Integrand size = 15, antiderivative size = 55 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=-\frac {25}{128 (1+2 x)^4}+\frac {7}{24 (1+2 x)^3}-\frac {3}{32 (1+2 x)^2}+\frac {1}{8 (1+2 x)}+\frac {1}{32} \log (1+2 x) \]
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Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1607, 1634} \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {1}{8 (2 x+1)}-\frac {3}{32 (2 x+1)^2}+\frac {7}{24 (2 x+1)^3}-\frac {25}{128 (2 x+1)^4}+\frac {1}{32} \log (2 x+1) \]
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Rule 1607
Rule 1634
Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (-3+x^3\right )}{(1+2 x)^5} \, dx \\ & = \int \left (\frac {25}{16 (1+2 x)^5}-\frac {7}{4 (1+2 x)^4}+\frac {3}{8 (1+2 x)^3}-\frac {1}{4 (1+2 x)^2}+\frac {1}{16 (1+2 x)}\right ) \, dx \\ & = -\frac {25}{128 (1+2 x)^4}+\frac {7}{24 (1+2 x)^3}-\frac {3}{32 (1+2 x)^2}+\frac {1}{8 (1+2 x)}+\frac {1}{32} \log (1+2 x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.75 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {49+368 x+432 x^2+384 x^3+12 (1+2 x)^4 \log (1+2 x)}{384 (1+2 x)^4} \]
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Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62
method | result | size |
risch | \(\frac {x^{3}+\frac {9}{8} x^{2}+\frac {23}{24} x +\frac {49}{384}}{\left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}\) | \(34\) |
norman | \(\frac {-\frac {37}{12} x^{3}-\frac {31}{16} x^{2}-\frac {1}{16} x -\frac {49}{24} x^{4}}{\left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}\) | \(37\) |
default | \(-\frac {25}{128 \left (1+2 x \right )^{4}}+\frac {7}{24 \left (1+2 x \right )^{3}}-\frac {3}{32 \left (1+2 x \right )^{2}}+\frac {1}{8+16 x}+\frac {\ln \left (1+2 x \right )}{32}\) | \(46\) |
meijerg | \(-\frac {x \left (1000 x^{3}+1040 x^{2}+420 x +60\right )}{960 \left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}-\frac {x^{2} \left (4 x^{2}+8 x +6\right )}{4 \left (1+2 x \right )^{4}}\) | \(57\) |
parallelrisch | \(\frac {48 \ln \left (x +\frac {1}{2}\right ) x^{4}+96 \ln \left (x +\frac {1}{2}\right ) x^{3}-196 x^{4}+72 \ln \left (x +\frac {1}{2}\right ) x^{2}-296 x^{3}+24 \ln \left (x +\frac {1}{2}\right ) x -186 x^{2}+3 \ln \left (x +\frac {1}{2}\right )-6 x}{96 \left (1+2 x \right )^{4}}\) | \(69\) |
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Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {384 \, x^{3} + 432 \, x^{2} + 12 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )} \log \left (2 \, x + 1\right ) + 368 \, x + 49}{384 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {384 x^{3} + 432 x^{2} + 368 x + 49}{6144 x^{4} + 12288 x^{3} + 9216 x^{2} + 3072 x + 384} + \frac {\log {\left (2 x + 1 \right )}}{32} \]
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Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {384 \, x^{3} + 432 \, x^{2} + 368 \, x + 49}{384 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )}} + \frac {1}{32} \, \log \left (2 \, x + 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {1}{8 \, {\left (2 \, x + 1\right )}} - \frac {3}{32 \, {\left (2 \, x + 1\right )}^{2}} + \frac {7}{24 \, {\left (2 \, x + 1\right )}^{3}} - \frac {25}{128 \, {\left (2 \, x + 1\right )}^{4}} - \frac {1}{32} \, \log \left (\frac {{\left | 2 \, x + 1 \right |}}{2 \, {\left (2 \, x + 1\right )}^{2}}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx=\frac {\ln \left (x+\frac {1}{2}\right )}{32}+\frac {\frac {x^3}{16}+\frac {9\,x^2}{128}+\frac {23\,x}{384}+\frac {49}{6144}}{x^4+2\,x^3+\frac {3\,x^2}{2}+\frac {x}{2}+\frac {1}{16}} \]
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