\(\int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 (-2 x+2 x^2)} (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)) \, dx\) [7444]

Optimal result

Integrand size = 62, antiderivative size = 19 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=e^{(-1+x) \left (4-\left (-e^8+x\right )^2\right )} \]

[Out]

exp((4-(x-exp(8))^2)*(-1+x))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6838} \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=\exp \left (-x^3+x^2-2 e^8 \left (x-x^2\right )+4 x+e^{16} (1-x)-4\right ) \]

[In]

Int[E^(-4 + E^16*(1 - x) + 4*x + x^2 - x^3 + E^8*(-2*x + 2*x^2))*(4 - E^16 + 2*x - 3*x^2 + E^8*(-2 + 4*x)),x]

[Out]

E^(-4 + E^16*(1 - x) + 4*x + x^2 - x^3 - 2*E^8*(x - 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}& = \exp \left (-4+e^{16} (1-x)+4 x+x^2-x^3-2 e^8 \left (x-x^2\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=e^{-\left (\left (-2+e^8-x\right ) \left (2+e^8-x\right ) (-1+x)\right )} \]

[In]

Integrate[E^(-4 + E^16*(1 - x) + 4*x + x^2 - x^3 + E^8*(-2*x + 2*x^2))*(4 - E^16 + 2*x - 3*x^2 + E^8*(-2 + 4*x
)),x]

[Out]

E^(-((-2 + E^8 - x)*(2 + E^8 - x)*(-1 + x)))

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
risch	\({\mathrm e}^{-\left (-1+x \right ) \left (-2 x \,{\mathrm e}^{8}+x^{2}+{\mathrm e}^{16}-4\right )}\)	\(19\)
derivativedivides	\({\mathrm e}^{\left (1-x \right ) {\mathrm e}^{16}+\left (2 x^{2}-2 x \right ) {\mathrm e}^{8}-x^{3}+x^{2}+4 x -4}\)	\(37\)
default	\({\mathrm e}^{\left (1-x \right ) {\mathrm e}^{16}+\left (2 x^{2}-2 x \right ) {\mathrm e}^{8}-x^{3}+x^{2}+4 x -4}\)	\(37\)
norman	\({\mathrm e}^{\left (1-x \right ) {\mathrm e}^{16}+\left (2 x^{2}-2 x \right ) {\mathrm e}^{8}-x^{3}+x^{2}+4 x -4}\)	\(37\)
parallelrisch	\({\mathrm e}^{\left (1-x \right ) {\mathrm e}^{16}+\left (2 x^{2}-2 x \right ) {\mathrm e}^{8}-x^{3}+x^{2}+4 x -4}\)	\(37\)
gosper	\({\mathrm e}^{2 x^{2} {\mathrm e}^{8}-x^{3}-2 x \,{\mathrm e}^{8}-x \,{\mathrm e}^{16}+x^{2}+{\mathrm e}^{16}+4 x -4}\)	\(38\)

[In]

int((-exp(8)^2+(4*x-2)*exp(8)-3*x^2+2*x+4)*exp((1-x)*exp(8)^2+(2*x^2-2*x)*exp(8)-x^3+x^2+4*x-4),x,method=_RETU
RNVERBOSE)

[Out]

exp(-(-1+x)*(-2*x*exp(8)+x^2+exp(16)-4))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=e^{\left (-x^{3} + x^{2} - {\left (x - 1\right )} e^{16} + 2 \, {\left (x^{2} - x\right )} e^{8} + 4 \, x - 4\right )} \]

[In]

integrate((-exp(8)^2+(4*x-2)*exp(8)-3*x^2+2*x+4)*exp((1-x)*exp(8)^2+(2*x^2-2*x)*exp(8)-x^3+x^2+4*x-4),x, algor
ithm="fricas")

[Out]

e^(-x^3 + x^2 - (x - 1)*e^16 + 2*(x^2 - x)*e^8 + 4*x - 4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (12) = 24\).

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=e^{- x^{3} + x^{2} + 4 x + \left (1 - x\right ) e^{16} + \left (2 x^{2} - 2 x\right ) e^{8} - 4} \]

[In]

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

[Out]

exp(-x**3 + x**2 + 4*x + (1 - x)*exp(16) + (2*x**2 - 2*x)*exp(8) - 4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx=e^{\left (-x^{3} + x^{2} - {\left (x - 1\right )} e^{16} + 2 \, {\left (x^{2} - x\right )} e^{8} + 4 \, x - 4\right )} \]

[In]

integrate((-exp(8)^2+(4*x-2)*exp(8)-3*x^2+2*x+4)*exp((1-x)*exp(8)^2+(2*x^2-2*x)*exp(8)-x^3+x^2+4*x-4),x, algor
ithm="maxima")

[Out]

e^(-x^3 + x^2 - (x - 1)*e^16 + 2*(x^2 - x)*e^8 + 4*x - 4)

Giac [F]

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

[In]

integrate((-exp(8)^2+(4*x-2)*exp(8)-3*x^2+2*x+4)*exp((1-x)*exp(8)^2+(2*x^2-2*x)*exp(8)-x^3+x^2+4*x-4),x, algor
ithm="giac")

[Out]

integrate(-(3*x^2 - 2*(2*x - 1)*e^8 - 2*x + e^16 - 4)*e^(-x^3 + x^2 - (x - 1)*e^16 + 2*(x^2 - x)*e^8 + 4*x - 4
), x)

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int e^{-4+e^{16} (1-x)+4 x+x^2-x^3+e^8 \left (-2 x+2 x^2\right )} \left (4-e^{16}+2 x-3 x^2+e^8 (-2+4 x)\right ) \, dx={\mathrm {e}}^{2\,x^2\,{\mathrm {e}}^8}\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^8}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{16}}\,{\mathrm {e}}^{{\mathrm {e}}^{16}} \]

[In]

int(exp(4*x - exp(8)*(2*x - 2*x^2) - exp(16)*(x - 1) + x^2 - x^3 - 4)*(2*x - exp(16) - 3*x^2 + exp(8)*(4*x - 2
) + 4),x)

[Out]

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