\(\int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} (40-5 x^2-60 x^4)}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 (8+x^2+4 x^4)}{x}} x^2} \, dx\) [3468]

Optimal result

Integrand size = 86, antiderivative size = 23 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5}{e^{-1+x+\frac {4 \left (2+x^4\right )}{x}}+\frac {1}{x}} \]

[Out]

5/(exp(4/x*(x^4+2)+x)/exp(1)+1/x)

Rubi [F]

\[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx \]

[In]

Int[(5*E^2 + E^(1 + (8 + x^2 + 4*x^4)/x)*(40 - 5*x^2 - 60*x^4))/(E^2 + 2*E^(1 + (8 + x^2 + 4*x^4)/x)*x + E^((2
*(8 + x^2 + 4*x^4))/x)*x^2),x]

[Out]

5*E^2*Defer[Int][(E + E^(8/x + x + 4*x^3)*x)^(-2), x] - 40*E^2*Defer[Int][1/(x*(E + E^(8/x + x + 4*x^3)*x)^2),
x] + 5*E^2*Defer[Int][x/(E + E^(8/x + x + 4*x^3)*x)^2, x] + 60*E^2*Defer[Int][x^3/(E + E^(8/x + x + 4*x^3)*x)
^2, x] + 40*E*Defer[Int][1/(x*(E + E^(8/x + x + 4*x^3)*x)), x] - 5*E*Defer[Int][x/(E + E^(8/x + x + 4*x^3)*x),
x] - 60*E*Defer[Int][x^3/(E + E^(8/x + x + 4*x^3)*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e \left (e-e^{\frac {8}{x}+x+4 x^3} \left (-8+x^2+12 x^4\right )\right )}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx \\ & = (5 e) \int \frac {e-e^{\frac {8}{x}+x+4 x^3} \left (-8+x^2+12 x^4\right )}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx \\ & = (5 e) \int \left (-\frac {-8+x^2+12 x^4}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )}+\frac {e \left (-8+x+x^2+12 x^4\right )}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2}\right ) \, dx \\ & = -\left ((5 e) \int \frac {-8+x^2+12 x^4}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )} \, dx\right )+\left (5 e^2\right ) \int \frac {-8+x+x^2+12 x^4}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx \\ & = -\left ((5 e) \int \left (-\frac {8}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )}+\frac {x}{e+e^{\frac {8}{x}+x+4 x^3} x}+\frac {12 x^3}{e+e^{\frac {8}{x}+x+4 x^3} x}\right ) \, dx\right )+\left (5 e^2\right ) \int \left (\frac {1}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2}-\frac {8}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2}+\frac {x}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2}+\frac {12 x^3}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2}\right ) \, dx \\ & = -\left ((5 e) \int \frac {x}{e+e^{\frac {8}{x}+x+4 x^3} x} \, dx\right )+(40 e) \int \frac {1}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )} \, dx-(60 e) \int \frac {x^3}{e+e^{\frac {8}{x}+x+4 x^3} x} \, dx+\left (5 e^2\right ) \int \frac {1}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx+\left (5 e^2\right ) \int \frac {x}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx-\left (40 e^2\right ) \int \frac {1}{x \left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx+\left (60 e^2\right ) \int \frac {x^3}{\left (e+e^{\frac {8}{x}+x+4 x^3} x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5 e x}{e+e^{\frac {8}{x}+x+4 x^3} x} \]

[In]

Integrate[(5*E^2 + E^(1 + (8 + x^2 + 4*x^4)/x)*(40 - 5*x^2 - 60*x^4))/(E^2 + 2*E^(1 + (8 + x^2 + 4*x^4)/x)*x +
E^((2*(8 + x^2 + 4*x^4))/x)*x^2),x]

[Out]

