Integrand size = 26, antiderivative size = 22 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=(d-e) \log (1+x)-(d-2 e) \log (2+x) \]
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Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1600, 646, 31} \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=(d-e) \log (x+1)-(d-2 e) \log (x+2) \]
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Rule 31
Rule 646
Rule 1600
Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x}{2+3 x+x^2} \, dx \\ & = -\left ((d-2 e) \int \frac {1}{2+x} \, dx\right )+(d-e) \int \frac {1}{1+x} \, dx \\ & = (d-e) \log (1+x)-(d-2 e) \log (2+x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=(d-e) \log (1+x)+(-d+2 e) \log (2+x) \]
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Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09
method | result | size |
default | \(\left (-d +2 e \right ) \ln \left (x +2\right )+\left (d -e \right ) \ln \left (x +1\right )\) | \(24\) |
norman | \(\left (-d +2 e \right ) \ln \left (x +2\right )+\left (d -e \right ) \ln \left (x +1\right )\) | \(24\) |
parallelrisch | \(\ln \left (x +1\right ) d -\ln \left (x +1\right ) e -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e\) | \(29\) |
risch | \(-\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e +\ln \left (-x -1\right ) d -\ln \left (-x -1\right ) e\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=-{\left (d - 2 \, e\right )} \log \left (x + 2\right ) + {\left (d - e\right )} \log \left (x + 1\right ) \]
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Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=\left (- d + 2 e\right ) \log {\left (x + \frac {4 d - 6 e}{2 d - 3 e} \right )} + \left (d - e\right ) \log {\left (x + 1 \right )} \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=-{\left (d - 2 \, e\right )} \log \left (x + 2\right ) + {\left (d - e\right )} \log \left (x + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=-{\left (d - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + {\left (d - e\right )} \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 7.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-3 x+x^2\right )}{4-5 x^2+x^4} \, dx=\ln \left (x+1\right )\,\left (d-e\right )-\ln \left (x+2\right )\,\left (d-2\,e\right ) \]
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