\(\int \frac {5}{2} e^{\frac {2+x}{2}} \, dx\) [3644]

Optimal result

Integrand size = 13, antiderivative size = 13 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=-5+5 e^{1+\frac {x}{2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 2225} \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 e^{\frac {x+2}{2}} \]

[In]

Int[(5*E^((2 + x)/2))/2,x]

[Out]

5*E^((2 + 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 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 {5}{2} \int e^{\frac {2+x}{2}} \, dx \\ & = 5 e^{\frac {2+x}{2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 e^{1+\frac {x}{2}} \]

[In]

Integrate[(5*E^((2 + x)/2))/2,x]

[Out]

5*E^(1 + x/2)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69

method	result	size
gosper	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
derivativedivides	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
default	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
norman	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
risch	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
parallelrisch	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
parts	\(5 \,{\mathrm e}^{1+\frac {x}{2}}\)	\(9\)
meijerg	\(-5 \,{\mathrm e} \left (1-{\mathrm e}^{\frac {x}{2}}\right )\)	\(13\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 \, e^{\left (\frac {1}{2} \, x + 1\right )} \]

[In]

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

[Out]

5*e^(1/2*x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 e^{\frac {x}{2} + 1} \]

[In]

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

[Out]

5*exp(x/2 + 1)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 \, e^{\left (\frac {1}{2} \, x + 1\right )} \]

[In]

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

[Out]

5*e^(1/2*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5 \, e^{\left (\frac {1}{2} \, x + 1\right )} \]

[In]

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

[Out]

5*e^(1/2*x + 1)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {5}{2} e^{\frac {2+x}{2}} \, dx=5\,{\mathrm {e}}^{x/2}\,\mathrm {e} \]

[In]

int((5*exp(x/2 + 1))/2,x)

[Out]

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