3.43.91 \(\int \frac {1}{5} (1+5 e^{-1+x}) \, dx\)

Optimal. Leaf size=14 \[ -7+e^{-1+x}+\frac {1}{5} (-5+x) \]

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Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2194} \begin {gather*} \frac {x}{5}+e^{x-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 5*E^(-1 + x))/5,x]

[Out]

E^(-1 + x) + x/5

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 {1}{5} \int \left (1+5 e^{-1+x}\right ) \, dx\\ &=\frac {x}{5}+\int e^{-1+x} \, dx\\ &=e^{-1+x}+\frac {x}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.79 \begin {gather*} e^{-1+x}+\frac {x}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 5*E^(-1 + x))/5,x]

[Out]

E^(-1 + x) + x/5

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fricas [A]  time = 0.53, size = 8, normalized size = 0.57 \begin {gather*} \frac {1}{5} \, x + e^{\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x + e^(x - 1)

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giac [A]  time = 0.12, size = 8, normalized size = 0.57 \begin {gather*} \frac {1}{5} \, x + e^{\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x + e^(x - 1)

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maple [A]  time = 0.01, size = 9, normalized size = 0.64

method	result	size
default	\(\frac {x}{5}+{\mathrm e}^{x -1}\)	\(9\)
norman	\(\frac {x}{5}+{\mathrm e}^{x -1}\)	\(9\)
risch	\(\frac {x}{5}+{\mathrm e}^{x -1}\)	\(9\)
derivativedivides	\({\mathrm e}^{x -1}+\frac {\ln \left ({\mathrm e}^{x -1}\right )}{5}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 8, normalized size = 0.57 \begin {gather*} \frac {1}{5} \, x + e^{\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x + e^(x - 1)

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mupad [B]  time = 3.22, size = 8, normalized size = 0.57 \begin {gather*} \frac {x}{5}+{\mathrm {e}}^{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x - 1) + 1/5,x)

[Out]

x/5 + exp(x - 1)

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sympy [A]  time = 0.07, size = 7, normalized size = 0.50 \begin {gather*} \frac {x}{5} + e^{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/5 + exp(x - 1)

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