\(\int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} (4-4 x^2+x^4)+e^{\frac {e^2}{-2+x^2}} (48-48 x^2+4 e^2 x^2+12 x^4)}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} (8-8 x^2+2 x^4)+e^{\frac {e^2}{-2+x^2}} (80-80 x^2+20 x^4)} \, dx\) [4654]

Optimal result

Integrand size = 139, antiderivative size = 32 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{5+e^{\frac {e^2}{x \left (-\frac {2}{x}+x\right )}}}+\frac {1+x}{2} \]

[Out]

1/2*x+1/2+x/(exp(exp(2)/(x-2/x)/x)+5)

Rubi [F]

\[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx \]

[In]

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))*(8
0 - 80*x^2 + 20*x^4)),x]

[Out]

x/2 + Defer[Int][(5 + E^(E^2/(-2 + x^2)))^(-1), x] + (5*E^2*Defer[Int][1/((5 + E^(E^2/(-2 + x^2)))^2*(Sqrt[2]
- x)), x])/Sqrt[2] - (E^2*Defer[Int][1/((5 + E^(E^2/(-2 + x^2)))*(Sqrt[2] - x)), x])/Sqrt[2] + (5*E^2*Defer[In
t][1/((5 + E^(E^2/(-2 + x^2)))^2*(Sqrt[2] + x)), x])/Sqrt[2] - (E^2*Defer[Int][1/((5 + E^(E^2/(-2 + x^2)))*(Sq
rt[2] + x)), x])/Sqrt[2] - 20*E^2*Defer[Int][1/((5 + E^(E^2/(-2 + x^2)))^2*(-2 + x^2)^2), x] + 4*E^2*Defer[Int
][1/((5 + E^(E^2/(-2 + x^2)))*(-2 + x^2)^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^{2+\frac {e^2}{-2+x^2}} x^2+35 \left (-2+x^2\right )^2+12 e^{\frac {e^2}{-2+x^2}} \left (-2+x^2\right )^2+e^{\frac {2 e^2}{-2+x^2}} \left (-2+x^2\right )^2}{2 \left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (2-x^2\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {4 e^{2+\frac {e^2}{-2+x^2}} x^2+35 \left (-2+x^2\right )^2+12 e^{\frac {e^2}{-2+x^2}} \left (-2+x^2\right )^2+e^{\frac {2 e^2}{-2+x^2}} \left (-2+x^2\right )^2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (2-x^2\right )^2} \, dx \\ & = \frac {1}{2} \int \left (1-\frac {20 e^2 x^2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2}+\frac {2 \left (4-2 \left (2-e^2\right ) x^2+x^4\right )}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (2-x^2\right )^2}\right ) \, dx \\ & = \frac {x}{2}-\left (10 e^2\right ) \int \frac {x^2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2} \, dx+\int \frac {4-2 \left (2-e^2\right ) x^2+x^4}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (2-x^2\right )^2} \, dx \\ & = \frac {x}{2}-\left (10 e^2\right ) \int \left (\frac {2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2}+\frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )}\right ) \, dx+\int \left (\frac {1}{5+e^{\frac {e^2}{-2+x^2}}}+\frac {4 e^2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )^2}+\frac {2 e^2}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )}\right ) \, dx \\ & = \frac {x}{2}+\left (2 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )} \, dx+\left (4 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )^2} \, dx-\left (10 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )} \, dx-\left (20 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2} \, dx+\int \frac {1}{5+e^{\frac {e^2}{-2+x^2}}} \, dx \\ & = \frac {x}{2}+\left (2 e^2\right ) \int \left (-\frac {1}{2 \sqrt {2} \left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (\sqrt {2}-x\right )}-\frac {1}{2 \sqrt {2} \left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (\sqrt {2}+x\right )}\right ) \, dx+\left (4 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )^2} \, dx-\left (10 e^2\right ) \int \left (-\frac {1}{2 \sqrt {2} \left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (\sqrt {2}-x\right )}-\frac {1}{2 \sqrt {2} \left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (\sqrt {2}+x\right )}\right ) \, dx-\left (20 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2} \, dx+\int \frac {1}{5+e^{\frac {e^2}{-2+x^2}}} \, dx \\ & = \frac {x}{2}+\left (4 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (-2+x^2\right )^2} \, dx-\left (20 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (-2+x^2\right )^2} \, dx-\frac {e^2 \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (\sqrt {2}-x\right )} \, dx}{\sqrt {2}}-\frac {e^2 \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right ) \left (\sqrt {2}+x\right )} \, dx}{\sqrt {2}}+\frac {\left (5 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (\sqrt {2}-x\right )} \, dx}{\sqrt {2}}+\frac {\left (5 e^2\right ) \int \frac {1}{\left (5+e^{\frac {e^2}{-2+x^2}}\right )^2 \left (\sqrt {2}+x\right )} \, dx}{\sqrt {2}}+\int \frac {1}{5+e^{\frac {e^2}{-2+x^2}}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {1}{2} \left (x+\frac {2 x}{5+e^{\frac {e^2}{-2+x^2}}}\right ) \]

