Integrand size = 11, antiderivative size = 30 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {x}{8}-\frac {x^2}{8}+\frac {x^3}{6}-\frac {1}{16} \log (1+2 x) \]
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Time = 0.01 (sec) , antiderivative size = 30, 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 = {45} \[ \int \frac {x^3}{1+2 x} \, dx=\frac {x^3}{6}-\frac {x^2}{8}+\frac {x}{8}-\frac {1}{16} \log (2 x+1) \]
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Rule 45
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8}-\frac {x}{4}+\frac {x^2}{2}-\frac {1}{8 (1+2 x)}\right ) \, dx \\ & = \frac {x}{8}-\frac {x^2}{8}+\frac {x^3}{6}-\frac {1}{16} \log (1+2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {1}{96} \left (11+12 x-12 x^2+16 x^3-6 \log (1+2 x)\right ) \]
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Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
parallelrisch | \(\frac {x^{3}}{6}-\frac {x^{2}}{8}+\frac {x}{8}-\frac {\ln \left (x +\frac {1}{2}\right )}{16}\) | \(21\) |
default | \(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\) | \(23\) |
norman | \(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\) | \(23\) |
meijerg | \(\frac {x \left (16 x^{2}-12 x +12\right )}{96}-\frac {\ln \left (1+2 x \right )}{16}\) | \(23\) |
risch | \(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\) | \(23\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left (2 \, x + 1\right ) \]
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Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {x^{3}}{6} - \frac {x^{2}}{8} + \frac {x}{8} - \frac {\log {\left (2 x + 1 \right )}}{16} \]
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Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left (2 \, x + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {x^3}{1+2 x} \, dx=\frac {x}{8}-\frac {\ln \left (x+\frac {1}{2}\right )}{16}-\frac {x^2}{8}+\frac {x^3}{6} \]
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