Integrand size = 38, antiderivative size = 76 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=h x+\frac {i x^2}{2}-\frac {1}{6} (d+4 f+16 h) \text {arctanh}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \text {arctanh}(x)-\frac {1}{6} (e+g+i) \log \left (1-x^2\right )+\frac {1}{6} (e+4 g+16 i) \log \left (4-x^2\right ) \]
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Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1687, 1690, 1180, 213, 1677, 1671, 646, 31} \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=-\frac {1}{6} \text {arctanh}\left (\frac {x}{2}\right ) (d+4 f+16 h)+\frac {1}{3} \text {arctanh}(x) (d+f+h)-\frac {1}{6} \log \left (1-x^2\right ) (e+g+i)+\frac {1}{6} \log \left (4-x^2\right ) (e+4 g+16 i)+h x+\frac {i x^2}{2} \]
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Rule 31
Rule 213
Rule 646
Rule 1180
Rule 1671
Rule 1677
Rule 1687
Rule 1690
Rubi steps \begin{align*} \text {integral}& = \int \frac {d+f x^2+h x^4}{4-5 x^2+x^4} \, dx+\int \frac {x \left (e+g x^2+i x^4\right )}{4-5 x^2+x^4} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {e+g x+i x^2}{4-5 x+x^2} \, dx,x,x^2\right )+\int \left (h+\frac {d-4 h+(f+5 h) x^2}{4-5 x^2+x^4}\right ) \, dx \\ & = h x+\frac {1}{2} \text {Subst}\left (\int \left (i+\frac {e-4 i+(g+5 i) x}{4-5 x+x^2}\right ) \, dx,x,x^2\right )+\int \frac {d-4 h+(f+5 h) x^2}{4-5 x^2+x^4} \, dx \\ & = h x+\frac {i x^2}{2}+\frac {1}{2} \text {Subst}\left (\int \frac {e-4 i+(g+5 i) x}{4-5 x+x^2} \, dx,x,x^2\right )-\frac {1}{3} (d+f+h) \int \frac {1}{-1+x^2} \, dx+\frac {1}{3} (d+4 f+16 h) \int \frac {1}{-4+x^2} \, dx \\ & = h x+\frac {i x^2}{2}-\frac {1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \tanh ^{-1}(x)+\frac {1}{6} (-e-g-i) \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,x^2\right )+\frac {1}{6} (e+4 g+16 i) \text {Subst}\left (\int \frac {1}{-4+x} \, dx,x,x^2\right ) \\ & = h x+\frac {i x^2}{2}-\frac {1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac {x}{2}\right )+\frac {1}{3} (d+f+h) \tanh ^{-1}(x)-\frac {1}{6} (e+g+i) \log \left (1-x^2\right )+\frac {1}{6} (e+4 g+16 i) \log \left (4-x^2\right ) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.29 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=\frac {1}{12} \left (12 h x+6 i x^2-2 (d+e+f+g+h+i) \log (1-x)+(d+2 e+4 (f+2 g+4 h+8 i)) \log (2-x)+2 (d-e+f-g+h-i) \log (1+x)-(d-2 (e-2 f+4 g-8 h+16 i)) \log (2+x)\right ) \]
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Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.41
method | result | size |
default | \(\frac {i \,x^{2}}{2}+h x +\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}-\frac {g}{6}-\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2 g}{3}+\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x -2\right )\) | \(107\) |
norman | \(\frac {i \,x^{2}}{2}+h x +\left (-\frac {d}{12}+\frac {e}{6}-\frac {f}{3}+\frac {2 g}{3}-\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x +2\right )+\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x +1\right )+\left (-\frac {d}{6}-\frac {e}{6}-\frac {f}{6}-\frac {g}{6}-\frac {h}{6}-\frac {i}{6}\right ) \ln \left (x -1\right )+\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2 g}{3}+\frac {4 h}{3}+\frac {8 i}{3}\right ) \ln \left (x -2\right )\) | \(107\) |
