3.89.5 \(\int \frac {4 \log (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x)))))}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx\)

Optimal. Leaf size=21 \[ \log ^2\left (2+\frac {1}{2} (1+\log (4))-\log (\log (-1+\log (x)))\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 3, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6684, 6686} \begin {gather*} \log ^2\left (\frac {1}{2} (-2 \log (\log (\log (x)-1))+5+\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2])/((5*x + x*Log[4] + (-5*x - x*Log[4])*Log[x])*Log[-1 + Lo
g[x]] + (-2*x + 2*x*Log[x])*Log[-1 + Log[x]]*Log[Log[-1 + Log[x]]]),x]

[Out]

Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2]^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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \int \frac {\log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {\log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+x)))\right )}{(-1+x) \log (-1+x) (5+\log (4)-2 \log (\log (-1+x)))} \, dx,x,\log (x)\right )\right )\\ &=\log ^2\left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 1.14 \begin {gather*} \log ^2\left (\frac {5}{2} \left (1+\frac {2 \log (2)}{5}\right )-\log (\log (-1+\log (x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2])/((5*x + x*Log[4] + (-5*x - x*Log[4])*Log[x])*Log[-
1 + Log[x]] + (-2*x + 2*x*Log[x])*Log[-1 + Log[x]]*Log[Log[-1 + Log[x]]]),x]

[Out]

Log[(5*(1 + (2*Log[2])/5))/2 - Log[Log[-1 + Log[x]]]]^2

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fricas [A]  time = 0.66, size = 15, normalized size = 0.71 \begin {gather*} \log \left (\log \relax (2) - \log \left (\log \left (\log \relax (x) - 1\right )\right ) + \frac {5}{2}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log(x)-1)*log(log(log(x)-1))+((-2*x*log(
2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log(x)-1)),x, algorithm="fricas")

[Out]

log(log(2) - log(log(log(x) - 1)) + 5/2)^2

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giac [A]  time = 0.18, size = 15, normalized size = 0.71 \begin {gather*} \log \left (\log \relax (2) - \log \left (\log \left (\log \relax (x) - 1\right )\right ) + \frac {5}{2}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log(x)-1)*log(log(log(x)-1))+((-2*x*log(
2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log(x)-1)),x, algorithm="giac")

[Out]

log(log(2) - log(log(log(x) - 1)) + 5/2)^2

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maple [A]  time = 0.04, size = 16, normalized size = 0.76

method	result	size
risch	\(\ln \left (-\ln \left (\ln \left (\ln \relax (x )-1\right )\right )+\ln \relax (2)+\frac {5}{2}\right )^{2}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*ln(-ln(ln(ln(x)-1))+ln(2)+5/2)/((2*x*ln(x)-2*x)*ln(ln(x)-1)*ln(ln(ln(x)-1))+((-2*x*ln(2)-5*x)*ln(x)+2*x*
ln(2)+5*x)*ln(ln(x)-1)),x,method=_RETURNVERBOSE)

[Out]

ln(-ln(ln(ln(x)-1))+ln(2)+5/2)^2

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maxima [B]  time = 0.46, size = 50, normalized size = 2.38 \begin {gather*} 2 \, \log \left (\log \relax (2) - \log \left (\log \left (\log \relax (x) - 1\right )\right ) + \frac {5}{2}\right ) \log \left (-2 \, \log \relax (2) + 2 \, \log \left (\log \left (\log \relax (x) - 1\right )\right ) - 5\right ) - \log \left (-2 \, \log \relax (2) + 2 \, \log \left (\log \left (\log \relax (x) - 1\right )\right ) - 5\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log(x)-1)*log(log(log(x)-1))+((-2*x*log(
2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log(x)-1)),x, algorithm="maxima")

[Out]

2*log(log(2) - log(log(log(x) - 1)) + 5/2)*log(-2*log(2) + 2*log(log(log(x) - 1)) - 5) - log(-2*log(2) + 2*log
(log(log(x) - 1)) - 5)^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {4\,\ln \left (\ln \relax (2)-\ln \left (\ln \left (\ln \relax (x)-1\right )\right )+\frac {5}{2}\right )}{\ln \left (\ln \relax (x)-1\right )\,\left (5\,x+2\,x\,\ln \relax (2)-\ln \relax (x)\,\left (5\,x+2\,x\,\ln \relax (2)\right )\right )-\ln \left (\ln \left (\ln \relax (x)-1\right )\right )\,\ln \left (\ln \relax (x)-1\right )\,\left (2\,x-2\,x\,\ln \relax (x)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(log(2) - log(log(log(x) - 1)) + 5/2))/(log(log(x) - 1)*(5*x + 2*x*log(2) - log(x)*(5*x + 2*x*log(2)
)) - log(log(log(x) - 1))*log(log(x) - 1)*(2*x - 2*x*log(x))),x)

[Out]

int((4*log(log(2) - log(log(log(x) - 1)) + 5/2))/(log(log(x) - 1)*(5*x + 2*x*log(2) - log(x)*(5*x + 2*x*log(2)
)) - log(log(log(x) - 1))*log(log(x) - 1)*(2*x - 2*x*log(x))), x)

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sympy [A]  time = 4.45, size = 17, normalized size = 0.81 \begin {gather*} \log {\left (- \log {\left (\log {\left (\log {\relax (x )} - 1 \right )} \right )} + \log {\relax (2 )} + \frac {5}{2} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4*ln(-ln(ln(ln(x)-1))+ln(2)+5/2)/((2*x*ln(x)-2*x)*ln(ln(x)-1)*ln(ln(ln(x)-1))+((-2*x*ln(2)-5*x)*ln(x
)+2*x*ln(2)+5*x)*ln(ln(x)-1)),x)

[Out]

log(-log(log(log(x) - 1)) + log(2) + 5/2)**2

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