\(\int \frac {1}{12} e^{-x} (48-12 e+e^x-48 x) \, dx\) [5394]

Optimal result

Integrand size = 20, antiderivative size = 22 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=\frac {1}{3} \left (\frac {x}{4}+3 e^{-x} (e+4 x)\right ) \]

[Out]

1/12*x+(exp(1)+4*x)/exp(x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 6873, 6874, 2225, 2207} \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=4 e^{-x} x+\frac {x}{12}-(4-e) e^{-x}+4 e^{-x} \]

[In]

Int[(48 - 12*E + E^x - 48*x)/(12*E^x),x]

[Out]

4/E^x - (4 - E)/E^x + x/12 + (4*x)/E^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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{12} \int e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx \\ & = \frac {1}{12} \int e^{-x} \left (48 \left (1-\frac {e}{4}\right )+e^x-48 x\right ) \, dx \\ & = \frac {1}{12} \int \left (1-12 (-4+e) e^{-x}-48 e^{-x} x\right ) \, dx \\ & = \frac {x}{12}-4 \int e^{-x} x \, dx+(4-e) \int e^{-x} \, dx \\ & = -\left ((4-e) e^{-x}\right )+\frac {x}{12}+4 e^{-x} x-4 \int e^{-x} \, dx \\ & = 4 e^{-x}-(4-e) e^{-x}+\frac {x}{12}+4 e^{-x} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=\frac {1}{12} \left (x+12 e^{-x} (e+4 x)\right ) \]

[In]

Integrate[(48 - 12*E + E^x - 48*x)/(12*E^x),x]

[Out]

(x + (12*(E + 4*x))/E^x)/12

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
norman	\(\left (4 x +\frac {{\mathrm e}^{x} x}{12}+{\mathrm e}\right ) {\mathrm e}^{-x}\)	\(17\)
risch	\(\frac {x}{12}+{\mathrm e}^{1-x}+4 x \,{\mathrm e}^{-x}\)	\(18\)
default	\(\frac {x}{12}+{\mathrm e}^{-x} {\mathrm e}+4 x \,{\mathrm e}^{-x}\)	\(19\)
parallelrisch	\(\frac {\left ({\mathrm e}^{x} x +12 \,{\mathrm e}+48 x \right ) {\mathrm e}^{-x}}{12}\)	\(19\)
parts	\(\frac {x}{12}+{\mathrm e}^{-x} {\mathrm e}+4 x \,{\mathrm e}^{-x}\)	\(19\)

[In]

int(1/12*(exp(x)-12*exp(1)-48*x+48)/exp(x),x,method=_RETURNVERBOSE)

[Out]

(4*x+1/12*exp(x)*x+exp(1))/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=\frac {1}{12} \, {\left (x e^{x} + 48 \, x + 12 \, e\right )} e^{\left (-x\right )} \]

[In]

integrate(1/12*(exp(x)-12*exp(1)-48*x+48)/exp(x),x, algorithm="fricas")

[Out]

1/12*(x*e^x + 48*x + 12*e)*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.55 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=\frac {x}{12} + \left (4 x + e\right ) e^{- x} \]

[In]

integrate(1/12*(exp(x)-12*exp(1)-48*x+48)/exp(x),x)

[Out]

x/12 + (4*x + E)*exp(-x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=4 \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {1}{12} \, x - 4 \, e^{\left (-x\right )} + e^{\left (-x + 1\right )} \]

[In]

integrate(1/12*(exp(x)-12*exp(1)-48*x+48)/exp(x),x, algorithm="maxima")

[Out]

4*(x + 1)*e^(-x) + 1/12*x - 4*e^(-x) + e^(-x + 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=4 \, x e^{\left (-x\right )} + \frac {1}{12} \, x + e^{\left (-x + 1\right )} \]

[In]

integrate(1/12*(exp(x)-12*exp(1)-48*x+48)/exp(x),x, algorithm="giac")

[Out]

4*x*e^(-x) + 1/12*x + e^(-x + 1)

Mupad [B] (verification not implemented)

Time = 11.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{12} e^{-x} \left (48-12 e+e^x-48 x\right ) \, dx=\frac {x}{12}+{\mathrm {e}}^{-x}\,\mathrm {e}+4\,x\,{\mathrm {e}}^{-x} \]

[In]

int(-exp(-x)*(4*x + exp(1) - exp(x)/12 - 4),x)

[Out]

x/12 + exp(-x)*exp(1) + 4*x*exp(-x)