Integrand size = 24, antiderivative size = 41 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{6}+6 \arctan (2-x)+\frac {3}{4} \log \left (5-4 x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1671, 648, 632, 210, 642} \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=6 \arctan (2-x)+\frac {x^3}{6}+\frac {x^2}{2}+\frac {3}{4} \log \left (x^2-4 x+5\right )+\frac {3 x}{2} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{2}+x+\frac {x^2}{2}-\frac {3 (6-x)}{10-8 x+2 x^2}\right ) \, dx \\ & = \frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{6}-3 \int \frac {6-x}{10-8 x+2 x^2} \, dx \\ & = \frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{6}+\frac {3}{4} \int \frac {-8+4 x}{10-8 x+2 x^2} \, dx-12 \int \frac {1}{10-8 x+2 x^2} \, dx \\ & = \frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{6}+\frac {3}{4} \log \left (5-4 x+x^2\right )+24 \text {Subst}\left (\int \frac {1}{-16-x^2} \, dx,x,-8+4 x\right ) \\ & = \frac {3 x}{2}+\frac {x^2}{2}+\frac {x^3}{6}+6 \tan ^{-1}(2-x)+\frac {3}{4} \log \left (5-4 x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {1}{2} \left (3 x+x^2+\frac {x^3}{3}+12 \arctan (2-x)+\frac {3}{2} \log \left (5-4 x+x^2\right )\right ) \]
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Time = 1.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {3 x}{2}+\frac {x^{2}}{2}+\frac {x^{3}}{6}-6 \arctan \left (x -2\right )+\frac {3 \ln \left (x^{2}-4 x +5\right )}{4}\) | \(32\) |
risch | \(\frac {3 x}{2}+\frac {x^{2}}{2}+\frac {x^{3}}{6}-6 \arctan \left (x -2\right )+\frac {3 \ln \left (x^{2}-4 x +5\right )}{4}\) | \(32\) |
parallelrisch | \(\frac {x^{3}}{6}+\frac {x^{2}}{2}+\frac {3 x}{2}+\frac {3 \ln \left (x -2-i\right )}{4}+3 i \ln \left (x -2-i\right )+\frac {3 \ln \left (x -2+i\right )}{4}-3 i \ln \left (x -2+i\right )\) | \(49\) |
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {1}{6} \, x^{3} + \frac {1}{2} \, x^{2} + \frac {3}{2} \, x - 6 \, \arctan \left (x - 2\right ) + \frac {3}{4} \, \log \left (x^{2} - 4 \, x + 5\right ) \]
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Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {x^{3}}{6} + \frac {x^{2}}{2} + \frac {3 x}{2} + \frac {3 \log {\left (x^{2} - 4 x + 5 \right )}}{4} - 6 \operatorname {atan}{\left (x - 2 \right )} \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {1}{6} \, x^{3} + \frac {1}{2} \, x^{2} + \frac {3}{2} \, x - 6 \, \arctan \left (x - 2\right ) + \frac {3}{4} \, \log \left (x^{2} - 4 \, x + 5\right ) \]
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {1}{6} \, x^{3} + \frac {1}{2} \, x^{2} + \frac {3}{2} \, x - 6 \, \arctan \left (x - 2\right ) + \frac {3}{4} \, \log \left (x^{2} - 4 \, x + 5\right ) \]
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Time = 9.57 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {-3+x-2 x^3+x^4}{10-8 x+2 x^2} \, dx=\frac {3\,x}{2}-6\,\mathrm {atan}\left (x-2\right )+\frac {3\,\ln \left (x^2-4\,x+5\right )}{4}+\frac {x^2}{2}+\frac {x^3}{6} \]
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