Integrand size = 41, antiderivative size = 66 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=(f-3 g+7 h) x+\frac {1}{2} (g-3 h) x^2+\frac {h x^3}{3}+(d-e+f-g+h) \log (1+x)-(d-2 e+4 f-8 g+16 h) \log (2+x) \]
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Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {1600, 1671, 646, 31} \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\log (x+1) (d-e+f-g+h)-\log (x+2) (d-2 e+4 f-8 g+16 h)+x (f-3 g+7 h)+\frac {1}{2} x^2 (g-3 h)+\frac {h x^3}{3} \]
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Rule 31
Rule 646
Rule 1600
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3+h x^4}{2+3 x+x^2} \, dx \\ & = \int \left (f-3 g+7 h+(g-3 h) x+h x^2+\frac {d-2 f+6 g-14 h+(e-3 f+7 g-15 h) x}{2+3 x+x^2}\right ) \, dx \\ & = (f-3 g+7 h) x+\frac {1}{2} (g-3 h) x^2+\frac {h x^3}{3}+\int \frac {d-2 f+6 g-14 h+(e-3 f+7 g-15 h) x}{2+3 x+x^2} \, dx \\ & = (f-3 g+7 h) x+\frac {1}{2} (g-3 h) x^2+\frac {h x^3}{3}+(d-e+f-g+h) \int \frac {1}{1+x} \, dx-(d-2 e+4 f-8 g+16 h) \int \frac {1}{2+x} \, dx \\ & = (f-3 g+7 h) x+\frac {1}{2} (g-3 h) x^2+\frac {h x^3}{3}+(d-e+f-g+h) \log (1+x)-(d-2 e+4 f-8 g+16 h) \log (2+x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=(f-3 g+7 h) x+\frac {1}{2} (g-3 h) x^2+\frac {h x^3}{3}+(d-e+f-g+h) \log (1+x)+(-d+2 e-4 f+8 g-16 h) \log (2+x) \]
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Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\left (\frac {g}{2}-\frac {3 h}{2}\right ) x^{2}+\left (f -3 g +7 h \right ) x +\frac {h \,x^{3}}{3}+\left (-d +2 e -4 f +8 g -16 h \right ) \ln \left (x +2\right )+\left (d -e +f -g +h \right ) \ln \left (x +1\right )\) | \(65\) |
default | \(\frac {h \,x^{3}}{3}+\frac {g \,x^{2}}{2}-\frac {3 h \,x^{2}}{2}+f x -3 g x +7 h x +\left (-d +2 e -4 f +8 g -16 h \right ) \ln \left (x +2\right )+\left (d -e +f -g +h \right ) \ln \left (x +1\right )\) | \(67\) |
parallelrisch | \(\frac {h \,x^{3}}{3}+\frac {g \,x^{2}}{2}-\frac {3 h \,x^{2}}{2}+f x -3 g x +7 h x +\ln \left (x +1\right ) d -\ln \left (x +1\right ) e +\ln \left (x +1\right ) f -\ln \left (x +1\right ) g +\ln \left (x +1\right ) h -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f +8 \ln \left (x +2\right ) g -16 \ln \left (x +2\right ) h\) | \(98\) |
risch | \(\frac {h \,x^{3}}{3}+\frac {g \,x^{2}}{2}-\frac {3 h \,x^{2}}{2}+f x -3 g x +7 h x +\ln \left (-x -1\right ) d -\ln \left (-x -1\right ) e +\ln \left (-x -1\right ) f -\ln \left (-x -1\right ) g +\ln \left (-x -1\right ) h -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f +8 \ln \left (x +2\right ) g -16 \ln \left (x +2\right ) h\) | \(108\) |
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Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, h x^{3} + \frac {1}{2} \, {\left (g - 3 \, h\right )} x^{2} + {\left (f - 3 \, g + 7 \, h\right )} x - {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) + {\left (d - e + f - g + h\right )} \log \left (x + 1\right ) \]
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Time = 0.85 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {h x^{3}}{3} + x^{2} \left (\frac {g}{2} - \frac {3 h}{2}\right ) + x \left (f - 3 g + 7 h\right ) + \left (- d + 2 e - 4 f + 8 g - 16 h\right ) \log {\left (x + \frac {4 d - 6 e + 10 f - 18 g + 34 h}{2 d - 3 e + 5 f - 9 g + 17 h} \right )} + \left (d - e + f - g + h\right ) \log {\left (x + 1 \right )} \]
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Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, h x^{3} + \frac {1}{2} \, {\left (g - 3 \, h\right )} x^{2} + {\left (f - 3 \, g + 7 \, h\right )} x - {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) + {\left (d - e + f - g + h\right )} \log \left (x + 1\right ) \]
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Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=\frac {1}{3} \, h x^{3} + \frac {1}{2} \, g x^{2} - \frac {3}{2} \, h x^{2} + f x - 3 \, g x + 7 \, h x - {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left ({\left | x + 2 \right |}\right ) + {\left (d - e + f - g + h\right )} \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3+h x^4\right )}{4-5 x^2+x^4} \, dx=x^2\,\left (\frac {g}{2}-\frac {3\,h}{2}\right )+x\,\left (f-3\,g+7\,h\right )-\ln \left (x+2\right )\,\left (d-2\,e+4\,f-8\,g+16\,h\right )+\frac {h\,x^3}{3}+\ln \left (x+1\right )\,\left (d-e+f-g+h\right ) \]
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