(5*E*x)/(E + E^(8/x + x + 4*x^3)*x)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22

method	result	size
norman	\(\frac {5 x \,{\mathrm e}}{x \,{\mathrm e}^{\frac {4 x^{4}+x^{2}+8}{x}}+{\mathrm e}}\)	\(28\)
risch	\(\frac {5 x \,{\mathrm e}}{x \,{\mathrm e}^{\frac {4 x^{4}+x^{2}+8}{x}}+{\mathrm e}}\)	\(28\)
parallelrisch	\(\frac {5 x \,{\mathrm e}}{x \,{\mathrm e}^{\frac {4 x^{4}+x^{2}+8}{x}}+{\mathrm e}}\)	\(28\)

[In]

int(((-60*x^4-5*x^2+40)*exp(1)*exp((4*x^4+x^2+8)/x)+5*exp(1)^2)/(x^2*exp((4*x^4+x^2+8)/x)^2+2*x*exp(1)*exp((4*
x^4+x^2+8)/x)+exp(1)^2),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(1)/(x*exp((4*x^4+x^2+8)/x)+exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5 \, x e^{2}}{x e^{\left (\frac {4 \, x^{4} + x^{2} + x + 8}{x}\right )} + e^{2}} \]

[In]

integrate(((-60*x^4-5*x^2+40)*exp(1)*exp((4*x^4+x^2+8)/x)+5*exp(1)^2)/(x^2*exp((4*x^4+x^2+8)/x)^2+2*x*exp(1)*e
xp((4*x^4+x^2+8)/x)+exp(1)^2),x, algorithm="fricas")

[Out]

5*x*e^2/(x*e^((4*x^4 + x^2 + x + 8)/x) + e^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5 e x}{x e^{\frac {4 x^{4} + x^{2} + 8}{x}} + e} \]

[In]

integrate(((-60*x**4-5*x**2+40)*exp(1)*exp((4*x**4+x**2+8)/x)+5*exp(1)**2)/(x**2*exp((4*x**4+x**2+8)/x)**2+2*x
*exp(1)*exp((4*x**4+x**2+8)/x)+exp(1)**2),x)

[Out]

5*E*x/(x*exp((4*x**4 + x**2 + 8)/x) + E)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5 \, x e}{x e^{\left (4 \, x^{3} + x + \frac {8}{x}\right )} + e} \]

[In]

integrate(((-60*x^4-5*x^2+40)*exp(1)*exp((4*x^4+x^2+8)/x)+5*exp(1)^2)/(x^2*exp((4*x^4+x^2+8)/x)^2+2*x*exp(1)*e
xp((4*x^4+x^2+8)/x)+exp(1)^2),x, algorithm="maxima")

[Out]

5*x*e/(x*e^(4*x^3 + x + 8/x) + e)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5 \, x e}{x e^{\left (\frac {4 \, x^{4} + x^{2} + 8}{x}\right )} + e} \]

[In]

integrate(((-60*x^4-5*x^2+40)*exp(1)*exp((4*x^4+x^2+8)/x)+5*exp(1)^2)/(x^2*exp((4*x^4+x^2+8)/x)^2+2*x*exp(1)*e
xp((4*x^4+x^2+8)/x)+exp(1)^2),x, algorithm="giac")

[Out]

5*x*e/(x*e^((4*x^4 + x^2 + 8)/x) + e)

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {5 e^2+e^{1+\frac {8+x^2+4 x^4}{x}} \left (40-5 x^2-60 x^4\right )}{e^2+2 e^{1+\frac {8+x^2+4 x^4}{x}} x+e^{\frac {2 \left (8+x^2+4 x^4\right )}{x}} x^2} \, dx=\frac {5\,x\,\mathrm {e}}{\mathrm {e}+x\,{\mathrm {e}}^{4\,x^3}\,{\mathrm {e}}^{8/x}\,{\mathrm {e}}^x} \]

[In]

int((5*exp(2) - exp((x^2 + 4*x^4 + 8)/x)*exp(1)*(5*x^2 + 60*x^4 - 40))/(exp(2) + x^2*exp((2*(x^2 + 4*x^4 + 8))
/x) + 2*x*exp((x^2 + 4*x^4 + 8)/x)*exp(1)),x)

[Out]

(5*x*exp(1))/(exp(1) + x*exp(4*x^3)*exp(8/x)*exp(x))