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

Optimal. Leaf size=22 \[ x+2 e^{-x} x \log \left (e^{12+2 e^2} x\right ) \]

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Rubi [B]  time = 0.25, antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {6742, 2194, 6688, 2176, 2554} \begin {gather*} -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) \end {gather*}

Antiderivative was successfully verified.

[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 2176

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] &&  !$UseGamma === True

Rule 2194

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 2554

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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 22, normalized size = 1.00 \begin {gather*} x+2 e^{-x} \left (2 \left (6+e^2\right ) x+x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.82, size = 23, normalized size = 1.05 \begin {gather*} {\left (x e^{x} + 2 \, x \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+2)*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)

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giac [A]  time = 0.19, size = 27, normalized size = 1.23 \begin {gather*} 2 \, x e^{\left (-x\right )} \log \relax (x) + 24 \, x e^{\left (-x\right )} + 4 \, x e^{\left (-x + 2\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+2)*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

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maple [A]  time = 0.06, size = 20, normalized size = 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	\(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\)	\(20\)
norman	\(\left ({\mathrm e}^{x} x +2 \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right ) x \right ) {\mathrm e}^{-x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x+2)*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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, {\left (x + 1\right )} e^{\left (-x\right )} \log \relax (x) - 2 \, e^{\left (-x\right )} \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right ) + x + 2 \, {\rm Ei}\left (-x\right ) - 2 \, e^{\left (-x\right )} - 2 \, \int \frac {{\left (2 \, x^{2} {\left (e^{2} + 6\right )} + x + 1\right )} e^{\left (-x\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+2)*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)

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mupad [B]  time = 4.46, size = 18, normalized size = 0.82 \begin {gather*} x\,{\mathrm {e}}^{-x}\,\left (4\,{\mathrm {e}}^2+{\mathrm {e}}^x+2\,\ln \relax (x)+24\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.30, size = 19, normalized size = 0.86 \begin {gather*} x + 2 x e^{- x} \log {\left (x e^{12 + 2 e^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+2)*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)))

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