3.89.34 \(\int \frac {e^{-4-e^{75}} (1+(-1-2 e^{4+e^{75}} x) \log (x))}{x \log (x)} \, dx\)

Optimal. Leaf size=26 \[ 3+e^5-2 x-e^{-4-e^{75}} \log \left (\frac {x}{\log (x)}\right ) \]

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Rubi [A]  time = 0.25, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {12, 6742, 43, 2302, 29} \begin {gather*} -2 x-e^{-4-e^{75}} \log (x)+e^{-4-e^{75}} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 - E^75)*(1 + (-1 - 2*E^(4 + E^75)*x)*Log[x]))/(x*Log[x]),x]

[Out]

-2*x - E^(-4 - E^75)*Log[x] + E^(-4 - E^75)*Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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} &=e^{-4-e^{75}} \int \frac {1+\left (-1-2 e^{4+e^{75}} x\right ) \log (x)}{x \log (x)} \, dx\\ &=e^{-4-e^{75}} \int \left (-\frac {1+2 e^{4+e^{75}} x}{x}+\frac {1}{x \log (x)}\right ) \, dx\\ &=-\left (e^{-4-e^{75}} \int \frac {1+2 e^{4+e^{75}} x}{x} \, dx\right )+e^{-4-e^{75}} \int \frac {1}{x \log (x)} \, dx\\ &=-\left (e^{-4-e^{75}} \int \left (2 e^{4+e^{75}}+\frac {1}{x}\right ) \, dx\right )+e^{-4-e^{75}} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-2 x-e^{-4-e^{75}} \log (x)+e^{-4-e^{75}} \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 29, normalized size = 1.12 \begin {gather*} -e^{-4-e^{75}} \left (2 e^{4+e^{75}} x+\log (x)-\log (\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 - E^75)*(1 + (-1 - 2*E^(4 + E^75)*x)*Log[x]))/(x*Log[x]),x]

[Out]

-(E^(-4 - E^75)*(2*E^(4 + E^75)*x + Log[x] - Log[Log[x]]))

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fricas [A]  time = 0.64, size = 25, normalized size = 0.96 \begin {gather*} -{\left (2 \, x e^{\left (e^{75} + 4\right )} + \log \relax (x) - \log \left (\log \relax (x)\right )\right )} e^{\left (-e^{75} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(exp(75)+4)-1)*log(x)+1)/x/exp(exp(75)+4)/log(x),x, algorithm="fricas")

[Out]

-(2*x*e^(e^75 + 4) + log(x) - log(log(x)))*e^(-e^75 - 4)

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giac [A]  time = 0.14, size = 25, normalized size = 0.96 \begin {gather*} -{\left (2 \, x e^{\left (e^{75} + 4\right )} + \log \relax (x) - \log \left (\log \relax (x)\right )\right )} e^{\left (-e^{75} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(exp(75)+4)-1)*log(x)+1)/x/exp(exp(75)+4)/log(x),x, algorithm="giac")

[Out]

-(2*x*e^(e^75 + 4) + log(x) - log(log(x)))*e^(-e^75 - 4)

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maple [A]  time = 0.06, size = 25, normalized size = 0.96

method	result	size
default	\({\mathrm e}^{-{\mathrm e}^{75}-4} \left (-2 \,{\mathrm e}^{{\mathrm e}^{75}} {\mathrm e}^{4} x -\ln \relax (x )+\ln \left (\ln \relax (x )\right )\right )\)	\(25\)
risch	\(-2 x -\ln \relax (x ) {\mathrm e}^{-{\mathrm e}^{75}-4}+\ln \left (\ln \relax (x )\right ) {\mathrm e}^{-{\mathrm e}^{75}-4}\)	\(27\)
norman	\(-{\mathrm e}^{-{\mathrm e}^{75}} {\mathrm e}^{-4} \ln \relax (x )-2 x +{\mathrm e}^{-{\mathrm e}^{75}} {\mathrm e}^{-4} \ln \left (\ln \relax (x )\right )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(exp(75)+4)-1)*ln(x)+1)/x/exp(exp(75)+4)/ln(x),x,method=_RETURNVERBOSE)

[Out]

1/exp(exp(75)+4)*(-2*exp(exp(75))*exp(4)*x-ln(x)+ln(ln(x)))

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maxima [A]  time = 0.36, size = 25, normalized size = 0.96 \begin {gather*} -{\left (2 \, x e^{\left (e^{75} + 4\right )} + \log \relax (x) - \log \left (\log \relax (x)\right )\right )} e^{\left (-e^{75} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(exp(75)+4)-1)*log(x)+1)/x/exp(exp(75)+4)/log(x),x, algorithm="maxima")

[Out]

-(2*x*e^(e^75 + 4) + log(x) - log(log(x)))*e^(-e^75 - 4)

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mupad [B]  time = 5.34, size = 25, normalized size = 0.96 \begin {gather*} -{\mathrm {e}}^{-{\mathrm {e}}^{75}-4}\,\left (\ln \relax (x)-\ln \left (\ln \relax (x)\right )+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{75}+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- exp(75) - 4)*(log(x)*(2*x*exp(exp(75) + 4) + 1) - 1))/(x*log(x)),x)

[Out]

-exp(- exp(75) - 4)*(log(x) - log(log(x)) + 2*x*exp(exp(75) + 4))

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sympy [A]  time = 0.14, size = 37, normalized size = 1.42 \begin {gather*} \frac {- 2 x e^{4} e^{e^{75}} - \log {\relax (x )}}{e^{4} e^{e^{75}}} + \frac {\log {\left (\log {\relax (x )} \right )}}{e^{4} e^{e^{75}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(exp(75)+4)-1)*ln(x)+1)/x/exp(exp(75)+4)/ln(x),x)

[Out]

(-2*x*exp(4)*exp(exp(75)) - log(x))*exp(-4)*exp(-exp(75)) + exp(-4)*exp(-exp(75))*log(log(x))

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