Integrand size = 32, antiderivative size = 63 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=-\frac {2 \log \left (2-\left (1-\sqrt {5}\right ) x+2 x^2\right )}{1-\sqrt {5}}-\frac {2 \log \left (2-\left (1+\sqrt {5}\right ) x+2 x^2\right )}{1+\sqrt {5}} \]
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Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2111, 642} \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=-\frac {2 \log \left (2 x^2-\left (1-\sqrt {5}\right ) x+2\right )}{1-\sqrt {5}}-\frac {2 \log \left (2 x^2-\left (1+\sqrt {5}\right ) x+2\right )}{1+\sqrt {5}} \]
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Rule 642
Rule 2111
Rubi steps \begin{align*} \text {integral}& = -\frac {\int \frac {-2 \sqrt {5}+\left (10-2 \sqrt {5}\right ) x}{2+\left (-1-\sqrt {5}\right ) x+2 x^2} \, dx}{\sqrt {5}}+\frac {\int \frac {2 \sqrt {5}+\left (10+2 \sqrt {5}\right ) x}{2+\left (-1+\sqrt {5}\right ) x+2 x^2} \, dx}{\sqrt {5}} \\ & = -\frac {2 \log \left (2-\left (1-\sqrt {5}\right ) x+2 x^2\right )}{1-\sqrt {5}}-\frac {2 \log \left (2-\left (1+\sqrt {5}\right ) x+2 x^2\right )}{1+\sqrt {5}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=\frac {1}{2} \left (-\left (\left (-1+\sqrt {5}\right ) \log \left (-2+x+\sqrt {5} x-2 x^2\right )\right )+\left (1+\sqrt {5}\right ) \log \left (2+\left (-1+\sqrt {5}\right ) x+2 x^2\right )\right ) \]
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Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84
method | result | size |
default | \(2 \left (\frac {\sqrt {5}}{4}+\frac {1}{4}\right ) \ln \left (x \sqrt {5}+2 x^{2}-x +2\right )-2 \left (\frac {\sqrt {5}}{4}-\frac {1}{4}\right ) \ln \left (-x \sqrt {5}+2 x^{2}-x +2\right )\) | \(53\) |
risch | \(\frac {\ln \left (2+2 x^{2}+\left (\sqrt {5}-1\right ) x \right )}{2}+\frac {\ln \left (2+2 x^{2}+\left (\sqrt {5}-1\right ) x \right ) \sqrt {5}}{2}+\frac {\ln \left (2+2 x^{2}+\left (-\sqrt {5}-1\right ) x \right )}{2}-\frac {\ln \left (2+2 x^{2}+\left (-\sqrt {5}-1\right ) x \right ) \sqrt {5}}{2}\) | \(80\) |
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Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=\frac {1}{2} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - 2 \, x^{3} + 7 \, x^{2} + \sqrt {5} {\left (2 \, x^{3} - x^{2} + 2 \, x\right )} - 2 \, x + 2}{x^{4} - x^{3} + x^{2} - x + 1}\right ) + \frac {1}{2} \, \log \left (x^{4} - x^{3} + x^{2} - x + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=\left (\frac {1}{2} + \frac {\sqrt {5}}{2}\right ) \log {\left (x^{2} + x \left (- \frac {1}{2} + \frac {\sqrt {5}}{2}\right ) + 1 \right )} + \left (\frac {1}{2} - \frac {\sqrt {5}}{2}\right ) \log {\left (x^{2} + x \left (- \frac {\sqrt {5}}{2} - \frac {1}{2}\right ) + 1 \right )} \]
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\[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=\int { \frac {2 \, x^{3} - 4 \, x^{2} + x + 2}{x^{4} - x^{3} + x^{2} - x + 1} \,d x } \]
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Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=-\frac {1}{2} \, \sqrt {5} \log \left (x^{2} - \frac {1}{2} \, x {\left (\sqrt {5} + 1\right )} + 1\right ) + \frac {1}{2} \, \sqrt {5} \log \left (x^{2} + \frac {1}{2} \, x {\left (\sqrt {5} - 1\right )} + 1\right ) + \frac {1}{2} \, \log \left (x^{4} - x^{3} + x^{2} - x + 1\right ) \]
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Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \frac {2+x-4 x^2+2 x^3}{1-x+x^2-x^3+x^4} \, dx=\frac {\ln \left (x^2-\frac {\sqrt {5}\,x}{2}-\frac {x}{2}+1\right )}{2}+\frac {\ln \left (\frac {\sqrt {5}\,x}{2}-\frac {x}{2}+x^2+1\right )}{2}-\frac {\sqrt {5}\,\ln \left (x^2-\frac {\sqrt {5}\,x}{2}-\frac {x}{2}+1\right )}{2}+\frac {\sqrt {5}\,\ln \left (\frac {\sqrt {5}\,x}{2}-\frac {x}{2}+x^2+1\right )}{2} \]
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