\(\int e^{e^{-2 x+2 e^{16} x}+8 x} (8+e^{-2 x+2 e^{16} x} (-2+2 e^{16})) \, dx\) [934]

Optimal result

Integrand size = 41, antiderivative size = 19 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{e^{2 \left (-x+e^{16} x\right )}+8 x} \]

[Out]

exp(exp(2*x*exp(4)^4-2*x)+8*x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6838} \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{8 x+e^{-2 \left (1-e^{16}\right ) x}} \]

[In]

Int[E^(E^(-2*x + 2*E^16*x) + 8*x)*(8 + E^(-2*x + 2*E^16*x)*(-2 + 2*E^16)),x]

[Out]

E^(E^(-2*(1 - E^16)*x) + 8*x)

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^{e^{-2 \left (1-e^{16}\right ) x}+8 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{e^{2 \left (-1+e^{16}\right ) x}+8 x} \]

[In]

Integrate[E^(E^(-2*x + 2*E^16*x) + 8*x)*(8 + E^(-2*x + 2*E^16*x)*(-2 + 2*E^16)),x]

[Out]

E^(E^(2*(-1 + E^16)*x) + 8*x)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{2 x \left (-1+{\mathrm e}^{16}\right )}+8 x}\)	\(14\)
derivativedivides	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{16}-2 x}+8 x}\)	\(18\)
default	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{16}-2 x}+8 x}\)	\(18\)
norman	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{16}-2 x}+8 x}\)	\(18\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{2 x \,{\mathrm e}^{16}-2 x}+8 x}\)	\(18\)

[In]

int(((2*exp(4)^4-2)*exp(2*x*exp(4)^4-2*x)+8)*exp(exp(2*x*exp(4)^4-2*x)+8*x),x,method=_RETURNVERBOSE)

[Out]

exp(exp(2*x*(-1+exp(16)))+8*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{\left (8 \, x + e^{\left (2 \, x e^{16} - 2 \, x\right )}\right )} \]

[In]

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

[Out]

e^(8*x + e^(2*x*e^16 - 2*x))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{8 x + e^{- 2 x + 2 x e^{16}}} \]

[In]

integrate(((2*exp(4)**4-2)*exp(2*x*exp(4)**4-2*x)+8)*exp(exp(2*x*exp(4)**4-2*x)+8*x),x)

[Out]

exp(8*x + exp(-2*x + 2*x*exp(16)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{\left (8 \, x + e^{\left (2 \, x e^{16} - 2 \, x\right )}\right )} \]

[In]

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

[Out]

e^(8*x + e^(2*x*e^16 - 2*x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx=e^{\left (8 \, x + e^{\left (2 \, x e^{16} - 2 \, x\right )}\right )} \]

[In]

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

[Out]

e^(8*x + e^(2*x*e^16 - 2*x))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int e^{e^{-2 x+2 e^{16} x}+8 x} \left (8+e^{-2 x+2 e^{16} x} \left (-2+2 e^{16}\right )\right ) \, dx={\mathrm {e}}^{8\,x+{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{16}-2\,x}} \]

[In]

int(exp(8*x + exp(2*x*exp(16) - 2*x))*(exp(2*x*exp(16) - 2*x)*(2*exp(16) - 2) + 8),x)

[Out]

exp(8*x + exp(2*x*exp(16) - 2*x))