\(\int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} (2+e^x-4 e^{16-4 e^x+x}-2 x) \, dx\) [7564]

Optimal result

Integrand size = 48, antiderivative size = 30 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {1}{2} e^{-5+e^{4 \left (4-e^x\right )}+e^x+2 x-x^2} \]

[Out]

exp(exp(-4*exp(x)+16)+exp(x)-ln(2)-x^2+2*x-5)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12, 6838} \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {1}{2} e^{-x^2+2 x+e^{16-4 e^x}+e^x-5} \]

[In]

Int[(E^(-5 + E^(16 - 4*E^x) + E^x + 2*x - x^2)*(2 + E^x - 4*E^(16 - 4*E^x + x) - 2*x))/2,x]

[Out]

E^(-5 + E^(16 - 4*E^x) + E^x + 2*x - x^2)/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}& = \frac {1}{2} \int e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx \\ & = \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \]

[In]

Integrate[(E^(-5 + E^(16 - 4*E^x) + E^x + 2*x - x^2)*(2 + E^x - 4*E^(16 - 4*E^x + x) - 2*x))/2,x]

[Out]

E^(-5 + E^(16 - 4*E^x) + E^x + 2*x - x^2)/2

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{x}+16}+{\mathrm e}^{x}-5-x^{2}+2 x}}{2}\)	\(23\)
derivativedivides	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{x}+16}+{\mathrm e}^{x}-\ln \left (2\right )-x^{2}+2 x -5}\)	\(25\)
default	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{x}+16}+{\mathrm e}^{x}-\ln \left (2\right )-x^{2}+2 x -5}\)	\(25\)
norman	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{x}+16}+{\mathrm e}^{x}-\ln \left (2\right )-x^{2}+2 x -5}\)	\(25\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{x}+16}+{\mathrm e}^{x}-\ln \left (2\right )-x^{2}+2 x -5}\)	\(25\)

[In]

int((-4*exp(x)*exp(-4*exp(x)+16)+exp(x)-2*x+2)*exp(exp(-4*exp(x)+16)+exp(x)-ln(2)-x^2+2*x-5),x,method=_RETURNV
ERBOSE)

[Out]

1/2*exp(exp(-4*exp(x)+16)+exp(x)-5-x^2+2*x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=e^{\left (-{\left ({\left (x^{2} - 2 \, x + \log \left (2\right ) + 5\right )} e^{x} - e^{\left (2 \, x\right )} - e^{\left (x - 4 \, e^{x} + 16\right )}\right )} e^{\left (-x\right )}\right )} \]

[In]

integrate((-4*exp(x)*exp(-4*exp(x)+16)+exp(x)-2*x+2)*exp(exp(-4*exp(x)+16)+exp(x)-log(2)-x^2+2*x-5),x, algorit
hm="fricas")

[Out]

e^(-((x^2 - 2*x + log(2) + 5)*e^x - e^(2*x) - e^(x - 4*e^x + 16))*e^(-x))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {e^{- x^{2} + 2 x + e^{x} + e^{16 - 4 e^{x}} - 5}}{2} \]

[In]

integrate((-4*exp(x)*exp(-4*exp(x)+16)+exp(x)-2*x+2)*exp(exp(-4*exp(x)+16)+exp(x)-ln(2)-x**2+2*x-5),x)

[Out]

exp(-x**2 + 2*x + exp(x) + exp(16 - 4*exp(x)) - 5)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {1}{2} \, e^{\left (-x^{2} + 2 \, x + e^{x} + e^{\left (-4 \, e^{x} + 16\right )} - 5\right )} \]

[In]

integrate((-4*exp(x)*exp(-4*exp(x)+16)+exp(x)-2*x+2)*exp(exp(-4*exp(x)+16)+exp(x)-log(2)-x^2+2*x-5),x, algorit
hm="maxima")

[Out]

1/2*e^(-x^2 + 2*x + e^x + e^(-4*e^x + 16) - 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=e^{\left (-x^{2} + 2 \, x + e^{x} + e^{\left (-4 \, e^{x} + 16\right )} - \log \left (2\right ) - 5\right )} \]

[In]

integrate((-4*exp(x)*exp(-4*exp(x)+16)+exp(x)-2*x+2)*exp(exp(-4*exp(x)+16)+exp(x)-log(2)-x^2+2*x-5),x, algorit
hm="giac")

[Out]

e^(-x^2 + 2*x + e^x + e^(-4*e^x + 16) - log(2) - 5)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{2} e^{-5+e^{16-4 e^x}+e^x+2 x-x^2} \left (2+e^x-4 e^{16-4 e^x+x}-2 x\right ) \, dx=\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{16}\,{\mathrm {e}}^{-4\,{\mathrm {e}}^x}}}{2} \]

[In]

int(-exp(2*x + exp(16 - 4*exp(x)) - log(2) + exp(x) - x^2 - 5)*(2*x - exp(x) + 4*exp(16 - 4*exp(x))*exp(x) - 2
),x)

[Out]

(exp(2*x)*exp(exp(x))*exp(-5)*exp(-x^2)*exp(exp(16)*exp(-4*exp(x))))/2