3.72.75 \(\int \frac {e^{-3-x} (-2+(1-x) \log (x))}{\log (5) \log ^3(x)} \, dx\)

Optimal. Leaf size=21 \[ e^4+\frac {e^{-3-x} x}{\log (5) \log ^2(x)} \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2201} \begin {gather*} \frac {e^{-x-3} x}{\log (5) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-3 - x)*(-2 + (1 - x)*Log[x]))/(Log[5]*Log[x]^3),x]

[Out]

(E^(-3 - x)*x)/(Log[5]*Log[x]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2201

Int[Log[(d_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((e_) + Log[(d_.)*(x_)]*(h_.)*((f_.) + (g_.)*(x_))
), x_Symbol] :> Simp[(e*x*F^(c*(a + b*x))*Log[d*x]^(n + 1))/(n + 1), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, n
}, x] && EqQ[e - f*h*(n + 1), 0] && EqQ[g*h*(n + 1) - b*c*e*Log[F], 0] && NeQ[n, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-3-x} (-2+(1-x) \log (x))}{\log ^3(x)} \, dx}{\log (5)}\\ &=\frac {e^{-3-x} x}{\log (5) \log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.81 \begin {gather*} \frac {e^{-3-x} x}{\log (5) \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 - x)*(-2 + (1 - x)*Log[x]))/(Log[5]*Log[x]^3),x]

[Out]

(E^(-3 - x)*x)/(Log[5]*Log[x]^2)

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fricas [A]  time = 0.60, size = 16, normalized size = 0.76 \begin {gather*} \frac {x e^{\left (-x - 3\right )}}{\log \relax (5) \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(x)-2)/log(5)/exp(3+x)/log(x)^3,x, algorithm="fricas")

[Out]

x*e^(-x - 3)/(log(5)*log(x)^2)

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giac [A]  time = 0.16, size = 16, normalized size = 0.76 \begin {gather*} \frac {x e^{\left (-x - 3\right )}}{\log \relax (5) \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(x)-2)/log(5)/exp(3+x)/log(x)^3,x, algorithm="giac")

[Out]

x*e^(-x - 3)/(log(5)*log(x)^2)

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maple [A]  time = 0.05, size = 17, normalized size = 0.81

method	result	size
norman	\(\frac {x \,{\mathrm e}^{-3-x}}{\ln \relax (x )^{2} \ln \relax (5)}\)	\(17\)
risch	\(\frac {x \,{\mathrm e}^{-3-x}}{\ln \relax (x )^{2} \ln \relax (5)}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*ln(x)-2)/ln(5)/exp(3+x)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x/ln(x)^2/ln(5)/exp(3+x)

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maxima [A]  time = 0.41, size = 16, normalized size = 0.76 \begin {gather*} \frac {x e^{\left (-x - 3\right )}}{\log \relax (5) \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*log(x)-2)/log(5)/exp(3+x)/log(x)^3,x, algorithm="maxima")

[Out]

x*e^(-x - 3)/(log(5)*log(x)^2)

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mupad [B]  time = 4.39, size = 16, normalized size = 0.76 \begin {gather*} \frac {x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}}{\ln \relax (5)\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- x - 3)*(log(x)*(x - 1) + 2))/(log(5)*log(x)^3),x)

[Out]

(x*exp(-x)*exp(-3))/(log(5)*log(x)^2)

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sympy [A]  time = 0.28, size = 15, normalized size = 0.71 \begin {gather*} \frac {x e^{- x - 3}}{\log {\relax (5 )} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*ln(x)-2)/ln(5)/exp(3+x)/ln(x)**3,x)

[Out]

x*exp(-x - 3)/(log(5)*log(x)**2)

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