\(\int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} (12-2 e^{1+e^{1+x}+x} x^3)}{x^3} \, dx\) [8999]

Optimal result

Integrand size = 39, antiderivative size = 18 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \]

[Out]

exp(-exp(exp(1+x))+6)^2/exp(2)^2/exp(3/x^2)^2

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6838} \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{-\frac {6}{x^2}-2 e^{e^{x+1}}+8} \]

[In]

Int[(E^(8 - 2*E^E^(1 + x) - 6/x^2)*(12 - 2*E^(1 + E^(1 + x) + x)*x^3))/x^3,x]

[Out]

E^(8 - 2*E^E^(1 + x) - 6/x^2)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \]

[In]

Integrate[(E^(8 - 2*E^E^(1 + x) - 6/x^2)*(12 - 2*E^(1 + E^(1 + x) + x)*x^3))/x^3,x]

[Out]

E^(8 - 2*E^E^(1 + x) - 6/x^2)

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left ({\mathrm e}^{{\mathrm e}^{1+x}} x^{2}-4 x^{2}+3\right )}{x^{2}}}\)	\(23\)
parallelrisch	\({\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{1+x}}+12} {\mathrm e}^{-4} {\mathrm e}^{-\frac {6}{x^{2}}}\)	\(26\)

[In]

int((-2*x^3*exp(1+x)*exp(exp(1+x))+12)*exp(-exp(exp(1+x))+6)^2/x^3/exp(2)^2/exp(3/x^2)^2,x,method=_RETURNVERBO
SE)

[Out]

exp(-2*(exp(exp(1+x))*x^2-4*x^2+3)/x^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.11 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{\left (-\frac {2 \, {\left (x^{2} e^{\left (x + e^{\left (x + 1\right )} + 1\right )} - {\left (4 \, x^{2} - 3\right )} e^{\left (x + 1\right )}\right )} e^{\left (-x - 1\right )}}{x^{2}}\right )} \]

[In]

integrate((-2*x^3*exp(1+x)*exp(exp(1+x))+12)*exp(-exp(exp(1+x))+6)^2/x^3/exp(2)^2/exp(3/x^2)^2,x, algorithm="f
ricas")

[Out]

e^(-2*(x^2*e^(x + e^(x + 1) + 1) - (4*x^2 - 3)*e^(x + 1))*e^(-x - 1)/x^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=\text {Timed out} \]

[In]

integrate((-2*x**3*exp(1+x)*exp(exp(1+x))+12)*exp(-exp(exp(1+x))+6)**2/x**3/exp(2)**2/exp(3/x**2)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{\left (-\frac {6}{x^{2}} - 2 \, e^{\left (e^{\left (x + 1\right )}\right )} + 8\right )} \]

[In]

integrate((-2*x^3*exp(1+x)*exp(exp(1+x))+12)*exp(-exp(exp(1+x))+6)^2/x^3/exp(2)^2/exp(3/x^2)^2,x, algorithm="m
axima")

[Out]

e^(-6/x^2 - 2*e^(e^(x + 1)) + 8)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx=e^{\left (-\frac {6}{x^{2}} - 2 \, e^{\left (e^{\left (x + 1\right )}\right )} + 8\right )} \]

[In]

integrate((-2*x^3*exp(1+x)*exp(exp(1+x))+12)*exp(-exp(exp(1+x))+6)^2/x^3/exp(2)^2/exp(3/x^2)^2,x, algorithm="g
iac")

[Out]

e^(-6/x^2 - 2*e^(e^(x + 1)) + 8)

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{8-2 e^{e^{1+x}}-\frac {6}{x^2}} \left (12-2 e^{1+e^{1+x}+x} x^3\right )}{x^3} \, dx={\mathrm {e}}^8\,{\mathrm {e}}^{-\frac {6}{x^2}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^x}} \]

[In]

int(-(exp(-4)*exp(-6/x^2)*exp(12 - 2*exp(exp(x + 1)))*(2*x^3*exp(x + 1)*exp(exp(x + 1)) - 12))/x^3,x)

[Out]

exp(8)*exp(-6/x^2)*exp(-2*exp(exp(1)*exp(x)))