\(\int e^{-x} (2+e^x+(2-2 x) \log (e^{12+2 e^2} x)) \, dx\) [6322]

Optimal result

Integrand size = 29, antiderivative size = 22 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x+2 e^{-x} x \log \left (e^{12+2 e^2} x\right ) \]

[Out]

x+2*ln(x*exp(exp(2)+6)^2)*x/exp(x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(22)=44\).

Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {6874, 2225, 6820, 2207, 2634} \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=-4 \left (6+e^2\right ) e^{-x} (1-x)+x+4 \left (6+e^2\right ) e^{-x}-2 e^{-x} (1-x) \log (x)+2 e^{-x} \log (x) \]

[In]

Int[(2 + E^x + (2 - 2*x)*Log[E^(12 + 2*E^2)*x])/E^x,x]

[Out]

(4*(6 + E^2))/E^x - (4*(6 + E^2)*(1 - x))/E^x + x + (2*Log[x])/E^x - (2*(1 - x)*Log[x])/E^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 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (1+2 e^{-x}+2 e^{-x} (1-x) \left (12 \left (1+\frac {e^2}{6}\right )+\log (x)\right )\right ) \, dx \\ & = x+2 \int e^{-x} \, dx+2 \int e^{-x} (1-x) \left (12 \left (1+\frac {e^2}{6}\right )+\log (x)\right ) \, dx \\ & = -2 e^{-x}+x+2 \int e^{-x} (1-x) \left (2 \left (6+e^2\right )+\log (x)\right ) \, dx \\ & = -2 e^{-x}+x+2 \int \left (-2 e^{-x} \left (6+e^2\right ) (-1+x)-e^{-x} (-1+x) \log (x)\right ) \, dx \\ & = -2 e^{-x}+x-2 \int e^{-x} (-1+x) \log (x) \, dx-\left (4 \left (6+e^2\right )\right ) \int e^{-x} (-1+x) \, dx \\ & = -2 e^{-x}-4 e^{-x} \left (6+e^2\right ) (1-x)+x+2 e^{-x} \log (x)-2 e^{-x} (1-x) \log (x)-2 \int e^{-x} \, dx-\left (4 \left (6+e^2\right )\right ) \int e^{-x} \, dx \\ & = 4 e^{-x} \left (6+e^2\right )-4 e^{-x} \left (6+e^2\right ) (1-x)+x+2 e^{-x} \log (x)-2 e^{-x} (1-x) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x+2 e^{-x} \left (2 \left (6+e^2\right ) x+x \log (x)\right ) \]

[In]

Integrate[(2 + E^x + (2 - 2*x)*Log[E^(12 + 2*E^2)*x])/E^x,x]

[Out]

x + (2*(2*(6 + E^2)*x + x*Log[x]))/E^x

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
default	\(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\)	\(20\)
risch	\(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\)	\(20\)
parts	\(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\)	\(20\)
norman	\(\left ({\mathrm e}^{x} x +2 \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right ) x \right ) {\mathrm e}^{-x}\)	\(24\)
parallelrisch	\(\left ({\mathrm e}^{x} x +2 \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right ) x \right ) {\mathrm e}^{-x}\)	\(24\)

[In]

int(((2-2*x)*ln(x*exp(exp(2)+6)^2)+exp(x)+2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

x+2*ln(x*exp(exp(2)+6)^2)*x/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx={\left (x e^{x} + 2 \, x \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right )\right )} e^{\left (-x\right )} \]

[In]

integrate(((2-2*x)*log(x*exp(exp(2)+6)^2)+exp(x)+2)/exp(x),x, algorithm="fricas")

[Out]

(x*e^x + 2*x*log(x*e^(2*e^2 + 12)))*e^(-x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x + 2 x e^{- x} \log {\left (x e^{12 + 2 e^{2}} \right )} \]

[In]

integrate(((2-2*x)*ln(x*exp(exp(2)+6)**2)+exp(x)+2)/exp(x),x)

[Out]

x + 2*x*exp(-x)*log(x*exp(12 + 2*exp(2)))

Maxima [F]

\[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=\int { -{\left (2 \, {\left (x - 1\right )} \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right ) - e^{x} - 2\right )} e^{\left (-x\right )} \,d x } \]

[In]

integrate(((2-2*x)*log(x*exp(exp(2)+6)^2)+exp(x)+2)/exp(x),x, algorithm="maxima")

[Out]

2*(x + 1)*e^(-x)*log(x) - 2*e^(-x)*log(x*e^(2*e^2 + 12)) + x + 2*Ei(-x) - 2*e^(-x) - 2*integrate((2*x^2*(e^2 +
6) + x + 1)*e^(-x)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=2 \, x e^{\left (-x\right )} \log \left (x\right ) + 24 \, x e^{\left (-x\right )} + 4 \, x e^{\left (-x + 2\right )} + x \]

[In]

integrate(((2-2*x)*log(x*exp(exp(2)+6)^2)+exp(x)+2)/exp(x),x, algorithm="giac")

[Out]

2*x*e^(-x)*log(x) + 24*x*e^(-x) + 4*x*e^(-x + 2) + x

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x\,{\mathrm {e}}^{-x}\,\left (4\,{\mathrm {e}}^2+{\mathrm {e}}^x+2\,\ln \left (x\right )+24\right ) \]

[In]

int(exp(-x)*(exp(x) - log(x*exp(2*exp(2) + 12))*(2*x - 2) + 2),x)

[Out]

x*exp(-x)*(4*exp(2) + exp(x) + 2*log(x) + 24)