∫ −7+(−7−7e2+35e3) log( 1 x)+7 log( 1 x) log(log ( 1x) _________x) ________________________________________________________________log( 1 _

Optimal. Leaf size=23 \[ -7 x \left (e^2-5 e^3-\log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right )\right ) \]

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Rubi [A] time = 0.12, antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 10, number of rules used = 5, integrand size = 41, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.122, Rules used = {6742, 2360, 2299, 2178, 2549} \begin {gather*} -7 \left (1+e^2-5 e^3\right ) x+7 x+7 x \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right ) \end {gather*}

Int[(-7 + (-7 - 7*E^2 + 35*E^3)*Log[x^(-1)] + 7*Log[x^(-1)]*Log[Log[x^(-1)]/x])/Log[x^(-1)],x]

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral

Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] && !$UseGamma === True

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p

, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E

xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {7 \left (-1-\left (1+(1-5 e) e^2\right ) \log \left (\frac {1}{x}\right )\right )}{\log \left (\frac {1}{x}\right )}+7 \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right )\right ) \, dx\\ &=7 \int \frac {-1+\left (-1-(1-5 e) e^2\right ) \log \left (\frac {1}{x}\right )}{\log \left (\frac {1}{x}\right )} \, dx+7 \int \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right ) \, dx\\ &=7 x \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right )-7 \int \left (-1-\frac {1}{\log \left (\frac {1}{x}\right )}\right ) \, dx+7 \int \left (-1-(1-5 e) e^2-\frac {1}{\log \left (\frac {1}{x}\right )}\right ) \, dx\\ &=7 x-7 \left (1+e^2-5 e^3\right ) x+7 x \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.02, size = 25, normalized size = 1.09 \begin {gather*} -7 e^2 x+35 e^3 x+7 x \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right ) \end {gather*}

Integrate[(-7 + (-7 - 7*E^2 + 35*E^3)*Log[x^(-1)] + 7*Log[x^(-1)]*Log[Log[x^(-1)]/x])/Log[x^(-1)],x]

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fricas [A] time = 0.65, size = 23, normalized size = 1.00 \begin {gather*} 35 \, x e^{3} - 7 \, x e^{2} + 7 \, x \log \left (\frac {\log \left (\frac {1}{x}\right )}{x}\right ) \end {gather*}

integrate((7*log(1/x)*log(log(1/x)/x)+(35*exp(3)-7*exp(2)-7)*log(1/x)-7)/log(1/x),x, algorithm="fricas")

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giac [A] time = 0.20, size = 24, normalized size = 1.04 \begin {gather*} 35 \, x e^{3} - 7 \, x e^{2} - 7 \, x \log \relax (x) + 7 \, x \log \left (-\log \relax (x)\right ) \end {gather*}

integrate((7*log(1/x)*log(log(1/x)/x)+(35*exp(3)-7*exp(2)-7)*log(1/x)-7)/log(1/x),x, algorithm="giac")

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method	result	size

default	$-7 \,{\mathrm e}^{2} x +35 x \,{\mathrm e}^{3}+7 \ln \left (\frac {\ln \left (\frac {1}{x}\right )}{x}\right ) x$	$24$
norman	$\left (35 \,{\mathrm e}^{3}-7 \,{\mathrm e}^{2}\right ) x +7 \ln \left (\frac {\ln \left (\frac {1}{x}\right )}{x}\right ) x$	$25$

int((7*ln(1/x)*ln(ln(1/x)/x)+(35*exp(3)-7*exp(2)-7)*ln(1/x)-7)/ln(1/x),x,method=_RETURNVERBOSE)

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 35 \, x e^{3} - 7 \, x e^{2} - 7 \, x \log \relax (x) + 7 \, x \log \left (-\log \relax (x)\right ) + 7 \, {\rm Ei}\left (\log \relax (x)\right ) - 7 \, \int \frac {1}{\log \relax (x)}\,{d x} \end {gather*}

integrate((7*log(1/x)*log(log(1/x)/x)+(35*exp(3)-7*exp(2)-7)*log(1/x)-7)/log(1/x),x, algorithm="maxima")

35*x*e^3 - 7*x*e^2 - 7*x*log(x) + 7*x*log(-log(x)) + 7*Ei(log(x)) - 7*integrate(1/log(x), x)

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mupad [B] time = 5.05, size = 21, normalized size = 0.91 \begin {gather*} 7\,x\,\left (5\,{\mathrm {e}}^3-{\mathrm {e}}^2+\ln \left (\frac {\ln \left (\frac {1}{x}\right )}{x}\right )\right ) \end {gather*}

int(-(log(1/x)*(7*exp(2) - 35*exp(3) + 7) - 7*log(1/x)*log(log(1/x)/x) + 7)/log(1/x),x)

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sympy [A] time = 0.33, size = 22, normalized size = 0.96 \begin {gather*} 7 x \log {\left (\frac {\log {\left (\frac {1}{x} \right )}}{x} \right )} + x \left (- 7 e^{2} + 35 e^{3}\right ) \end {gather*}

integrate((7*ln(1/x)*ln(ln(1/x)/x)+(35*exp(3)-7*exp(2)-7)*ln(1/x)-7)/ln(1/x),x)

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3.83.30 \(\int \frac {-7+(-7-7 e^2+35 e^3) \log (\frac {1}{x})+7 \log (\frac {1}{x}) \log (\frac {\log (\frac {1}{x})}{x})}{\log (\frac {1}{x})} \, dx\)