\(\int \frac {1}{9} e^{e^{\frac {1}{9} (4 e^{16} x-8 e^8 x^2+4 x^3)} (3+x)+\frac {1}{9} (4 e^{16} x-8 e^8 x^2+4 x^3)} (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 (-48 x-16 x^2)) \, dx\) [7467]

Optimal result

Integrand size = 95, antiderivative size = 24 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=1+e^{e^{\frac {4}{9} x \left (-e^8+x\right )^2} (3+x)} \]

[Out]

1+exp((3+x)*exp(4/3*(x-exp(8))*(1/3*x-1/3*exp(8))*x))

Rubi [F]

\[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=\int \frac {1}{9} \exp \left (e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )\right ) \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx \]

[In]

Int[(E^(E^((4*E^16*x - 8*E^8*x^2 + 4*x^3)/9)*(3 + x) + (4*E^16*x - 8*E^8*x^2 + 4*x^3)/9)*(9 + 36*x^2 + 12*x^3
+ E^16*(12 + 4*x) + E^8*(-48*x - 16*x^2)))/9,x]

[Out]

((3 + 4*E^16)*Defer[Int][E^((4*(E^8 - x)^2*x)/9 + E^((4*(E^8 - x)^2*x)/9)*(3 + x)), x])/3 - (4*(12 - E^8)*Defe
r[Int][E^(8 + (4*(E^8 - x)^2*x)/9 + E^((4*(E^8 - x)^2*x)/9)*(3 + x))*x, x])/9 + (4*(9 - 4*E^8)*Defer[Int][E^((
4*(E^8 - x)^2*x)/9 + E^((4*(E^8 - x)^2*x)/9)*(3 + x))*x^2, x])/9 + (4*Defer[Int][E^((4*(E^8 - x)^2*x)/9 + E^((
4*(E^8 - x)^2*x)/9)*(3 + x))*x^3, x])/3

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \exp \left (e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )\right ) \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx \\ & = \frac {1}{9} \int \exp \left (e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )\right ) \left (3 \left (3+4 e^{16}\right )-4 e^8 \left (12-e^8\right ) x+4 \left (9-4 e^8\right ) x^2+12 x^3\right ) \, dx \\ & = \frac {1}{9} \int \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) \left (3 \left (3+4 e^{16}\right )-4 e^8 \left (12-e^8\right ) x+4 \left (9-4 e^8\right ) x^2+12 x^3\right ) \, dx \\ & = \frac {1}{9} \int \left (3 \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) \left (3+4 e^{16}\right )+4 \exp \left (8+\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) \left (-12+e^8\right ) x-4 \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) \left (-9+4 e^8\right ) x^2+12 \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) x^3\right ) \, dx \\ & = \frac {4}{3} \int \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) x^3 \, dx+\frac {1}{9} \left (4 \left (9-4 e^8\right )\right ) \int \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) x^2 \, dx-\frac {1}{9} \left (4 \left (12-e^8\right )\right ) \int \exp \left (8+\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) x \, dx+\frac {1}{3} \left (3+4 e^{16}\right ) \int \exp \left (\frac {4}{9} \left (e^8-x\right )^2 x+e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)\right ) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=e^{e^{\frac {4}{9} \left (e^8-x\right )^2 x} (3+x)} \]

[In]

Integrate[(E^(E^((4*E^16*x - 8*E^8*x^2 + 4*x^3)/9)*(3 + x) + (4*E^16*x - 8*E^8*x^2 + 4*x^3)/9)*(9 + 36*x^2 + 1
2*x^3 + E^16*(12 + 4*x) + E^8*(-48*x - 16*x^2)))/9,x]

[Out]

E^(E^((4*(E^8 - x)^2*x)/9)*(3 + x))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
risch	\({\mathrm e}^{\left (3+x \right ) {\mathrm e}^{\frac {4 x \left (-2 x \,{\mathrm e}^{8}+x^{2}+{\mathrm e}^{16}\right )}{9}}}\)	\(21\)
norman	\({\mathrm e}^{\left (3+x \right ) {\mathrm e}^{\frac {4 x \,{\mathrm e}^{16}}{9}-\frac {8 x^{2} {\mathrm e}^{8}}{9}+\frac {4 x^{3}}{9}}}\)	\(27\)
parallelrisch	\({\mathrm e}^{\left (3+x \right ) {\mathrm e}^{\frac {4 x \,{\mathrm e}^{16}}{9}-\frac {8 x^{2} {\mathrm e}^{8}}{9}+\frac {4 x^{3}}{9}}}\)	\(27\)

[In]

int(1/9*((4*x+12)*exp(8)^2+(-16*x^2-48*x)*exp(8)+12*x^3+36*x^2+9)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*x^3)*e
xp((3+x)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*x^3)),x,method=_RETURNVERBOSE)

