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

Optimal. Leaf size=19 \[ \log \left (\frac {-21+x-\log (x)+\frac {1}{5} \log (\log (x))}{x}\right ) \]

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Rubi [A]  time = 0.44, antiderivative size = 20, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6741, 6742, 6684} \begin {gather*} \log (-5 x+5 \log (x)-\log (\log (x))+105)-\log (x) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 20, normalized size = 1.05 \begin {gather*} -\log (x)+\log (105-5 x+5 \log (x)-\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[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]]]

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fricas [A]  time = 0.72, size = 18, normalized size = 0.95 \begin {gather*} \log \left (5 \, x - 5 \, \log \relax (x) + \log \left (\log \relax (x)\right ) - 105\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.42, size = 20, normalized size = 1.05 \begin {gather*} -\log \relax (x) + \log \left (-5 \, x + 5 \, \log \relax (x) - \log \left (\log \relax (x)\right ) + 105\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 19, normalized size = 1.00

method	result	size
risch	\(-\ln \relax (x )+\ln \left (5 x -5 \ln \relax (x )+\ln \left (\ln \relax (x )\right )-105\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.53, size = 18, normalized size = 0.95 \begin {gather*} \log \left (5 \, x - 5 \, \log \relax (x) + \log \left (\log \relax (x)\right ) - 105\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.76, size = 18, normalized size = 0.95 \begin {gather*} \ln \left (5\,x+\ln \left (\ln \relax (x)\right )-5\,\ln \relax (x)-105\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.32, size = 19, normalized size = 1.00 \begin {gather*} - \log {\relax (x )} + \log {\left (5 x - 5 \log {\relax (x )} + \log {\left (\log {\relax (x )} \right )} - 105 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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