\(\int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} (e^x (-1-9 x+x^2+x^3)+e^{3 x} (4 x+8 x^2+4 x^3))}{x^2+2 x^3+x^4} \, dx\) [5330]

Optimal result

Integrand size = 90, antiderivative size = 24 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=\int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx \]

[In]

Int[(E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/(1 + x))/x + 8/(1 + x))*(E^x*(-1 - 9*x + x^2 + x^3) + E^(3*x)*(4*x +
8*x^2 + 4*x^3)))/(x^2 + 2*x^3 + x^4),x]

[Out]

-Defer[Int][E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/(1 + x))/x + x + 8/(1 + x))/x^2, x] - 7*Defer[Int][E^(2*E^(2*x
) + E^(2*E^(2*x) + x + 8/(1 + x))/x + x + 8/(1 + x))/x, x] + 4*Defer[Int][E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/
(1 + x))/x + 3*x + 8/(1 + x))/x, x] + 8*Defer[Int][E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/(1 + x))/x + x + 8/(1 +
x))/(1 + x)^2, x] + 8*Defer[Int][E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/(1 + x))/x + x + 8/(1 + x))/(1 + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2 \left (1+2 x+x^2\right )} \, dx \\ & = \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2 (1+x)^2} \, dx \\ & = \int \left (\frac {4 \exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+3 x+\frac {8}{1+x}\right )}{x}+\frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right ) \left (-1-9 x+x^2+x^3\right )}{x^2 (1+x)^2}\right ) \, dx \\ & = 4 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+3 x+\frac {8}{1+x}\right )}{x} \, dx+\int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right ) \left (-1-9 x+x^2+x^3\right )}{x^2 (1+x)^2} \, dx \\ & = 4 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+3 x+\frac {8}{1+x}\right )}{x} \, dx+\int \left (-\frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{x^2}-\frac {7 \exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{x}+\frac {8 \exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{(1+x)^2}+\frac {8 \exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{1+x}\right ) \, dx \\ & = 4 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+3 x+\frac {8}{1+x}\right )}{x} \, dx-7 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{x} \, dx+8 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{(1+x)^2} \, dx+8 \int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{1+x} \, dx-\int \frac {\exp \left (2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+x+\frac {8}{1+x}\right )}{x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}} \]

[In]

Integrate[(E^(2*E^(2*x) + E^(2*E^(2*x) + x + 8/(1 + x))/x + 8/(1 + x))*(E^x*(-1 - 9*x + x^2 + x^3) + E^(3*x)*(
4*x + 8*x^2 + 4*x^3)))/(x^2 + 2*x^3 + x^4),x]

[Out]

E^(E^(2*E^(2*x) + x + 8/(1 + x))/x)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

\[{\mathrm e}^{\frac {{\mathrm e}^{\frac {2 x \,{\mathrm e}^{2 x}+x^{2}+2 \,{\mathrm e}^{2 x}+x +8}{1+x}}}{x}}\]

[In]

int(((4*x^3+8*x^2+4*x)*exp(x)*exp(2*x)+(x^3+x^2-9*x-1)*exp(x))*exp(exp(2*x))^2*exp(exp(x)*exp(exp(2*x))^2/x/ex
p(-4/(1+x))^2)/(x^4+2*x^3+x^2)/exp(-4/(1+x))^2,x)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (21) = 42\).

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.08 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (x + \frac {2 \, {\left (x^{2} + x\right )} e^{\left (2 \, x\right )} + {\left (x + 1\right )} e^{\left (\frac {x^{2} + 2 \, {\left (x + 1\right )} e^{\left (2 \, x\right )} + x + 8}{x + 1}\right )} + 8 \, x}{x^{2} + x} - \frac {x^{2} + 2 \, {\left (x + 1\right )} e^{\left (2 \, x\right )} + x + 8}{x + 1}\right )} \]

[In]

integrate(((4*x^3+8*x^2+4*x)*exp(x)*exp(2*x)+(x^3+x^2-9*x-1)*exp(x))*exp(exp(2*x))^2*exp(exp(x)*exp(exp(2*x))^
2/x/exp(-4/(1+x))^2)/(x^4+2*x^3+x^2)/exp(-4/(1+x))^2,x, algorithm="fricas")

[Out]

e^(x + (2*(x^2 + x)*e^(2*x) + (x + 1)*e^((x^2 + 2*(x + 1)*e^(2*x) + x + 8)/(x + 1)) + 8*x)/(x^2 + x) - (x^2 +
2*(x + 1)*e^(2*x) + x + 8)/(x + 1))

Sympy [A] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\frac {e^{x} e^{\frac {8}{x + 1}} e^{2 e^{2 x}}}{x}} \]

[In]

integrate(((4*x**3+8*x**2+4*x)*exp(x)*exp(2*x)+(x**3+x**2-9*x-1)*exp(x))*exp(exp(2*x))**2*exp(exp(x)*exp(exp(2
*x))**2/x/exp(-4/(1+x))**2)/(x**4+2*x**3+x**2)/exp(-4/(1+x))**2,x)

[Out]

exp(exp(x)*exp(8/(x + 1))*exp(2*exp(2*x))/x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (x + \frac {8}{x + 1} + 2 \, e^{\left (2 \, x\right )}\right )}}{x}\right )} \]

[In]

integrate(((4*x^3+8*x^2+4*x)*exp(x)*exp(2*x)+(x^3+x^2-9*x-1)*exp(x))*exp(exp(2*x))^2*exp(exp(x)*exp(exp(2*x))^
2/x/exp(-4/(1+x))^2)/(x^4+2*x^3+x^2)/exp(-4/(1+x))^2,x, algorithm="maxima")

[Out]

e^(e^(x + 8/(x + 1) + 2*e^(2*x))/x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx=e^{\left (\frac {e^{\left (x + \frac {8}{x + 1} + 2 \, e^{\left (2 \, x\right )}\right )}}{x}\right )} \]

[In]

integrate(((4*x^3+8*x^2+4*x)*exp(x)*exp(2*x)+(x^3+x^2-9*x-1)*exp(x))*exp(exp(2*x))^2*exp(exp(x)*exp(exp(2*x))^
2/x/exp(-4/(1+x))^2)/(x^4+2*x^3+x^2)/exp(-4/(1+x))^2,x, algorithm="giac")

[Out]

e^(e^(x + 8/(x + 1) + 2*e^(2*x))/x)

Mupad [B] (verification not implemented)

Time = 10.63 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{2 e^{2 x}+\frac {e^{2 e^{2 x}+x+\frac {8}{1+x}}}{x}+\frac {8}{1+x}} \left (e^x \left (-1-9 x+x^2+x^3\right )+e^{3 x} \left (4 x+8 x^2+4 x^3\right )\right )}{x^2+2 x^3+x^4} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{\frac {8}{x+1}}}{x}} \]

[In]

int((exp(2*exp(2*x))*exp((exp(2*exp(2*x))*exp(x)*exp(8/(x + 1)))/x)*exp(8/(x + 1))*(exp(3*x)*(4*x + 8*x^2 + 4*
x^3) - exp(x)*(9*x - x^2 - x^3 + 1)))/(x^2 + 2*x^3 + x^4),x)

[Out]

exp((exp(2*exp(2*x))*exp(x)*exp(8/(x + 1)))/x)