Integrand size = 139, antiderivative size = 32 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{5+e^{\frac {e^2}{x \left (-\frac {2}{x}+x\right )}}}+\frac {1+x}{2} \]
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {1}{2} \left (x+\frac {2 x}{5+e^{\frac {e^2}{-2+x^2}}}\right ) \]
Integrate[(140 - 140*x^2 + 35*x^4 + E^((2*E^2)/(-2 + x^2))*(4 - 4*x^2 + x^ 4) + E^(E^2/(-2 + x^2))*(48 - 48*x^2 + 4*E^2*x^2 + 12*x^4))/(200 - 200*x^2 + 50*x^4 + E^((2*E^2)/(-2 + x^2))*(8 - 8*x^2 + 2*x^4) + E^(E^2/(-2 + x^2) )*(80 - 80*x^2 + 20*x^4)),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {35 x^4-140 x^2+e^{\frac {2 e^2}{x^2-2}} \left (x^4-4 x^2+4\right )+e^{\frac {e^2}{x^2-2}} \left (12 x^4+4 e^2 x^2-48 x^2+48\right )+140}{50 x^4-200 x^2+e^{\frac {2 e^2}{x^2-2}} \left (2 x^4-8 x^2+8\right )+e^{\frac {e^2}{x^2-2}} \left (20 x^4-80 x^2+80\right )+200} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {35 x^4-140 x^2+e^{\frac {2 e^2}{x^2-2}} \left (x^4-4 x^2+4\right )+e^{\frac {e^2}{x^2-2}} \left (12 x^4+4 e^2 x^2-48 x^2+48\right )+140}{2 \left (e^{\frac {e^2}{x^2-2}}+5\right )^2 \left (2-x^2\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {35 x^4-140 x^2+e^{-\frac {2 e^2}{2-x^2}} \left (x^4-4 x^2+4\right )+4 e^{-\frac {e^2}{2-x^2}} \left (3 x^4+e^2 x^2-12 x^2+12\right )+140}{\left (5+e^{-\frac {e^2}{2-x^2}}\right )^2 \left (2-x^2\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{2} \int \left (-\frac {4 e^{2+\frac {2 e^2}{x^2-2}} x^2}{5 \left (5+e^{\frac {e^2}{x^2-2}}\right )^2 \left (x^2-2\right )^2}+\frac {2 e^{\frac {e^2}{x^2-2}} \left (-x^4+2 \left (2+e^2\right ) x^2-4\right )}{5 \left (5+e^{\frac {e^2}{x^2-2}}\right ) \left (2-x^2\right )^2}+\frac {7}{5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{5+e^{\frac {e^2}{x^2-2}}}dx+\frac {1}{5} \sqrt {2} \int \frac {e^{2+\frac {2 e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right )^2 \left (\sqrt {2}-x\right )}dx-\frac {1}{5} \sqrt {2} \int \frac {e^{2+\frac {e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right ) \left (\sqrt {2}-x\right )}dx+\frac {1}{5} \sqrt {2} \int \frac {e^{2+\frac {2 e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right )^2 \left (x+\sqrt {2}\right )}dx-\frac {1}{5} \sqrt {2} \int \frac {e^{2+\frac {e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right ) \left (x+\sqrt {2}\right )}dx-\frac {8}{5} \int \frac {e^{2+\frac {2 e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right )^2 \left (x^2-2\right )^2}dx+\frac {8}{5} \int \frac {e^{2+\frac {e^2}{x^2-2}}}{\left (5+e^{\frac {e^2}{x^2-2}}\right ) \left (x^2-2\right )^2}dx+x\right )\) |
Int[(140 - 140*x^2 + 35*x^4 + E^((2*E^2)/(-2 + x^2))*(4 - 4*x^2 + x^4) + E ^(E^2/(-2 + x^2))*(48 - 48*x^2 + 4*E^2*x^2 + 12*x^4))/(200 - 200*x^2 + 50* x^4 + E^((2*E^2)/(-2 + x^2))*(8 - 8*x^2 + 2*x^4) + E^(E^2/(-2 + x^2))*(80 - 80*x^2 + 20*x^4)),x]
3.17.54.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69
| method | result | size |
| risch | \(\frac {x}{2}+\frac {x}{{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}}+5}\) | \(22\) |
