\(\int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{(-105 x+5 x^2) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx\) [759]

Optimal result

Integrand size = 49, antiderivative size = 19 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=\log \left (\frac {-21+x-\log (x)+\frac {1}{5} \log (\log (x))}{x}\right ) \]

[Out]

ln((1/5*ln(ln(x))+x-21-ln(x))/x)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6873, 6874, 6816} \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=\log (-5 x+5 \log (x)-\log (\log (x))+105)-\log (x) \]

[In]

Int[(1 + 100*Log[x] + 5*Log[x]^2 - Log[x]*Log[Log[x]])/((-105*x + 5*x^2)*Log[x] - 5*x*Log[x]^2 + x*Log[x]*Log[
Log[x]]),x]

[Out]

-Log[x] + Log[105 - 5*x + 5*Log[x] - Log[Log[x]]]

Rule 6816

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

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1-100 \log (x)-5 \log ^2(x)+\log (x) \log (\log (x))}{x \log (x) (105-5 x+5 \log (x)-\log (\log (x)))} \, dx \\ & = \int \left (-\frac {1}{x}+\frac {1-5 \log (x)+5 x \log (x)}{x \log (x) (-105+5 x-5 \log (x)+\log (\log (x)))}\right ) \, dx \\ & = -\log (x)+\int \frac {1-5 \log (x)+5 x \log (x)}{x \log (x) (-105+5 x-5 \log (x)+\log (\log (x)))} \, dx \\ & = -\log (x)+\log (105-5 x+5 \log (x)-\log (\log (x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=-\log (x)+\log (105-5 x+5 \log (x)-\log (\log (x))) \]

[In]

Integrate[(1 + 100*Log[x] + 5*Log[x]^2 - Log[x]*Log[Log[x]])/((-105*x + 5*x^2)*Log[x] - 5*x*Log[x]^2 + x*Log[x
]*Log[Log[x]]),x]

[Out]

-Log[x] + Log[105 - 5*x + 5*Log[x] - Log[Log[x]]]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\ln \left (x \right )+\ln \left (5 x -5 \ln \left (x \right )+\ln \left (\ln \left (x \right )\right )-105\right )\)	\(19\)
parallelrisch	\(\ln \left (\frac {\ln \left (\ln \left (x \right )\right )}{5}+x -21-\ln \left (x \right )\right )-\ln \left (x \right )\)	\(19\)
default	\(-\ln \left (x \right )+\ln \left (5 \ln \left (x \right )-\ln \left (\ln \left (x \right )\right )-5 x +105\right )\)	\(21\)

[In]

int((-ln(x)*ln(ln(x))+5*ln(x)^2+100*ln(x)+1)/(x*ln(x)*ln(ln(x))-5*x*ln(x)^2+(5*x^2-105*x)*ln(x)),x,method=_RET
URNVERBOSE)

[Out]

-ln(x)+ln(5*x-5*ln(x)+ln(ln(x))-105)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=\log \left (5 \, x - 5 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) - 105\right ) - \log \left (x\right ) \]

[In]

integrate((-log(x)*log(log(x))+5*log(x)^2+100*log(x)+1)/(x*log(x)*log(log(x))-5*x*log(x)^2+(5*x^2-105*x)*log(x
)),x, algorithm="fricas")

[Out]

log(5*x - 5*log(x) + log(log(x)) - 105) - log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=- \log {\left (x \right )} + \log {\left (5 x - 5 \log {\left (x \right )} + \log {\left (\log {\left (x \right )} \right )} - 105 \right )} \]

[In]

integrate((-ln(x)*ln(ln(x))+5*ln(x)**2+100*ln(x)+1)/(x*ln(x)*ln(ln(x))-5*x*ln(x)**2+(5*x**2-105*x)*ln(x)),x)

[Out]

-log(x) + log(5*x - 5*log(x) + log(log(x)) - 105)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=\log \left (5 \, x - 5 \, \log \left (x\right ) + \log \left (\log \left (x\right )\right ) - 105\right ) - \log \left (x\right ) \]

[In]

integrate((-log(x)*log(log(x))+5*log(x)^2+100*log(x)+1)/(x*log(x)*log(log(x))-5*x*log(x)^2+(5*x^2-105*x)*log(x
)),x, algorithm="maxima")

[Out]

log(5*x - 5*log(x) + log(log(x)) - 105) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=-\log \left (x\right ) + \log \left (-5 \, x + 5 \, \log \left (x\right ) - \log \left (\log \left (x\right )\right ) + 105\right ) \]

[In]

integrate((-log(x)*log(log(x))+5*log(x)^2+100*log(x)+1)/(x*log(x)*log(log(x))-5*x*log(x)^2+(5*x^2-105*x)*log(x
)),x, algorithm="giac")

[Out]

-log(x) + log(-5*x + 5*log(x) - log(log(x)) + 105)

Mupad [B] (verification not implemented)

Time = 8.40 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {1+100 \log (x)+5 \log ^2(x)-\log (x) \log (\log (x))}{\left (-105 x+5 x^2\right ) \log (x)-5 x \log ^2(x)+x \log (x) \log (\log (x))} \, dx=\ln \left (5\,x+\ln \left (\ln \left (x\right )\right )-5\,\ln \left (x\right )-105\right )-\ln \left (x\right ) \]

[In]

int(-(100*log(x) + 5*log(x)^2 - log(log(x))*log(x) + 1)/(5*x*log(x)^2 + log(x)*(105*x - 5*x^2) - x*log(log(x))
*log(x)),x)

[Out]

log(5*x + log(log(x)) - 5*log(x) - 105) - log(x)