parallelrisch | \(\frac {i \,x^{2}}{2}+\frac {\ln \left (x -2\right ) d}{12}+\frac {\ln \left (x -2\right ) e}{6}-\frac {\ln \left (x -1\right ) d}{6}-\frac {\ln \left (x -1\right ) e}{6}-\frac {\ln \left (x +1\right ) i}{6}-\frac {\ln \left (x +2\right ) f}{3}+\frac {\ln \left (x +1\right ) f}{6}-\frac {\ln \left (x +2\right ) d}{12}+h x +\frac {\ln \left (x +2\right ) e}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {8 \ln \left (x +2\right ) i}{3}-\frac {\ln \left (x +1\right ) g}{6}+\frac {2 \ln \left (x +2\right ) g}{3}+\frac {8 \ln \left (x -2\right ) i}{3}-\frac {\ln \left (x -1\right ) i}{6}+\frac {2 \ln \left (x -2\right ) g}{3}-\frac {\ln \left (x -1\right ) g}{6}+\frac {\ln \left (x -2\right ) f}{3}-\frac {\ln \left (x -1\right ) f}{6}+\frac {4 \ln \left (x -2\right ) h}{3}-\frac {\ln \left (x -1\right ) h}{6}+\frac {\ln \left (x +1\right ) h}{6}-\frac {4 \ln \left (x +2\right ) h}{3}\) | \(179\) |
risch | \(\frac {i \,x^{2}}{2}-\frac {\ln \left (x +1\right ) i}{6}-\frac {\ln \left (x +2\right ) f}{3}-\frac {\ln \left (1-x \right ) f}{6}+\frac {\ln \left (x +1\right ) f}{6}+\frac {\ln \left (2-x \right ) f}{3}+\frac {\ln \left (2-x \right ) d}{12}+\frac {\ln \left (2-x \right ) e}{6}-\frac {\ln \left (x +2\right ) d}{12}+h x +\frac {\ln \left (x +2\right ) e}{6}-\frac {\ln \left (1-x \right ) d}{6}-\frac {\ln \left (1-x \right ) e}{6}+\frac {\ln \left (x +1\right ) d}{6}-\frac {\ln \left (x +1\right ) e}{6}+\frac {8 \ln \left (2-x \right ) i}{3}-\frac {\ln \left (1-x \right ) i}{6}+\frac {8 \ln \left (x +2\right ) i}{3}-\frac {\ln \left (1-x \right ) g}{6}-\frac {\ln \left (x +1\right ) g}{6}+\frac {2 \ln \left (x +2\right ) g}{3}+\frac {2 \ln \left (2-x \right ) g}{3}-\frac {\ln \left (1-x \right ) h}{6}+\frac {4 \ln \left (2-x \right ) h}{3}+\frac {\ln \left (x +1\right ) h}{6}-\frac {4 \ln \left (x +2\right ) h}{3}\) | \(203\) |
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Time = 5.00 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, i x^{2} + h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \]
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Timed out. \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, i x^{2} + h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \]
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Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.21 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, i x^{2} + h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d - e + f - g + h - i\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, {\left (d + e + f + g + h + i\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 8.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.42 \[ \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{4-5 x^2+x^4} \, dx=h\,x+\frac {i\,x^2}{2}-\ln \left (x-1\right )\,\left (\frac {d}{6}+\frac {e}{6}+\frac {f}{6}+\frac {g}{6}+\frac {h}{6}+\frac {i}{6}\right )+\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}-\frac {i}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2\,g}{3}+\frac {4\,h}{3}+\frac {8\,i}{3}\right )-\ln \left (x+2\right )\,\left (\frac {d}{12}-\frac {e}{6}+\frac {f}{3}-\frac {2\,g}{3}+\frac {4\,h}{3}-\frac {8\,i}{3}\right ) \]
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