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

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

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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} $$\begin{array}{c}{\log }^2(\frac{1}{2}(-2\log (\log (\log (x)-1))+5+\log (4)))\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =4\int \frac{\log (\frac{1}{2}(5+\log (4)-2\log (\log (-1+\log (x)))))}{(5x+x\log (4)+(-5x-x\log (4))\log (x))\log (-1+\log (x))+(-2x+2x\log (x))\log (-1+\log (x))\log (\log (-1+\log (x)))}dx\\ & =-\left(4\mathrm{Subst}(\int \frac{\log (\frac{1}{2}(5+\log (4)-2\log (\log (-1+x))))}{(-1+x)\log (-1+x)(5+\log (4)-2\log (\log (-1+x)))}dx,x,\log (x))\right)\\ & ={\log }^2(\frac{1}{2}(5+\log (4)-2\log (\log (-1+\log (x)))))\end{array}\end{array}$$

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

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 $$\begin{array}{c}\log {(\log (2)-\log (\log (\log (x)-1))+\frac{5}{2})}^2\end{array}$$

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 $$\begin{array}{c}\log {(\log (2)-\log (\log (\log (x)-1))+\frac{5}{2})}^2\end{array}$$

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	$\ln {(-\ln (\ln (\ln (x)-1))+\ln (2)+\frac{5}{2})}^2$	$16$

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 $$\begin{array}{c}2\log (\log (2)-\log (\log (\log (x)-1))+\frac{5}{2})\log (-2\log (2)+2\log (\log (\log (x)-1))-5)-\log {(-2\log (2)+2\log (\log (\log (x)-1))-5)}^2\end{array}$$

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 $$\begin{array}{c}\int \frac{4\ln (\ln (2)-\ln (\ln (\ln (x)-1))+\frac{5}{2})}{\ln (\ln (x)-1)(5x+2x\ln (2)-\ln (x)(5x+2x\ln (2)))-\ln (\ln (\ln (x)-1))\ln (\ln (x)-1)(2x-2x\ln (x))}dx\end{array}$$

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 $$\begin{array}{c}\log {(-\log (\log (\log (x)-1))+\log (2)+\frac{5}{2})}^2\end{array}$$

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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