\(\int -\frac {1}{2} e^{5+\frac {1}{2} (2-e^5 x)} \, dx\) [6457]

Optimal result

Integrand size = 20, antiderivative size = 12 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{1-\frac {e^5 x}{2}} \]

[Out]

exp(-1/2*x*exp(5)+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2234} \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{\frac {1}{2} \left (2-e^5 x\right )} \]

[In]

Int[-1/2*E^(5 + (2 - E^5*x)/2),x]

[Out]

E^((2 - E^5*x)/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 2234

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{1-\frac {e^5 x}{2}} \]

[In]

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

[Out]

E^(1 - (E^5*x)/2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75

method	result	size
gosper	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
derivativedivides	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
default	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
norman	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
risch	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
parallelrisch	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
parts	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}+1}\)	\(9\)
meijerg	\(-{\mathrm e} \left (1-{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{2}}\right )\)	\(15\)

[In]

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

[Out]

exp(-1/2*x*exp(5)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \]

[In]

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

[Out]

e^(-1/2*x*e^5 + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{- \frac {x e^{5}}{2} + 1} \]

[In]

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

[Out]

exp(-x*exp(5)/2 + 1)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \]

[In]

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

[Out]

e^(-1/2*x*e^5 + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=e^{\left (-\frac {1}{2} \, x e^{5} + 1\right )} \]

[In]

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

[Out]

e^(-1/2*x*e^5 + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int -\frac {1}{2} e^{5+\frac {1}{2} \left (2-e^5 x\right )} \, dx=\mathrm {e}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{2}} \]

[In]

int(-(exp(1 - (x*exp(5))/2)*exp(5))/2,x)

[Out]

exp(1)*exp(-(x*exp(5))/2)