\(\int \frac {1}{12} e^{-2+x} (30+5 e^4) \, dx\) [7600]

Optimal result

Integrand size = 16, antiderivative size = 14 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} e^{-2+x} \left (6+e^4\right ) \]

[Out]

5/12*(exp(4)+6)/exp(2-x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2225} \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} \left (6+e^4\right ) e^{x-2} \]

[In]

Int[(E^(-2 + x)*(30 + 5*E^4))/12,x]

[Out]

(5*E^(-2 + x)*(6 + E^4))/12

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}{12} \left (5 \left (6+e^4\right )\right ) \int e^{-2+x} \, dx \\ & = \frac {5}{12} e^{-2+x} \left (6+e^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} e^{-2+x} \left (6+e^4\right ) \]

[In]

Integrate[(E^(-2 + x)*(30 + 5*E^4))/12,x]

[Out]

(5*E^(-2 + x)*(6 + E^4))/12

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
gosper	\(\frac {5 \left ({\mathrm e}^{4}+6\right ) {\mathrm e}^{-2+x}}{12}\)	\(15\)
default	\(\left (\frac {5 \,{\mathrm e}^{4}}{12}+\frac {5}{2}\right ) {\mathrm e}^{-2+x}\)	\(16\)
norman	\(\left (\frac {5 \,{\mathrm e}^{4}}{12}+\frac {5}{2}\right ) {\mathrm e}^{-2+x}\)	\(16\)
risch	\(\frac {5 \,{\mathrm e}^{-2+x} {\mathrm e}^{4}}{12}+\frac {5 \,{\mathrm e}^{-2+x}}{2}\)	\(16\)
parallelrisch	\(\left (\frac {5 \,{\mathrm e}^{4}}{12}+\frac {5}{2}\right ) {\mathrm e}^{-2+x}\)	\(16\)
derivativedivides	\(-\left (-\frac {5 \,{\mathrm e}^{4}}{12}-\frac {5}{2}\right ) {\mathrm e}^{-2+x}\)	\(17\)

[In]

int(1/12*(5*exp(4)+30)/exp(2-x),x,method=_RETURNVERBOSE)

[Out]

5/12*(exp(4)+6)/exp(2-x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} \, {\left (e^{4} + 6\right )} e^{\left (x - 2\right )} \]

[In]

integrate(1/12*(5*exp(4)+30)/exp(2-x),x, algorithm="fricas")

[Out]

5/12*(e^4 + 6)*e^(x - 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {\left (30 + 5 e^{4}\right ) e^{x - 2}}{12} \]

[In]

integrate(1/12*(5*exp(4)+30)/exp(2-x),x)

[Out]

(30 + 5*exp(4))*exp(x - 2)/12

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} \, {\left (e^{4} + 6\right )} e^{\left (x - 2\right )} \]

[In]

integrate(1/12*(5*exp(4)+30)/exp(2-x),x, algorithm="maxima")

[Out]

5/12*(e^4 + 6)*e^(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5}{12} \, {\left (e^{4} + 6\right )} e^{\left (x - 2\right )} \]

[In]

integrate(1/12*(5*exp(4)+30)/exp(2-x),x, algorithm="giac")

[Out]

5/12*(e^4 + 6)*e^(x - 2)

Mupad [B] (verification not implemented)

Time = 13.39 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {1}{12} e^{-2+x} \left (30+5 e^4\right ) \, dx=\frac {5\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^4+6\right )}{12} \]

[In]

int(exp(x - 2)*((5*exp(4))/12 + 5/2),x)

[Out]

(5*exp(-2)*exp(x)*(exp(4) + 6))/12