[Out]

exp((3+x)*exp(4/9*x*(-2*x*exp(8)+x^2+exp(16))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=e^{\left ({\left (x + 3\right )} e^{\left (\frac {4}{9} \, x^{3} - \frac {8}{9} \, x^{2} e^{8} + \frac {4}{9} \, x e^{16}\right )}\right )} \]

[In]

integrate(1/9*((4*x+12)*exp(8)^2+(-16*x^2-48*x)*exp(8)+12*x^3+36*x^2+9)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*
x^3)*exp((3+x)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*x^3)),x, algorithm="fricas")

[Out]

e^((x + 3)*e^(4/9*x^3 - 8/9*x^2*e^8 + 4/9*x*e^16))

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=e^{\left (x + 3\right ) e^{\frac {4 x^{3}}{9} - \frac {8 x^{2} e^{8}}{9} + \frac {4 x e^{16}}{9}}} \]

[In]

integrate(1/9*((4*x+12)*exp(8)**2+(-16*x**2-48*x)*exp(8)+12*x**3+36*x**2+9)*exp(4/9*x*exp(8)**2-8/9*x**2*exp(8
)+4/9*x**3)*exp((3+x)*exp(4/9*x*exp(8)**2-8/9*x**2*exp(8)+4/9*x**3)),x)

[Out]

exp((x + 3)*exp(4*x**3/9 - 8*x**2*exp(8)/9 + 4*x*exp(16)/9))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=e^{\left (x e^{\left (\frac {4}{9} \, x^{3} - \frac {8}{9} \, x^{2} e^{8} + \frac {4}{9} \, x e^{16}\right )} + 3 \, e^{\left (\frac {4}{9} \, x^{3} - \frac {8}{9} \, x^{2} e^{8} + \frac {4}{9} \, x e^{16}\right )}\right )} \]

[In]

integrate(1/9*((4*x+12)*exp(8)^2+(-16*x^2-48*x)*exp(8)+12*x^3+36*x^2+9)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*
x^3)*exp((3+x)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*x^3)),x, algorithm="maxima")

[Out]

e^(x*e^(4/9*x^3 - 8/9*x^2*e^8 + 4/9*x*e^16) + 3*e^(4/9*x^3 - 8/9*x^2*e^8 + 4/9*x*e^16))

Giac [F]

\[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx=\int { \frac {1}{9} \, {\left (12 \, x^{3} + 36 \, x^{2} + 4 \, {\left (x + 3\right )} e^{16} - 16 \, {\left (x^{2} + 3 \, x\right )} e^{8} + 9\right )} e^{\left (\frac {4}{9} \, x^{3} - \frac {8}{9} \, x^{2} e^{8} + \frac {4}{9} \, x e^{16} + {\left (x + 3\right )} e^{\left (\frac {4}{9} \, x^{3} - \frac {8}{9} \, x^{2} e^{8} + \frac {4}{9} \, x e^{16}\right )}\right )} \,d x } \]

[In]

integrate(1/9*((4*x+12)*exp(8)^2+(-16*x^2-48*x)*exp(8)+12*x^3+36*x^2+9)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*
x^3)*exp((3+x)*exp(4/9*x*exp(8)^2-8/9*x^2*exp(8)+4/9*x^3)),x, algorithm="giac")

[Out]

integrate(1/9*(12*x^3 + 36*x^2 + 4*(x + 3)*e^16 - 16*(x^2 + 3*x)*e^8 + 9)*e^(4/9*x^3 - 8/9*x^2*e^8 + 4/9*x*e^1
6 + (x + 3)*e^(4/9*x^3 - 8/9*x^2*e^8 + 4/9*x*e^16)), x)

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {1}{9} e^{e^{\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} (3+x)+\frac {1}{9} \left (4 e^{16} x-8 e^8 x^2+4 x^3\right )} \left (9+36 x^2+12 x^3+e^{16} (12+4 x)+e^8 \left (-48 x-16 x^2\right )\right ) \, dx={\mathrm {e}}^{3\,{\mathrm {e}}^{-\frac {8\,x^2\,{\mathrm {e}}^8}{9}}\,{\mathrm {e}}^{\frac {4\,x^3}{9}}\,{\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{16}}{9}}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-\frac {8\,x^2\,{\mathrm {e}}^8}{9}}\,{\mathrm {e}}^{\frac {4\,x^3}{9}}\,{\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{16}}{9}}} \]

[In]

int((exp(exp((4*x*exp(16))/9 - (8*x^2*exp(8))/9 + (4*x^3)/9)*(x + 3))*exp((4*x*exp(16))/9 - (8*x^2*exp(8))/9 +
(4*x^3)/9)*(36*x^2 - exp(8)*(48*x + 16*x^2) + 12*x^3 + exp(16)*(4*x + 12) + 9))/9,x)

[Out]

exp(3*exp(-(8*x^2*exp(8))/9)*exp((4*x^3)/9)*exp((4*x*exp(16))/9))*exp(x*exp(-(8*x^2*exp(8))/9)*exp((4*x^3)/9)*
exp((4*x*exp(16))/9))