\(\int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx\) [5673]

Optimal result

Integrand size = 24, antiderivative size = 21 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=2+\frac {4}{e^2}+e^{-5-\frac {9}{e^2}-2 x} x \]

[Out]

2+x/exp(2*x+5+9/exp(2))+4/exp(2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2218, 2207, 2225} \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=\frac {1}{2} e^{-2 x-\frac {9}{e^2}-5}-\frac {1}{2} e^{-2 x-\frac {9}{e^2}-5} (1-2 x) \]

[In]

Int[(1 - 2*x)/E^((9 + E^2*(5 + 2*x))/E^2),x]

[Out]

E^(-5 - 9/E^2 - 2*x)/2 - (E^(-5 - 9/E^2 - 2*x)*(1 - 2*x))/2

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2218

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int e^{-5-\frac {9}{e^2}-2 x} (1-2 x) \, dx \\ & = -\frac {1}{2} e^{-5-\frac {9}{e^2}-2 x} (1-2 x)-\int e^{-5-\frac {9}{e^2}-2 x} \, dx \\ & = \frac {1}{2} e^{-5-\frac {9}{e^2}-2 x}-\frac {1}{2} e^{-5-\frac {9}{e^2}-2 x} (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=e^{-5-\frac {9}{e^2}-2 x} x \]

[In]

Integrate[(1 - 2*x)/E^((9 + E^2*(5 + 2*x))/E^2),x]

[Out]

E^(-5 - 9/E^2 - 2*x)*x

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(x \,{\mathrm e}^{-\left (2 \,{\mathrm e}^{2} x +5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}\)	\(19\)
norman	\(x \,{\mathrm e}^{-\left (\left (5+2 x \right ) {\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}\)	\(21\)
parallelrisch	\(x \,{\mathrm e}^{-\left (\left (5+2 x \right ) {\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}\)	\(21\)
gosper	\(x \,{\mathrm e}^{-\left (2 \,{\mathrm e}^{2} x +5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}\)	\(22\)
meijerg	\(\frac {{\mathrm e}^{-2 x -9 \,{\mathrm e}^{-2}+2 \,{\mathrm e}^{-5} x} \left (1-{\mathrm e}^{-2 \,{\mathrm e}^{-5} x}\right )}{2}-\frac {{\mathrm e}^{-2 x -9 \,{\mathrm e}^{-2}+2 \,{\mathrm e}^{-5} x +5} \left (1-\frac {\left (2+4 \,{\mathrm e}^{-5} x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{-5} x}}{2}\right )}{2}\)	\(62\)
derivativedivides	\(-\frac {5 \,{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}}{2}+\frac {{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}} \left (2 x +\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}\right )}{2}-\frac {9 \,{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}} {\mathrm e}^{-2}}{2}\)	\(81\)
default	\(-\frac {5 \,{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}}}{2}+\frac {{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}} \left (2 x +\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}\right )}{2}-\frac {9 \,{\mathrm e}^{-2 x -\left (5 \,{\mathrm e}^{2}+9\right ) {\mathrm e}^{-2}} {\mathrm e}^{-2}}{2}\)	\(81\)

[In]

int((1-2*x)/exp(((5+2*x)*exp(2)+9)/exp(2)),x,method=_RETURNVERBOSE)

[Out]

x*exp(-(2*exp(2)*x+5*exp(2)+9)*exp(-2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=x e^{\left (-{\left ({\left (2 \, x + 5\right )} e^{2} + 9\right )} e^{\left (-2\right )}\right )} \]

[In]

integrate((1-2*x)/exp(((5+2*x)*exp(2)+9)/exp(2)),x, algorithm="fricas")

[Out]

x*e^(-((2*x + 5)*e^2 + 9)*e^(-2))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=x e^{- \frac {\left (2 x + 5\right ) e^{2} + 9}{e^{2}}} \]

[In]

integrate((1-2*x)/exp(((5+2*x)*exp(2)+9)/exp(2)),x)

[Out]

x*exp(-((2*x + 5)*exp(2) + 9)*exp(-2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=\frac {1}{2} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x - 9 \, e^{\left (-2\right )} - 5\right )} - \frac {1}{2} \, e^{\left (-2 \, x - 9 \, e^{\left (-2\right )} - 5\right )} \]

[In]

integrate((1-2*x)/exp(((5+2*x)*exp(2)+9)/exp(2)),x, algorithm="maxima")

[Out]

1/2*(2*x + 1)*e^(-2*x - 9*e^(-2) - 5) - 1/2*e^(-2*x - 9*e^(-2) - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=x e^{\left (-{\left (2 \, x e^{2} + 5 \, e^{2} + 9\right )} e^{\left (-2\right )}\right )} \]

[In]

integrate((1-2*x)/exp(((5+2*x)*exp(2)+9)/exp(2)),x, algorithm="giac")

[Out]

x*e^(-(2*x*e^2 + 5*e^2 + 9)*e^(-2))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int e^{-\frac {9+e^2 (5+2 x)}{e^2}} (1-2 x) \, dx=x\,{\mathrm {e}}^{-9\,{\mathrm {e}}^{-2}}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-5} \]

[In]

int(-exp(-exp(-2)*(exp(2)*(2*x + 5) + 9))*(2*x - 1),x)

[Out]

x*exp(-9*exp(-2))*exp(-2*x)*exp(-5)