\(\int e^{-3-2 x} (4 e^{2+2 x}+e (1+4 x-4 x^2)+e (1-2 x) \log (x)) \, dx\) [6820]

Optimal result

Integrand size = 39, antiderivative size = 23 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=-2+\frac {4 x}{e}+e^{-2-2 x} x (2 x+\log (x)) \]

[Out]

4*x/exp(1)-2+x/exp(1+x)^2*(2*x+ln(x))

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 18, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {6873, 6874, 2225, 2207, 2634, 12, 2209, 2230} \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=2 e^{-2 x-2} x^2+\frac {4 x}{e}+e^{-2 x-2} x \log (x) \]

[In]

Int[E^(-3 - 2*x)*(4*E^(2 + 2*x) + E*(1 + 4*x - 4*x^2) + E*(1 - 2*x)*Log[x]),x]

[Out]

(4*x)/E + 2*E^(-2 - 2*x)*x^2 + E^(-2 - 2*x)*x*Log[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 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !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 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

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 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}& = \int e^{-2-2 x} \left (1+4 e^{1+2 x}+4 x-4 x^2+\log (x)-2 x \log (x)\right ) \, dx \\ & = \int \left (\frac {4}{e}+e^{-2-2 x}+4 e^{-2-2 x} x-4 e^{-2-2 x} x^2+e^{-2-2 x} \log (x)-2 e^{-2-2 x} x \log (x)\right ) \, dx \\ & = \frac {4 x}{e}-2 \int e^{-2-2 x} x \log (x) \, dx+4 \int e^{-2-2 x} x \, dx-4 \int e^{-2-2 x} x^2 \, dx+\int e^{-2-2 x} \, dx+\int e^{-2-2 x} \log (x) \, dx \\ & = -\frac {1}{2} e^{-2-2 x}+\frac {4 x}{e}-2 e^{-2-2 x} x+2 e^{-2-2 x} x^2+e^{-2-2 x} x \log (x)+2 \int e^{-2-2 x} \, dx+2 \int \frac {e^{-2-2 x} (-1-2 x)}{4 x} \, dx-4 \int e^{-2-2 x} x \, dx-\int -\frac {e^{-2-2 x}}{2 x} \, dx \\ & = -\frac {3}{2} e^{-2-2 x}+\frac {4 x}{e}+2 e^{-2-2 x} x^2+e^{-2-2 x} x \log (x)+\frac {1}{2} \int \frac {e^{-2-2 x}}{x} \, dx+\frac {1}{2} \int \frac {e^{-2-2 x} (-1-2 x)}{x} \, dx-2 \int e^{-2-2 x} \, dx \\ & = -\frac {1}{2} e^{-2-2 x}+\frac {4 x}{e}+2 e^{-2-2 x} x^2+\frac {\text {Ei}(-2 x)}{2 e^2}+e^{-2-2 x} x \log (x)+\frac {1}{2} \int \left (-2 e^{-2-2 x}-\frac {e^{-2-2 x}}{x}\right ) \, dx \\ & = -\frac {1}{2} e^{-2-2 x}+\frac {4 x}{e}+2 e^{-2-2 x} x^2+\frac {\text {Ei}(-2 x)}{2 e^2}+e^{-2-2 x} x \log (x)-\frac {1}{2} \int \frac {e^{-2-2 x}}{x} \, dx-\int e^{-2-2 x} \, dx \\ & = \frac {4 x}{e}+2 e^{-2-2 x} x^2+e^{-2-2 x} x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=e^{-2 (1+x)} x \left (4 e^{1+2 x}+2 x+\log (x)\right ) \]

[In]