[In]

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]

[Out]

(x + (2*x)/(5 + E^(E^2/(-2 + x^2))))/2

Maple [A] (verified)

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	\(\frac {-7 x +\frac {7 x^{3}}{2}-{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}} x +\frac {{\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}} x^{3}}{2}}{\left ({\mathrm e}^{\frac {{\mathrm e}^{2}}{x^{2}-2}}+5\right ) \left (x^{2}-2\right )}\)	\(63\)

[In]

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)*exp(exp(2)/(x^2-2))+35*x^4-140*x^2+14
0)/((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=
_RETURNVERBOSE)

[Out]

1/2*x+x/(exp(exp(2)/(x^2-2))+5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 7 \, x}{2 \, {\left (e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 5\right )}} \]

[In]

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,
algorithm="fricas")

[Out]

1/2*(x*e^(e^2/(x^2 - 2)) + 7*x)/(e^(e^2/(x^2 - 2)) + 5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.47 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{2} + \frac {x}{e^{\frac {e^{2}}{x^{2} - 2}} + 5} \]

[In]

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)

[Out]

x/2 + x/(exp(exp(2)/(x**2 - 2)) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 7 \, x}{2 \, {\left (e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 5\right )}} \]

[In]

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,
algorithm="maxima")

[Out]

1/2*(x*e^(e^2/(x^2 - 2)) + 7*x)/(e^(e^2/(x^2 - 2)) + 5)

Giac [F]

\[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\int { \frac {35 \, x^{4} - 140 \, x^{2} + {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {2 \, e^{2}}{x^{2} - 2}\right )} + 4 \, {\left (3 \, x^{4} + x^{2} e^{2} - 12 \, x^{2} + 12\right )} e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 140}{2 \, {\left (25 \, x^{4} - 100 \, x^{2} + {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {2 \, e^{2}}{x^{2} - 2}\right )} + 10 \, {\left (x^{4} - 4 \, x^{2} + 4\right )} e^{\left (\frac {e^{2}}{x^{2} - 2}\right )} + 100\right )}} \,d x } \]

[In]

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,
algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.66 \[ \int \frac {140-140 x^2+35 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (4-4 x^2+x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (48-48 x^2+4 e^2 x^2+12 x^4\right )}{200-200 x^2+50 x^4+e^{\frac {2 e^2}{-2+x^2}} \left (8-8 x^2+2 x^4\right )+e^{\frac {e^2}{-2+x^2}} \left (80-80 x^2+20 x^4\right )} \, dx=\frac {x}{2}+\frac {x}{{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x^2-2}}+5} \]

[In]

int((exp(exp(2)/(x^2 - 2))*(4*x^2*exp(2) - 48*x^2 + 12*x^4 + 48) + exp((2*exp(2))/(x^2 - 2))*(x^4 - 4*x^2 + 4)
- 140*x^2 + 35*x^4 + 140)/(50*x^4 - 200*x^2 + exp((2*exp(2))/(x^2 - 2))*(2*x^4 - 8*x^2 + 8) + exp(exp(2)/(x^2
- 2))*(20*x^4 - 80*x^2 + 80) + 200),x)

[Out]

x/2 + x/(exp(exp(2)/(x^2 - 2)) + 5)