3.29.40 \(\int \frac {5 e^{-5+x}}{4} \, dx\)

Optimal. Leaf size=15 \[ 15+5 \left (-2+\frac {e^{-5+x}}{4}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 9, normalized size of antiderivative = 0.60, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 2194} \begin {gather*} \frac {5 e^{x-5}}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^(-5 + x))/4,x]

[Out]

(5*E^(-5 + x))/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

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 {gather*} \begin {aligned} \text {integral} &=\frac {5}{4} \int e^{-5+x} \, dx\\ &=\frac {5 e^{-5+x}}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 9, normalized size = 0.60 \begin {gather*} \frac {5 e^{-5+x}}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(-5 + x))/4,x]

[Out]

(5*E^(-5 + x))/4

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fricas [A]  time = 0.71, size = 10, normalized size = 0.67 \begin {gather*} 5 \, e^{\left (x - 2 \, \log \relax (2) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*e^(x - 2*log(2) - 5)

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giac [A]  time = 0.28, size = 10, normalized size = 0.67 \begin {gather*} 5 \, e^{\left (x - 2 \, \log \relax (2) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*e^(x - 2*log(2) - 5)

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maple [A]  time = 0.03, size = 7, normalized size = 0.47

method	result	size
risch	\(\frac {5 \,{\mathrm e}^{x -5}}{4}\)	\(7\)
meijerg	\(-\frac {5 \,{\mathrm e}^{-5} \left (1-{\mathrm e}^{x}\right )}{4}\)	\(11\)
gosper	\(5 \,{\mathrm e}^{x -2 \ln \relax (2)} {\mathrm e}^{-5}\)	\(14\)
derivativedivides	\(5 \,{\mathrm e}^{x -2 \ln \relax (2)} {\mathrm e}^{-5}\)	\(14\)
default	\(5 \,{\mathrm e}^{x -2 \ln \relax (2)} {\mathrm e}^{-5}\)	\(14\)
norman	\(5 \,{\mathrm e}^{x -2 \ln \relax (2)} {\mathrm e}^{-5}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*exp(x-5)

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maxima [A]  time = 0.58, size = 6, normalized size = 0.40 \begin {gather*} \frac {5}{4} \, e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*e^(x - 5)

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mupad [B]  time = 0.03, size = 6, normalized size = 0.40 \begin {gather*} \frac {5\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*exp(x - 2*log(2))*exp(-5),x)

[Out]

(5*exp(-5)*exp(x))/4

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sympy [A]  time = 0.08, size = 8, normalized size = 0.53 \begin {gather*} \frac {5 e^{x}}{4 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(5*exp(x-2*ln(2))/exp(5),x)

[Out]

5*exp(-5)*exp(x)/4

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