3.45.33 \(\int \frac {-e^{e^{4-x}}+e^{4+e^{4-x}-x} x \log (x) \log ((-20+5 e^2+5 \log (5)) \log (x))}{x \log (x)} \, dx\)

Optimal. Leaf size=23 \[ -e^{e^{4-x}} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \]

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Rubi [A]  time = 1.09, antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6741, 6742, 2282, 2194, 2555} \begin {gather*} -e^{e^{4-x}} \log \left (-5 \left (4-e^2-\log (5)\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^E^(4 - x) + E^(4 + E^(4 - x) - x)*x*Log[x]*Log[(-20 + 5*E^2 + 5*Log[5])*Log[x]])/(x*Log[x]),x]

[Out]

-(E^E^(4 - x)*Log[-5*(4 - E^2 - Log[5])*Log[x]])

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2555

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

Rule 6741

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

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 \frac {e^{e^{4-x}-x} \left (-e^x+e^4 x \log (x) \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right )\right )}{x \log (x)} \, dx\\ &=\int \left (-\frac {e^{e^{4-x}}}{x \log (x)}+e^{4+e^{4-x}-x} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right )\right ) \, dx\\ &=-\int \frac {e^{e^{4-x}}}{x \log (x)} \, dx+\int e^{4+e^{4-x}-x} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \, dx\\ &=-e^{e^{4-x}} \log \left (-5 \left (4-e^2-\log (5)\right ) \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 23, normalized size = 1.00 \begin {gather*} -e^{e^{4-x}} \log \left (5 \left (-4+e^2+\log (5)\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^E^(4 - x) + E^(4 + E^(4 - x) - x)*x*Log[x]*Log[(-20 + 5*E^2 + 5*Log[5])*Log[x]])/(x*Log[x]),x]

[Out]

-(E^E^(4 - x)*Log[5*(-4 + E^2 + Log[5])*Log[x]])

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fricas [A]  time = 0.73, size = 20, normalized size = 0.87 \begin {gather*} -e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (5 \, {\left (e^{2} + \log \relax (5) - 4\right )} \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(-x+4)*log(x)*exp(exp(-x+4))*log((5*log(5)+5*exp(2)-20)*log(x))-exp(exp(-x+4)))/x/log(x),x, al
gorithm="fricas")

[Out]

-e^(e^(-x + 4))*log(5*(e^2 + log(5) - 4)*log(x))

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giac [A]  time = 0.26, size = 40, normalized size = 1.74 \begin {gather*} -e^{\left (e^{\left (-x + 4\right )}\right )} \log \relax (5) - e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (e^{2} + \log \relax (5) - 4\right ) - e^{\left (e^{\left (-x + 4\right )}\right )} \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(-x+4)*log(x)*exp(exp(-x+4))*log((5*log(5)+5*exp(2)-20)*log(x))-exp(exp(-x+4)))/x/log(x),x, al
gorithm="giac")

[Out]

-e^(e^(-x + 4))*log(5) - e^(e^(-x + 4))*log(e^2 + log(5) - 4) - e^(e^(-x + 4))*log(log(x))

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maple [A]  time = 0.08, size = 24, normalized size = 1.04

method	result	size
risch	\(-{\mathrm e}^{{\mathrm e}^{-x +4}} \ln \left (\left (5 \ln \relax (5)+5 \,{\mathrm e}^{2}-20\right ) \ln \relax (x )\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(-x+4)*ln(x)*exp(exp(-x+4))*ln((5*ln(5)+5*exp(2)-20)*ln(x))-exp(exp(-x+4)))/x/ln(x),x,method=_RETURN
VERBOSE)

[Out]

-exp(exp(-x+4))*ln((5*ln(5)+5*exp(2)-20)*ln(x))

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maxima [A]  time = 0.49, size = 22, normalized size = 0.96 \begin {gather*} -{\left (\log \relax (5) + \log \left (e^{2} + \log \relax (5) - 4\right ) + \log \left (\log \relax (x)\right )\right )} e^{\left (e^{\left (-x + 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(-x+4)*log(x)*exp(exp(-x+4))*log((5*log(5)+5*exp(2)-20)*log(x))-exp(exp(-x+4)))/x/log(x),x, al
gorithm="maxima")

[Out]

-(log(5) + log(e^2 + log(5) - 4) + log(log(x)))*e^(e^(-x + 4))

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mupad [B]  time = 3.37, size = 25, normalized size = 1.09 \begin {gather*} -{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^4}\,\left (\ln \left (\ln \relax (x)\right )+\ln \left (5\,{\mathrm {e}}^2+5\,\ln \relax (5)-20\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(4 - x)) - x*exp(exp(4 - x))*exp(4 - x)*log(log(x)*(5*exp(2) + 5*log(5) - 20))*log(x))/(x*log(x))
,x)

[Out]

-exp(exp(-x)*exp(4))*(log(log(x)) + log(5*exp(2) + 5*log(5) - 20))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(-x+4)*ln(x)*exp(exp(-x+4))*ln((5*ln(5)+5*exp(2)-20)*ln(x))-exp(exp(-x+4)))/x/ln(x),x)

[Out]

Timed out

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