\(\int \frac {1}{3} e^{-x} (-3+3 e^x+3 x+(2-2 x) \log (3)) \, dx\) [7696]

Optimal result

Integrand size = 27, antiderivative size = 23 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=5+x-e^{-x} x+\frac {2}{3} e^{-x} x \log (3) \]

[Out]

x-x/exp(x)+5+2/3*x*ln(3)/exp(x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 6820, 2207, 2225} \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=x+\frac {1}{3} e^{-x} (1-x) (3-\log (9))-\frac {1}{3} e^{-x} (3-\log (9)) \]

[In]

Int[(-3 + 3*E^x + 3*x + (2 - 2*x)*Log[3])/(3*E^x),x]

[Out]

x - (3 - Log[9])/(3*E^x) + ((1 - x)*(3 - Log[9]))/(3*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 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx \\ & = \frac {1}{3} \int \left (3-e^{-x} (-1+x) (-3+\log (9))\right ) \, dx \\ & = x+\frac {1}{3} (3-\log (9)) \int e^{-x} (-1+x) \, dx \\ & = x+\frac {1}{3} e^{-x} (1-x) (3-\log (9))+\frac {1}{3} (3-\log (9)) \int e^{-x} \, dx \\ & = x-\frac {1}{3} e^{-x} (3-\log (9))+\frac {1}{3} e^{-x} (1-x) (3-\log (9)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=x+\frac {1}{3} e^{-x} x (-3+\log (9)) \]

[In]

Integrate[(-3 + 3*E^x + 3*x + (2 - 2*x)*Log[3])/(3*E^x),x]

[Out]

x + (x*(-3 + Log[9]))/(3*E^x)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
risch	\(x +\frac {\left (2 \ln \left (3\right )-3\right ) x \,{\mathrm e}^{-x}}{3}\)	\(16\)
norman	\(\left (\left (\frac {2 \ln \left (3\right )}{3}-1\right ) x +{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}\)	\(19\)
parts	\(x +\frac {2 x \,{\mathrm e}^{-x} \ln \left (3\right )}{3}-x \,{\mathrm e}^{-x}\)	\(19\)
parallelrisch	\(\frac {\left (2 x \ln \left (3\right )+3 \,{\mathrm e}^{x} x -3 x \right ) {\mathrm e}^{-x}}{3}\)	\(21\)
default	\(x -x \,{\mathrm e}^{-x}-\frac {2 \,{\mathrm e}^{-x} \ln \left (3\right )}{3}-\frac {2 \ln \left (3\right ) \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )}{3}\)	\(36\)

[In]

int(1/3*(3*exp(x)+(2-2*x)*ln(3)+3*x-3)/exp(x),x,method=_RETURNVERBOSE)

[Out]

x+1/3*(2*ln(3)-3)*x*exp(-x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=\frac {1}{3} \, {\left (3 \, x e^{x} + 2 \, x \log \left (3\right ) - 3 \, x\right )} e^{\left (-x\right )} \]

[In]

integrate(1/3*(3*exp(x)+(2-2*x)*log(3)+3*x-3)/exp(x),x, algorithm="fricas")

[Out]

1/3*(3*x*e^x + 2*x*log(3) - 3*x)*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=x + \frac {\left (- 3 x + 2 x \log {\left (3 \right )}\right ) e^{- x}}{3} \]

[In]

integrate(1/3*(3*exp(x)+(2-2*x)*ln(3)+3*x-3)/exp(x),x)

[Out]

x + (-3*x + 2*x*log(3))*exp(-x)/3

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=\frac {2}{3} \, {\left (x + 1\right )} e^{\left (-x\right )} \log \left (3\right ) - {\left (x + 1\right )} e^{\left (-x\right )} - \frac {2}{3} \, e^{\left (-x\right )} \log \left (3\right ) + x + e^{\left (-x\right )} \]

[In]

integrate(1/3*(3*exp(x)+(2-2*x)*log(3)+3*x-3)/exp(x),x, algorithm="maxima")

[Out]

2/3*(x + 1)*e^(-x)*log(3) - (x + 1)*e^(-x) - 2/3*e^(-x)*log(3) + x + e^(-x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=\frac {1}{3} \, {\left (2 \, x \log \left (3\right ) - 3 \, x\right )} e^{\left (-x\right )} + x \]

[In]

integrate(1/3*(3*exp(x)+(2-2*x)*log(3)+3*x-3)/exp(x),x, algorithm="giac")

[Out]

1/3*(2*x*log(3) - 3*x)*e^(-x) + x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{3} e^{-x} \left (-3+3 e^x+3 x+(2-2 x) \log (3)\right ) \, dx=\frac {x\,\left (2\,{\mathrm {e}}^{-x}\,\ln \left (3\right )-3\,{\mathrm {e}}^{-x}+3\right )}{3} \]

[In]

int(exp(-x)*(x + exp(x) - (log(3)*(2*x - 2))/3 - 1),x)

[Out]

(x*(2*exp(-x)*log(3) - 3*exp(-x) + 3))/3