\(\int \frac {1}{4} (4+5 e^{-5 x/4}) \, dx\) [6668]

Optimal result

Integrand size = 15, antiderivative size = 14 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=\frac {1}{e^3}-e^{-5 x/4}+x \]

[Out]

exp(-3)+x-exp(-5/4*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 2225} \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x-e^{-5 x/4} \]

[In]

Int[(4 + 5/E^((5*x)/4))/4,x]

[Out]

-E^((-5*x)/4) + x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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}& = \frac {1}{4} \int \left (4+5 e^{-5 x/4}\right ) \, dx \\ & = x+\frac {5}{4} \int e^{-5 x/4} \, dx \\ & = -e^{-5 x/4}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=-e^{-5 x/4}+x \]

[In]

Integrate[(4 + 5/E^((5*x)/4))/4,x]

[Out]

-E^((-5*x)/4) + x

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64

method	result	size
default	\(x -{\mathrm e}^{-\frac {5 x}{4}}\)	\(9\)
norman	\(x -{\mathrm e}^{-\frac {5 x}{4}}\)	\(9\)
risch	\(x -{\mathrm e}^{-\frac {5 x}{4}}\)	\(9\)
parallelrisch	\(x -{\mathrm e}^{-\frac {5 x}{4}}\)	\(9\)
parts	\(x -{\mathrm e}^{-\frac {5 x}{4}}\)	\(9\)
derivativedivides	\(-{\mathrm e}^{-\frac {5 x}{4}}-\frac {4 \ln \left ({\mathrm e}^{-\frac {5 x}{4}}\right )}{5}\)	\(15\)

[In]

int(5/4*exp(-5/4*x)+1,x,method=_RETURNVERBOSE)

[Out]

x-exp(-5/4*x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x - e^{\left (-\frac {5}{4} \, x\right )} \]

[In]

integrate(5/4*exp(-5/4*x)+1,x, algorithm="fricas")

[Out]

x - e^(-5/4*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x - e^{- \frac {5 x}{4}} \]

[In]

integrate(5/4*exp(-5/4*x)+1,x)

[Out]

x - exp(-5*x/4)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x - e^{\left (-\frac {5}{4} \, x\right )} \]

[In]

integrate(5/4*exp(-5/4*x)+1,x, algorithm="maxima")

[Out]

x - e^(-5/4*x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x - e^{\left (-\frac {5}{4} \, x\right )} \]

[In]

integrate(5/4*exp(-5/4*x)+1,x, algorithm="giac")

[Out]

x - e^(-5/4*x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{4} \left (4+5 e^{-5 x/4}\right ) \, dx=x-{\mathrm {e}}^{-\frac {5\,x}{4}} \]

[In]

int((5*exp(-(5*x)/4))/4 + 1,x)

[Out]

x - exp(-(5*x)/4)