Integrand size = 39, antiderivative size = 23 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=\frac {20 \left (2+e^{4 x^2}\right ) x}{-\frac {1}{x}-x} \]
[Out]
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {28, 6874, 267, 6847, 2230, 2225, 2208, 2209} \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=-20 e^{4 x^2}+\frac {20 e^{4 x^2}}{x^2+1}+\frac {40}{x^2+1} \]
[In]
[Out]
Rule 28
Rule 267
Rule 2208
Rule 2209
Rule 2225
Rule 2230
Rule 6847
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{\left (1+x^2\right )^2} \, dx \\ & = \int \left (-\frac {80 x}{\left (1+x^2\right )^2}-\frac {40 e^{4 x^2} x \left (1+2 x^2\right )^2}{\left (1+x^2\right )^2}\right ) \, dx \\ & = -\left (40 \int \frac {e^{4 x^2} x \left (1+2 x^2\right )^2}{\left (1+x^2\right )^2} \, dx\right )-80 \int \frac {x}{\left (1+x^2\right )^2} \, dx \\ & = \frac {40}{1+x^2}-20 \text {Subst}\left (\int \frac {e^{4 x} (1+2 x)^2}{(1+x)^2} \, dx,x,x^2\right ) \\ & = \frac {40}{1+x^2}-20 \text {Subst}\left (\int \left (4 e^{4 x}+\frac {e^{4 x}}{(1+x)^2}-\frac {4 e^{4 x}}{1+x}\right ) \, dx,x,x^2\right ) \\ & = \frac {40}{1+x^2}-20 \text {Subst}\left (\int \frac {e^{4 x}}{(1+x)^2} \, dx,x,x^2\right )-80 \text {Subst}\left (\int e^{4 x} \, dx,x,x^2\right )+80 \text {Subst}\left (\int \frac {e^{4 x}}{1+x} \, dx,x,x^2\right ) \\ & = -20 e^{4 x^2}+\frac {40}{1+x^2}+\frac {20 e^{4 x^2}}{1+x^2}+\frac {80 \text {Ei}\left (4 \left (1+x^2\right )\right )}{e^4}-80 \text {Subst}\left (\int \frac {e^{4 x}}{1+x} \, dx,x,x^2\right ) \\ & = -20 e^{4 x^2}+\frac {40}{1+x^2}+\frac {20 e^{4 x^2}}{1+x^2} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=-\frac {20 \left (-2+e^{4 x^2} x^2\right )}{1+x^2} \]
[In]
[Out]
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {-20 x^{2} {\mathrm e}^{4 x^{2}}+40}{x^{2}+1}\) | \(22\) |
parallelrisch | \(-\frac {20 x^{2} {\mathrm e}^{4 x^{2}}-40}{x^{2}+1}\) | \(23\) |
risch | \(\frac {40}{x^{2}+1}-\frac {20 x^{2} {\mathrm e}^{4 x^{2}}}{x^{2}+1}\) | \(29\) |
parts | \(\frac {40}{x^{2}+1}-\frac {20 x^{2} {\mathrm e}^{4 x^{2}}}{x^{2}+1}\) | \(29\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=-\frac {20 \, {\left (x^{2} e^{\left (4 \, x^{2}\right )} - 2\right )}}{x^{2} + 1} \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=- \frac {20 x^{2} e^{4 x^{2}}}{x^{2} + 1} + \frac {80}{2 x^{2} + 2} \]
[In]
[Out]
\[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=\int { -\frac {40 \, {\left ({\left (4 \, x^{5} + 4 \, x^{3} + x\right )} e^{\left (4 \, x^{2}\right )} + 2 \, x\right )}}{x^{4} + 2 \, x^{2} + 1} \,d x } \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=-\frac {20 \, {\left (x^{2} e^{\left (4 \, x^{2}\right )} - 2\right )}}{x^{2} + 1} \]
[In]
[Out]
Time = 12.76 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-80 x+e^{4 x^2} \left (-40 x-160 x^3-160 x^5\right )}{1+2 x^2+x^4} \, dx=-\frac {20\,x^2\,\left ({\mathrm {e}}^{4\,x^2}+2\right )}{x^2+1} \]
[In]
[Out]