Integrand size = 51, antiderivative size = 92 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=(e-2 f+4 g-8 h+16 i) x+\frac {1}{2} (f-2 g+4 h-8 i) x^2+\frac {1}{3} (g-2 h+4 i) x^3+\frac {1}{4} (h-2 i) x^4+\frac {i x^5}{5}+(d-2 e+4 f-8 g+16 h-32 i) \log (2+x) \]
[Out]
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {1600, 1864} \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\log (x+2) (d-2 e+4 f-8 g+16 h-32 i)+x (e-2 f+4 g-8 h+16 i)+\frac {1}{2} x^2 (f-2 g+4 h-8 i)+\frac {1}{3} x^3 (g-2 h+4 i)+\frac {1}{4} x^4 (h-2 i)+\frac {i x^5}{5} \]
[In]
[Out]
Rule 1600
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{2+x} \, dx \\ & = \int \left (e \left (1-\frac {2 (f-2 g+4 h-8 i)}{e}\right )+(f-2 g+4 h-8 i) x+(g-2 h+4 i) x^2+(h-2 i) x^3+i x^4+\frac {d-2 e+4 f-8 g+16 h-32 i}{2+x}\right ) \, dx \\ & = (e-2 f+4 g-8 h+16 i) x+\frac {1}{2} (f-2 g+4 h-8 i) x^2+\frac {1}{3} (g-2 h+4 i) x^3+\frac {1}{4} (h-2 i) x^4+\frac {i x^5}{5}+(d-2 e+4 f-8 g+16 h-32 i) \log (2+x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=(e-2 f+4 g-8 h+16 i) x+\frac {1}{2} (f-2 g+4 h-8 i) x^2+\frac {1}{3} (g-2 h+4 i) x^3+\frac {1}{4} (h-2 i) x^4+\frac {i x^5}{5}+(d-2 e+4 f-8 g+16 h-32 i) \log (2+x) \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\left (\frac {h}{4}-\frac {i}{2}\right ) x^{4}+\left (\frac {g}{3}-\frac {2 h}{3}+\frac {4 i}{3}\right ) x^{3}+\left (\frac {f}{2}-g +2 h -4 i \right ) x^{2}+\left (e -2 f +4 g -8 h +16 i \right ) x +\frac {i \,x^{5}}{5}+\left (d -2 e +4 f -8 g +16 h -32 i \right ) \ln \left (x +2\right )\) | \(88\) |
default | \(\frac {i \,x^{5}}{5}+\frac {h \,x^{4}}{4}-\frac {i \,x^{4}}{2}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {4 i \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}-4 i \,x^{2}+e x -2 f x +4 g x -8 h x +16 i x +\left (d -2 e +4 f -8 g +16 h -32 i \right ) \ln \left (x +2\right )\) | \(103\) |
risch | \(\frac {i \,x^{5}}{5}+\frac {h \,x^{4}}{4}-\frac {i \,x^{4}}{2}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {4 i \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}-4 i \,x^{2}+e x -2 f x +4 g x -8 h x +16 i x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e +4 \ln \left (x +2\right ) f -8 \ln \left (x +2\right ) g +16 \ln \left (x +2\right ) h -32 \ln \left (x +2\right ) i\) | \(122\) |
parallelrisch | \(\frac {i \,x^{5}}{5}+\frac {h \,x^{4}}{4}-\frac {i \,x^{4}}{2}+\frac {g \,x^{3}}{3}-\frac {2 h \,x^{3}}{3}+\frac {4 i \,x^{3}}{3}+\frac {f \,x^{2}}{2}-g \,x^{2}+2 h \,x^{2}-4 i \,x^{2}+e x -2 f x +4 g x -8 h x +16 i x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e +4 \ln \left (x +2\right ) f -8 \ln \left (x +2\right ) g +16 \ln \left (x +2\right ) h -32 \ln \left (x +2\right ) i\) | \(122\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{5} \, i x^{5} + \frac {1}{4} \, {\left (h - 2 \, i\right )} x^{4} + \frac {1}{3} \, {\left (g - 2 \, h + 4 \, i\right )} x^{3} + \frac {1}{2} \, {\left (f - 2 \, g + 4 \, h - 8 \, i\right )} x^{2} + {\left (e - 2 \, f + 4 \, g - 8 \, h + 16 \, i\right )} x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) \]
[In]
[Out]
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {i x^{5}}{5} + x^{4} \left (\frac {h}{4} - \frac {i}{2}\right ) + x^{3} \left (\frac {g}{3} - \frac {2 h}{3} + \frac {4 i}{3}\right ) + x^{2} \left (\frac {f}{2} - g + 2 h - 4 i\right ) + x \left (e - 2 f + 4 g - 8 h + 16 i\right ) + \left (d - 2 e + 4 f - 8 g + 16 h - 32 i\right ) \log {\left (x + 2 \right )} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{5} \, i x^{5} + \frac {1}{4} \, {\left (h - 2 \, i\right )} x^{4} + \frac {1}{3} \, {\left (g - 2 \, h + 4 \, i\right )} x^{3} + \frac {1}{2} \, {\left (f - 2 \, g + 4 \, h - 8 \, i\right )} x^{2} + {\left (e - 2 \, f + 4 \, g - 8 \, h + 16 \, i\right )} x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=\frac {1}{5} \, i x^{5} + \frac {1}{4} \, h x^{4} - \frac {1}{2} \, i x^{4} + \frac {1}{3} \, g x^{3} - \frac {2}{3} \, h x^{3} + \frac {4}{3} \, i x^{3} + \frac {1}{2} \, f x^{2} - g x^{2} + 2 \, h x^{2} - 4 \, i x^{2} + e x - 2 \, f x + 4 \, g x - 8 \, h x + 16 \, i x + {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left ({\left | x + 2 \right |}\right ) \]
[In]
[Out]
Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\left (2-x-2 x^2+x^3\right ) \left (d+e x+f x^2+g x^3+h x^4+i x^5\right )}{4-5 x^2+x^4} \, dx=x^4\,\left (\frac {h}{4}-\frac {i}{2}\right )+\ln \left (x+2\right )\,\left (d-2\,e+4\,f-8\,g+16\,h-32\,i\right )+\frac {i\,x^5}{5}+x^2\,\left (\frac {f}{2}-g+2\,h-4\,i\right )+x\,\left (e-2\,f+4\,g-8\,h+16\,i\right )+x^3\,\left (\frac {g}{3}-\frac {2\,h}{3}+\frac {4\,i}{3}\right ) \]
[In]
[Out]