Integrand size = 31, antiderivative size = 57 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=g x-\frac {1}{2} (d+e+f+g) \log (1-x)+\frac {1}{3} (d+2 e+4 f+8 g) \log (2-x)+\frac {1}{6} (d-e+f-g) \log (1+x) \]
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Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1600, 2099} \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=-\frac {1}{2} \log (1-x) (d+e+f+g)+\frac {1}{3} \log (2-x) (d+2 e+4 f+8 g)+\frac {1}{6} \log (x+1) (d-e+f-g)+g x \]
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Rule 1600
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3}{2-x-2 x^2+x^3} \, dx \\ & = \int \left (g+\frac {d+2 e+4 f+8 g}{3 (-2+x)}+\frac {-d-e-f-g}{2 (-1+x)}+\frac {d-e+f-g}{6 (1+x)}\right ) \, dx \\ & = g x-\frac {1}{2} (d+e+f+g) \log (1-x)+\frac {1}{3} (d+2 e+4 f+8 g) \log (2-x)+\frac {1}{6} (d-e+f-g) \log (1+x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\frac {1}{6} (6 g x-3 (d+e+f+g) \log (1-x)+2 (d+2 e+4 f+8 g) \log (2-x)+(d-e+f-g) \log (1+x)) \]
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Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04
method | result | size |
default | \(g x +\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{2}-\frac {e}{2}-\frac {f}{2}-\frac {g}{2}\right ) \ln \left (x -1\right )+\left (\frac {d}{3}+\frac {2 e}{3}+\frac {4 f}{3}+\frac {8 g}{3}\right ) \ln \left (x -2\right )\) | \(59\) |
norman | \(g x +\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{2}-\frac {e}{2}-\frac {f}{2}-\frac {g}{2}\right ) \ln \left (x -1\right )+\left (\frac {d}{3}+\frac {2 e}{3}+\frac {4 f}{3}+\frac {8 g}{3}\right ) \ln \left (x -2\right )\) | \(59\) |
parallelrisch | \(g x +\frac {\ln \left (x -2\right ) d}{3}+\frac {2 \ln \left (x -2\right ) e}{3}+\frac {4 \ln \left (x -2\right ) f}{3}+\frac {8 \ln \left (x -2\right ) g}{3}-\frac {\ln \left (x -1\right ) d}{2}-\frac {\ln \left (x -1\right ) e}{2}-\frac {\ln \left (x -1\right ) f}{2}-\frac {\ln \left (x -1\right ) g}{2}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {\ln \left (x +1\right ) f}{6}-\frac {\ln \left (x +1\right ) g}{6}\) | \(89\) |
risch | \(g x +\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {\ln \left (x +1\right ) f}{6}-\frac {\ln \left (x +1\right ) g}{6}-\frac {\ln \left (1-x \right ) d}{2}-\frac {\ln \left (1-x \right ) e}{2}-\frac {\ln \left (1-x \right ) f}{2}-\frac {\ln \left (1-x \right ) g}{2}+\frac {\ln \left (2-x \right ) d}{3}+\frac {2 \ln \left (2-x \right ) e}{3}+\frac {4 \ln \left (2-x \right ) f}{3}+\frac {8 \ln \left (2-x \right ) g}{3}\) | \(105\) |
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Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=g x + \frac {1}{6} \, {\left (d - e + f - g\right )} \log \left (x + 1\right ) - \frac {1}{2} \, {\left (d + e + f + g\right )} \log \left (x - 1\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g\right )} \log \left (x - 2\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 1389 vs. \(2 (63) = 126\).
Time = 54.94 (sec) , antiderivative size = 1389, normalized size of antiderivative = 24.37 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\text {Too large to display} \]
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Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=g x + \frac {1}{6} \, {\left (d - e + f - g\right )} \log \left (x + 1\right ) - \frac {1}{2} \, {\left (d + e + f + g\right )} \log \left (x - 1\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g\right )} \log \left (x - 2\right ) \]
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Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=g x + \frac {1}{6} \, {\left (d - e + f - g\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{2} \, {\left (d + e + f + g\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{3} \, {\left (d + 2 \, e + 4 \, f + 8 \, g\right )} \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 7.87 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {(2+x) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}\right )-\ln \left (x-1\right )\,\left (\frac {d}{2}+\frac {e}{2}+\frac {f}{2}+\frac {g}{2}\right )+\ln \left (x-2\right )\,\left (\frac {d}{3}+\frac {2\,e}{3}+\frac {4\,f}{3}+\frac {8\,g}{3}\right )+g\,x \]
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