3.35.14 \(\int \frac {e^{18+e^4-3 x-x \log (x)} (-4-\log (x))}{2+\log (2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {e^{18+e^4-3 x-x \log (x)}}{2+\log (2)} \]

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 6706} \begin {gather*} \frac {e^{-3 x+e^4+18} x^{-x}}{2+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(18 + E^4 - 3*x - x*Log[x])*(-4 - Log[x]))/(2 + Log[2]),x]

[Out]

E^(18 + E^4 - 3*x)/(x^x*(2 + Log[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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int e^{18+e^4-3 x-x \log (x)} (-4-\log (x)) \, dx}{2+\log (2)}\\ &=\frac {e^{18+e^4-3 x} x^{-x}}{2+\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 1.00 \begin {gather*} \frac {e^{18+e^4-3 x} x^{-x}}{2+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(18 + E^4 - 3*x - x*Log[x])*(-4 - Log[x]))/(2 + Log[2]),x]

[Out]

E^(18 + E^4 - 3*x)/(x^x*(2 + Log[2]))

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fricas [A]  time = 0.68, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-x \log \relax (x) - 3 \, x + e^{4} - \log \left (\log \relax (2) + 2\right ) + 18\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)*exp(-log(log(2)+2)-x*log(x)+exp(4)-3*x+18),x, algorithm="fricas")

[Out]

e^(-x*log(x) - 3*x + e^4 - log(log(2) + 2) + 18)

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giac [A]  time = 0.22, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-x \log \relax (x) - 3 \, x + e^{4} - \log \left (\log \relax (2) + 2\right ) + 18\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)*exp(-log(log(2)+2)-x*log(x)+exp(4)-3*x+18),x, algorithm="giac")

[Out]

e^(-x*log(x) - 3*x + e^4 - log(log(2) + 2) + 18)

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maple [A]  time = 0.02, size = 21, normalized size = 0.95

method	result	size
derivativedivides	\({\mathrm e}^{-\ln \left (\ln \relax (2)+2\right )-x \ln \relax (x )+{\mathrm e}^{4}-3 x +18}\)	\(21\)
default	\({\mathrm e}^{-\ln \left (\ln \relax (2)+2\right )-x \ln \relax (x )+{\mathrm e}^{4}-3 x +18}\)	\(21\)
norman	\({\mathrm e}^{-\ln \left (\ln \relax (2)+2\right )-x \ln \relax (x )+{\mathrm e}^{4}-3 x +18}\)	\(21\)
risch	\(\frac {x^{-x} {\mathrm e}^{18+{\mathrm e}^{4}-3 x}}{\ln \relax (2)+2}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)-4)*exp(-ln(ln(2)+2)-x*ln(x)+exp(4)-3*x+18),x,method=_RETURNVERBOSE)

[Out]

exp(-ln(ln(2)+2)-x*ln(x)+exp(4)-3*x+18)

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maxima [A]  time = 0.56, size = 20, normalized size = 0.91 \begin {gather*} \frac {e^{\left (-x \log \relax (x) - 3 \, x + e^{4} + 18\right )}}{\log \relax (2) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)*exp(-log(log(2)+2)-x*log(x)+exp(4)-3*x+18),x, algorithm="maxima")

[Out]

e^(-x*log(x) - 3*x + e^4 + 18)/(log(2) + 2)

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mupad [B]  time = 2.02, size = 21, normalized size = 0.95 \begin {gather*} \frac {{\mathrm {e}}^{-3\,x}\,{\mathrm {e}}^{18}\,{\mathrm {e}}^{{\mathrm {e}}^4}}{x^x\,\left (\ln \relax (2)+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(exp(4) - log(log(2) + 2) - 3*x - x*log(x) + 18)*(log(x) + 4),x)

[Out]

(exp(-3*x)*exp(18)*exp(exp(4)))/(x^x*(log(2) + 2))

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sympy [A]  time = 0.30, size = 19, normalized size = 0.86 \begin {gather*} \frac {e^{- x \log {\relax (x )} - 3 x + 18 + e^{4}}}{\log {\relax (2 )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)-4)*exp(-ln(ln(2)+2)-x*ln(x)+exp(4)-3*x+18),x)

[Out]

exp(-x*log(x) - 3*x + 18 + exp(4))/(log(2) + 2)

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