\,\mathrm {atanh}\left (3\,x+1\right )}{64}-\frac {\frac {3\,x^2}{64}+\frac {3\,x}{64}+\frac {1}{144}}{x^3+\frac {4\,x^2}{3}+\frac {4\,x}{9}} \]
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