Integrate[E^(-3 - 2*x)*(4*E^(2 + 2*x) + E*(1 + 4*x - 4*x^2) + E*(1 - 2*x)*Log[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
parts	\(\left (x \ln \left (x \right )+2 x^{2}\right ) {\mathrm e}^{-2-2 x}+4 \,{\mathrm e}^{-1} x\)	\(26\)
risch	\(2 \,{\mathrm e}^{-2-2 x} x^{2}+x \,{\mathrm e}^{-2-2 x} \ln \left (x \right )+4 \,{\mathrm e}^{-1} x\)	\(28\)
default	\({\mathrm e}^{-1} \left (4 x +\left (x \,{\mathrm e} \ln \left (x \right )+2 x^{2} {\mathrm e}\right ) {\mathrm e}^{-2-2 x}\right )\)	\(31\)
norman	\(\left (x \ln \left (x \right )+2 x^{2}+4 x \,{\mathrm e}^{-1} {\mathrm e}^{2+2 x}\right ) {\mathrm e}^{-2-2 x}\)	\(31\)
parallelrisch	\({\mathrm e}^{-1} \left (2 x^{2} {\mathrm e}+x \,{\mathrm e} \ln \left (x \right )+4 x \,{\mathrm e}^{2+2 x}\right ) {\mathrm e}^{-2-2 x}\)	\(35\)

[In]

int(((1-2*x)*exp(1)*ln(x)+4*exp(1+x)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(1+x)^2,x,method=_RETURNVERBOSE)

[Out]

(x*ln(x)+2*x^2)/exp(1+x)^2+4*x/exp(1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx={\left (2 \, x^{2} e + x e \log \left (x\right ) + 4 \, x e^{\left (2 \, x + 2\right )}\right )} e^{\left (-2 \, x - 3\right )} \]

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(1+x)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(1+x)^2,x, algorithm="fricas")

[Out]

(2*x^2*e + x*e*log(x) + 4*x*e^(2*x + 2))*e^(-2*x - 3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=\frac {4 x}{e} + \left (2 x^{2} + x \log {\left (x \right )}\right ) e^{- 2 x - 2} \]

[In]

integrate(((1-2*x)*exp(1)*ln(x)+4*exp(1+x)**2+(-4*x**2+4*x+1)*exp(1))/exp(1)/exp(1+x)**2,x)

[Out]

4*x*exp(-1) + (2*x**2 + x*log(x))*exp(-2*x - 2)

Maxima [F]

\[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=\int { -{\left ({\left (2 \, x - 1\right )} e \log \left (x\right ) + {\left (4 \, x^{2} - 4 \, x - 1\right )} e - 4 \, e^{\left (2 \, x + 2\right )}\right )} e^{\left (-2 \, x - 3\right )} \,d x } \]

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(1+x)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(1+x)^2,x, algorithm="maxima")

[Out]

1/2*(2*x + 1)*e^(-2*x - 2)*log(x) + 4*x*e^(-1) + 1/2*Ei(-2*x)*e^(-2) + (2*x^2 + 2*x + 1)*e^(-2*x - 2) - (2*x +
1)*e^(-2*x - 2) - 1/2*e^(-2*x - 2)*log(x) - 1/2*e^(-2*x - 2) - 1/2*integrate((2*x + 1)*e^(-2*x - 2)/x, x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx={\left (2 \, x^{2} e^{\left (-2 \, x + 1\right )} + x e^{\left (-2 \, x + 1\right )} \log \left (x\right ) + 4 \, x e^{2}\right )} e^{\left (-3\right )} \]

[In]

integrate(((1-2*x)*exp(1)*log(x)+4*exp(1+x)^2+(-4*x^2+4*x+1)*exp(1))/exp(1)/exp(1+x)^2,x, algorithm="giac")

[Out]

(2*x^2*e^(-2*x + 1) + x*e^(-2*x + 1)*log(x) + 4*x*e^2)*e^(-3)

Mupad [B] (verification not implemented)

Time = 12.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int e^{-3-2 x} \left (4 e^{2+2 x}+e \left (1+4 x-4 x^2\right )+e (1-2 x) \log (x)\right ) \, dx=4\,x\,{\mathrm {e}}^{-1}+2\,x^2\,{\mathrm {e}}^{-2\,x-2}+x\,{\mathrm {e}}^{-2\,x-2}\,\ln \left (x\right ) \]

[In]

int(exp(-1)*exp(- 2*x - 2)*(4*exp(2*x + 2) + exp(1)*(4*x - 4*x^2 + 1) - exp(1)*log(x)*(2*x - 1)),x)

[Out]

4*x*exp(-1) + 2*x^2*exp(- 2*x - 2) + x*exp(- 2*x - 2)*log(x)