3.69.68 $\int \frac{10-5\log (\log (6))}{x{\log }^2(x)}dx$

Optimal. Leaf size=11 $$\frac{5(-2+\log (\log (6)))}{\log (x)}$$

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, integrand size = 15, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.200, Rules used = {12, 2302, 30} $$\begin{array}{c}-\frac{5(2-\log (\log (6)))}{\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(10 - 5*Log[Log[6]])/(x*Log[x]^2),x]

[Out]

(-5*(2 - Log[Log[6]]))/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =(5(2-\log (\log (6))))\int \frac{1}{x{\log }^2(x)}dx\\ & =(5(2-\log (\log (6))))\mathrm{Subst}(\int \frac{1}{x^2}dx,x,\log (x))\\ & =-\frac{5(2-\log (\log (6)))}{\log (x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.45 $$\begin{array}{c}-\frac{10}{\log (x)}+\frac{5\log (\log (6))}{\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(10 - 5*Log[Log[6]])/(x*Log[x]^2),x]

[Out]

-10/Log[x] + (5*Log[Log[6]])/Log[x]

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fricas [A]  time = 1.44, size = 11, normalized size = 1.00 $$\begin{array}{c}\frac{5(\log (\log (6))-2)}{\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="fricas")

[Out]

5*(log(log(6)) - 2)/log(x)

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giac [A]  time = 0.18, size = 11, normalized size = 1.00 $$\begin{array}{c}\frac{5(\log (\log (6))-2)}{\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="giac")

[Out]

5*(log(log(6)) - 2)/log(x)

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maple [A]  time = 0.03, size = 13, normalized size = 1.18

method	result	size
norman	$\frac{5\ln (\ln (6))-10}{\ln (x)}$	$13$
derivativedivides	$-\frac{-5\ln (\ln (6))+10}{\ln (x)}$	$14$
default	$-\frac{-5\ln (\ln (6))+10}{\ln (x)}$	$14$
risch	$\frac{5\ln (\ln (2)+\ln (3))}{\ln (x)}-\frac{10}{\ln (x)}$	$20$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(ln(6))+10)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(5*ln(ln(6))-10)/ln(x)

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maxima [A]  time = 0.35, size = 11, normalized size = 1.00 $$\begin{array}{c}\frac{5(\log (\log (6))-2)}{\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="maxima")

[Out]

5*(log(log(6)) - 2)/log(x)

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mupad [B]  time = 0.03, size = 12, normalized size = 1.09 $$\begin{array}{c}\frac{5\ln (\ln (6))-10}{\ln (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(log(6)) - 10)/(x*log(x)^2),x)

[Out]

(5*log(log(6)) - 10)/log(x)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.91 $$\begin{array}{c}\frac{-10+5\log (\log (6))}{\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(ln(6))+10)/x/ln(x)**2,x)

[Out]

(-10 + 5*log(log(6)))/log(x)

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