\(\int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx\) [6868]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 11 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {5 (-2+\log (\log (6)))}{\log (x)} \]

[Out]

5*(ln(ln(6))-2)/ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2339, 30} \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=-\frac {5 (2-\log (\log (6)))}{\log (x)} \]

[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 2339

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{align*} \text {integral}& = (5 (2-\log (\log (6)))) \int \frac {1}{x \log ^2(x)} \, dx \\ & = (5 (2-\log (\log (6)))) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = -\frac {5 (2-\log (\log (6)))}{\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=-\frac {10}{\log (x)}+\frac {5 \log (\log (6))}{\log (x)} \]

[In]

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

[Out]

-10/Log[x] + (5*Log[Log[6]])/Log[x]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.18

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {5 \, {\left (\log \left (\log \left (6\right )\right ) - 2\right )}}{\log \left (x\right )} \]

[In]

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

[Out]

5*(log(log(6)) - 2)/log(x)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.91 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {-10 + 5 \log {\left (\log {\left (6 \right )} \right )}}{\log {\left (x \right )}} \]

[In]

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

[Out]

(-10 + 5*log(log(6)))/log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {5 \, {\left (\log \left (\log \left (6\right )\right ) - 2\right )}}{\log \left (x\right )} \]

[In]

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

[Out]

5*(log(log(6)) - 2)/log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {5 \, {\left (\log \left (\log \left (6\right )\right ) - 2\right )}}{\log \left (x\right )} \]

[In]

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

[Out]

5*(log(log(6)) - 2)/log(x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.09 \[ \int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx=\frac {5\,\ln \left (\ln \left (6\right )\right )-10}{\ln \left (x\right )} \]

[In]

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

[Out]

(5*log(log(6)) - 10)/log(x)