∫ (25+x) log(x)+log(x) log(log(x))+(25+x+log(log(x)))(12+(300+24x) log(x)+12 log(x) log(log(x))) e5__________________

Optimal. Leaf size=17 \[ \frac {17}{5}+x+\frac {12 x (25+x+\log (\log (x)))}{e^5} \]

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Rubi [A] time = 0.13, antiderivative size = 28, normalized size of antiderivative = 1.65, number of steps used = 9, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {6688, 12, 6742, 2298, 2520} \begin {gather*} \frac {12 x^2}{e^5}+\frac {\left (300+e^5\right ) x}{e^5}+\frac {12 x \log (\log (x))}{e^5} \end {gather*}

Int[((25 + x)*Log[x] + Log[x]*Log[Log[x]] + ((25 + x + Log[Log[x]])*(12 + (300 + 24*x)*Log[x] + 12*Log[x]*Log[

Log[x]]))/E^5)/((25 + x)*Log[x] + Log[x]*Log[Log[x]]),x]

((300 + E^5)*x)/E^5 + (12*x^2)/E^5 + (12*x*Log[Log[x]])/E^5

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

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[

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

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12+\log (x) \left (300+e^5+24 x+12 \log (\log (x))\right )}{e^5 \log (x)} \, dx\\ &=\frac {\int \frac {12+\log (x) \left (300+e^5+24 x+12 \log (\log (x))\right )}{\log (x)} \, dx}{e^5}\\ &=\frac {\int \left (\frac {12+300 \left (1+\frac {e^5}{300}\right ) \log (x)+24 x \log (x)}{\log (x)}+12 \log (\log (x))\right ) \, dx}{e^5}\\ &=\frac {\int \frac {12+300 \left (1+\frac {e^5}{300}\right ) \log (x)+24 x \log (x)}{\log (x)} \, dx}{e^5}+\frac {12 \int \log (\log (x)) \, dx}{e^5}\\ &=\frac {12 x \log (\log (x))}{e^5}+\frac {\int \left (300+e^5+24 x+\frac {12}{\log (x)}\right ) \, dx}{e^5}-\frac {12 \int \frac {1}{\log (x)} \, dx}{e^5}\\ &=\frac {\left (300+e^5\right ) x}{e^5}+\frac {12 x^2}{e^5}+\frac {12 x \log (\log (x))}{e^5}-\frac {12 \text {li}(x)}{e^5}+\frac {12 \int \frac {1}{\log (x)} \, dx}{e^5}\\ &=\frac {\left (300+e^5\right ) x}{e^5}+\frac {12 x^2}{e^5}+\frac {12 x \log (\log (x))}{e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.05, size = 19, normalized size = 1.12 \begin {gather*} \frac {x \left (e^5+12 (25+x)+12 \log (\log (x))\right )}{e^5} \end {gather*}

Integrate[((25 + x)*Log[x] + Log[x]*Log[Log[x]] + ((25 + x + Log[Log[x]])*(12 + (300 + 24*x)*Log[x] + 12*Log[x

]*Log[Log[x]]))/E^5)/((25 + x)*Log[x] + Log[x]*Log[Log[x]]),x]

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fricas [A] time = 0.72, size = 22, normalized size = 1.29 \begin {gather*} {\left (12 \, x^{2} + x e^{5} + 12 \, x \log \left (\log \relax (x)\right ) + 300 \, x\right )} e^{\left (-5\right )} \end {gather*}

integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x))+x+25)-5)+log(x)*log(log(x))+(x+25)

*log(x))/(log(x)*log(log(x))+(x+25)*log(x)),x, algorithm="fricas")

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giac [A] time = 0.49, size = 22, normalized size = 1.29 \begin {gather*} 12 \, x^{2} e^{\left (-5\right )} + 12 \, x e^{\left (-5\right )} \log \left (\log \relax (x)\right ) + 300 \, x e^{\left (-5\right )} + x \end {gather*}

integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x))+x+25)-5)+log(x)*log(log(x))+(x+25)

*log(x))/(log(x)*log(log(x))+(x+25)*log(x)),x, algorithm="giac")

12*x^2*e^(-5) + 12*x*e^(-5)*log(log(x)) + 300*x*e^(-5) + x

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method	result	size

default	\(x +12 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-5} x +12 \,{\mathrm e}^{-5} x^{2}+300 x \,{\mathrm e}^{-5}\)	\(23\)
risch	\(x +12 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-5} x +12 \,{\mathrm e}^{-5} x^{2}+300 x \,{\mathrm e}^{-5}\)	\(23\)

int(((12*ln(x)*ln(ln(x))+(24*x+300)*ln(x)+12)*exp(ln(ln(ln(x))+x+25)-5)+ln(x)*ln(ln(x))+(x+25)*ln(x))/(ln(x)*l

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maxima [A] time = 0.93, size = 21, normalized size = 1.24 \begin {gather*} {\left (12 \, x^{2} + x {\left (e^{5} + 300\right )} + 12 \, x \log \left (\log \relax (x)\right )\right )} e^{\left (-5\right )} \end {gather*}

integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x))+x+25)-5)+log(x)*log(log(x))+(x+25)

*log(x))/(log(x)*log(log(x))+(x+25)*log(x)),x, algorithm="maxima")

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mupad [B] time = 0.58, size = 16, normalized size = 0.94 \begin {gather*} x\,{\mathrm {e}}^{-5}\,\left (12\,x+12\,\ln \left (\ln \relax (x)\right )+{\mathrm {e}}^5+300\right ) \end {gather*}

int((log(x)*(x + 25) + exp(log(x + log(log(x)) + 25) - 5)*(log(x)*(24*x + 300) + 12*log(log(x))*log(x) + 12) +

log(log(x))*log(x))/(log(x)*(x + 25) + log(log(x))*log(x)),x)

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sympy [A] time = 0.33, size = 29, normalized size = 1.71 \begin {gather*} \frac {12 x^{2}}{e^{5}} + \frac {12 x \log {\left (\log {\relax (x )} \right )}}{e^{5}} + \frac {x \left (e^{5} + 300\right )}{e^{5}} \end {gather*}

integrate(((12*ln(x)*ln(ln(x))+(24*x+300)*ln(x)+12)*exp(ln(ln(ln(x))+x+25)-5)+ln(x)*ln(ln(x))+(x+25)*ln(x))/(l

12*x**2*exp(-5) + 12*x*exp(-5)*log(log(x)) + x*(exp(5) + 300)*exp(-5)

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3.6.61 \(\int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx\)