| parallelrisch | \(\frac {{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}} x +7 x}{2 \,{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}}+10}\) | \(35\) |
| norman | \(\frac {-7 x +\frac {7 x^{3}}{2}-{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}} x +\frac {{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}} x^{3}}{2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}}+5\right ) \left (x^{2}-2\right )}\) | \(63\) |
int(((x^4-4*x^2+4)*exp(exp(2)/(x^2-2))^2+(4*x^2*exp(2)+12*x^4-48*x^2+48)*e xp(exp(2)/(x^2-2))+35*x^4-140*x^2+140)/((2*x^4-8*x^2+8)*exp(exp(2)/(x^2-2) )^2+(20*x^4-80*x^2+80)*exp(exp(2)/(x^2-2))+50*x^4-200*x^2+200),x,method=_R ETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 7 \, x}{2 \, {\left (e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 5\right )}} \]
integrate(((x^4-4*x^2+4)*exp(exp(2)/(x^2-2))^2+(4*x^2*exp(2)+12*x^4-48*x^2 +48)*exp(exp(2)/(x^2-2))+35*x^4-140*x^2+140)/((2*x^4-8*x^2+8)*exp(exp(2)/( x^2-2))^2+(20*x^4-80*x^2+80)*exp(exp(2)/(x^2-2))+50*x^4-200*x^2+200),x, al gorithm=\
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.47 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{2} + \frac {x}{e^{\frac {e^{2}}{x^{2} - 2}} + 5} \]
integrate(((x**4-4*x**2+4)*exp(exp(2)/(x**2-2))**2+(4*x**2*exp(2)+12*x**4- 48*x**2+48)*exp(exp(2)/(x**2-2))+35*x**4-140*x**2+140)/((2*x**4-8*x**2+8)* exp(exp(2)/(x**2-2))**2+(20*x**4-80*x**2+80)*exp(exp(2)/(x**2-2))+50*x**4- 200*x**2+200),x)
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 7 \, x}{2 \, {\left (e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 5\right )}} \]
integrate(((x^4-4*x^2+4)*exp(exp(2)/(x^2-2))^2+(4*x^2*exp(2)+12*x^4-48*x^2 +48)*exp(exp(2)/(x^2-2))+35*x^4-140*x^2+140)/((2*x^4-8*x^2+8)*exp(exp(2)/( x^2-2))^2+(20*x^4-80*x^2+80)*exp(exp(2)/(x^2-2))+50*x^4-200*x^2+200),x, al gorithm=\
\[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\int { \frac {35 \, x^{4} - 140 \, x^{2} + {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {2 \, e^{2}}{x^{2} - 2}\right )} + 4 \, {\left (3 \, x^{4} + x^{2} e^{2} - 12 \, x^{2} + 12\right )} e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 140}{2 \, {\left (25 \, x^{4} - 100 \, x^{2} + {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {2 \, e^{2}}{x^{2} - 2}\right )} + 10 \, {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 100\right )}} \,d x } \]
integrate(((x^4-4*x^2+4)*exp(exp(2)/(x^2-2))^2+(4*x^2*exp(2)+12*x^4-48*x^2 +48)*exp(exp(2)/(x^2-2))+35*x^4-140*x^2+140)/((2*x^4-8*x^2+8)*exp(exp(2)/( x^2-2))^2+(20*x^4-80*x^2+80)*exp(exp(2)/(x^2-2))+50*x^4-200*x^2+200),x, al gorithm=\
integrate(1/2*(35*x^4 - 140*x^2 + (x^4 - 4*x^2 + 4)*e^(2*e^2/(x^2 - 2)) + 4*(3*x^4 + x^2*e^2 - 12*x^2 + 12)*e^(e^2/(x^2 - 2)) + 140)/(25*x^4 - 100*x ^2 + (x^4 - 4*x^2 + 4)*e^(2*e^2/(x^2 - 2)) + 10*(x^4 - 4*x^2 + 4)*e^(e^2/( x^2 - 2)) + 100), x)
Time = 11.72 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{2}+\frac {x}{{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x^2-2}}+5} \]