\(\int \frac {1}{9} e^{3^{e^{-x} x}} (e^x (2 x+4 x^2+x^3)+3^{e^{-x} x} (x^2-x^4) \log (3)) \, dx\) [6089]

Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} e^{3^{e^{-x} x}+x} x^2 (1+x) \]

[Out]

1/9*x^2*exp(x)*exp(exp(x*ln(3)/exp(x)))*(1+x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {e^{3^{e^{-x} x}} \left (x^2-x^4\right )}{9 \left (e^{-x}-e^{-x} x\right )} \]

[In]

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

[Out]

(E^3^(x/E^x)*(x^2 - x^4))/(9*(E^(-x) - x/E^x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} e^{3^{e^{-x} x}+x} x^2 (1+x) \]

[In]

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

[Out]

(E^(3^(x/E^x) + x)*x^2*(1 + x))/9

Maple [A] (verified)

Time = 4.79 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {x^{2} \left (1+x \right ) {\mathrm e}^{x +3^{x \,{\mathrm e}^{-x}}}}{9}\)	\(20\)
parallelrisch	\(\frac {x^{3} {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x \ln \left (3\right ) {\mathrm e}^{-x}}}}{9}+\frac {{\mathrm e}^{{\mathrm e}^{x \ln \left (3\right ) {\mathrm e}^{-x}}} {\mathrm e}^{x} x^{2}}{9}\)	\(36\)

[In]

int(1/9*((-x^4+x^2)*ln(3)*exp(x*ln(3)/exp(x))+(x^3+4*x^2+2*x)*exp(x))*exp(exp(x*ln(3)/exp(x))),x,method=_RETUR
NVERBOSE)

[Out]

1/9*x^2*(1+x)*exp(x+3^(x*exp(-x)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} \, {\left (x^{3} + x^{2}\right )} e^{\left (3^{x e^{\left (-x\right )}} + x\right )} \]

[In]

integrate(1/9*((-x^4+x^2)*log(3)*exp(x*log(3)/exp(x))+(x^3+4*x^2+2*x)*exp(x))*exp(exp(x*log(3)/exp(x))),x, alg
orithm="fricas")

[Out]

1/9*(x^3 + x^2)*e^(3^(x*e^(-x)) + x)

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {\left (x^{3} e^{x} + x^{2} e^{x}\right ) e^{e^{x e^{- x} \log {\left (3 \right )}}}}{9} \]

[In]

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

[Out]

(x**3*exp(x) + x**2*exp(x))*exp(exp(x*exp(-x)*log(3)))/9

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {1}{9} \, {\left (x^{3} + x^{2}\right )} e^{\left (3^{x e^{\left (-x\right )}} + x\right )} \]

[In]

integrate(1/9*((-x^4+x^2)*log(3)*exp(x*log(3)/exp(x))+(x^3+4*x^2+2*x)*exp(x))*exp(exp(x*log(3)/exp(x))),x, alg
orithm="maxima")

[Out]

1/9*(x^3 + x^2)*e^(3^(x*e^(-x)) + x)

Giac [F]

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

[In]

integrate(1/9*((-x^4+x^2)*log(3)*exp(x*log(3)/exp(x))+(x^3+4*x^2+2*x)*exp(x))*exp(exp(x*log(3)/exp(x))),x, alg
orithm="giac")

[Out]

integrate(-1/9*((x^4 - x^2)*3^(x*e^(-x))*log(3) - (x^3 + 4*x^2 + 2*x)*e^x)*e^(3^(x*e^(-x))), x)

Mupad [B] (verification not implemented)

Time = 12.60 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{9} e^{3^{e^{-x} x}} \left (e^x \left (2 x+4 x^2+x^3\right )+3^{e^{-x} x} \left (x^2-x^4\right ) \log (3)\right ) \, dx=\frac {x^2\,{\mathrm {e}}^{3^{x\,{\mathrm {e}}^{-x}}}\,{\mathrm {e}}^x\,\left (x+1\right )}{9} \]

[In]

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

[Out]

(x^2*exp(3^(x*exp(-x)))*exp(x)*(x + 1))/9