\(\int e^{-5+x} \, dx\) [9520]

Optimal result

Integrand size = 5, antiderivative size = 5 \[ \int e^{-5+x} \, dx=e^{-5+x} \]

[Out]

exp(-5+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2225} \[ \int e^{-5+x} \, dx=e^{x-5} \]

[In]

Int[E^(-5 + x),x]

[Out]

E^(-5 + x)

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}& = e^{-5+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int e^{-5+x} \, dx=e^{-5+x} \]

[In]

Integrate[E^(-5 + x),x]

[Out]

E^(-5 + x)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00

method	result	size
gosper	\({\mathrm e}^{-5+x}\)	\(5\)
derivativedivides	\({\mathrm e}^{-5+x}\)	\(5\)
default	\({\mathrm e}^{-5+x}\)	\(5\)
norman	\({\mathrm e}^{-5+x}\)	\(5\)
risch	\({\mathrm e}^{-5+x}\)	\(5\)
parallelrisch	\({\mathrm e}^{-5+x}\)	\(5\)
meijerg	\(-{\mathrm e}^{-5} \left (1-{\mathrm e}^{x}\right )\)	\(11\)

[In]

int(exp(-5+x),x,method=_RETURNVERBOSE)

[Out]

exp(-5+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 4, normalized size of antiderivative = 0.80 \[ \int e^{-5+x} \, dx=e^{\left (x - 5\right )} \]

[In]

integrate(exp(-5+x),x, algorithm="fricas")

[Out]

e^(x - 5)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.60 \[ \int e^{-5+x} \, dx=e^{x - 5} \]

[In]

integrate(exp(-5+x),x)

[Out]

exp(x - 5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 4, normalized size of antiderivative = 0.80 \[ \int e^{-5+x} \, dx=e^{\left (x - 5\right )} \]

[In]

integrate(exp(-5+x),x, algorithm="maxima")

[Out]

e^(x - 5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 4, normalized size of antiderivative = 0.80 \[ \int e^{-5+x} \, dx=e^{\left (x - 5\right )} \]

[In]

integrate(exp(-5+x),x, algorithm="giac")

[Out]

e^(x - 5)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int e^{-5+x} \, dx={\mathrm {e}}^{-5}\,{\mathrm {e}}^x \]

[In]

int(exp(x - 5),x)

[Out]

exp(-5)*exp(x)