3.53.99 \(\int \frac {-1-x \log (x)+(2-x \log ^2(x)) \log (\log (x))+(-1+x \log ^2(x)) \log ^2(\log (x))}{x \log ^2(x)-2 x \log ^2(x) \log (\log (x))+x \log ^2(x) \log ^2(\log (x))} \, dx\)

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

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Rubi [F]  time = 0.14, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-x \log (x)+\left (2-x \log ^2(x)\right ) \log (\log (x))+\left (-1+x \log ^2(x)\right ) \log ^2(\log (x))}{x \log ^2(x)-2 x \log ^2(x) \log (\log (x))+x \log ^2(x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x + Log[x]^(-1) - Defer[Int][1/(Log[x]*(-1 + Log[Log[x]])^2), x] + Defer[Int][(-1 + Log[Log[x]])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x \log ^2(x)}-\frac {1}{\log (x) (-1+\log (\log (x)))^2}+\frac {\log (\log (x))}{-1+\log (\log (x))}\right ) \, dx\\ &=-\int \frac {1}{x \log ^2(x)} \, dx-\int \frac {1}{\log (x) (-1+\log (\log (x)))^2} \, dx+\int \frac {\log (\log (x))}{-1+\log (\log (x))} \, dx\\ &=\int \left (1+\frac {1}{-1+\log (\log (x))}\right ) \, dx-\int \frac {1}{\log (x) (-1+\log (\log (x)))^2} \, dx-\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=x+\frac {1}{\log (x)}-\int \frac {1}{\log (x) (-1+\log (\log (x)))^2} \, dx+\int \frac {1}{-1+\log (\log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 15, normalized size = 0.71 \begin {gather*} x+\frac {1}{\log (x)}+\frac {x}{-1+\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - x*Log[x] + (2 - x*Log[x]^2)*Log[Log[x]] + (-1 + x*Log[x]^2)*Log[Log[x]]^2)/(x*Log[x]^2 - 2*x*L
og[x]^2*Log[Log[x]] + x*Log[x]^2*Log[Log[x]]^2),x]

[Out]

x + Log[x]^(-1) + x/(-1 + Log[Log[x]])

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fricas [A]  time = 0.64, size = 26, normalized size = 1.24 \begin {gather*} \frac {{\left (x \log \relax (x) + 1\right )} \log \left (\log \relax (x)\right ) - 1}{\log \relax (x) \log \left (\log \relax (x)\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*log(x) + 1)*log(log(x)) - 1)/(log(x)*log(log(x)) - log(x))

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giac [A]  time = 0.23, size = 15, normalized size = 0.71 \begin {gather*} x + \frac {x}{\log \left (\log \relax (x)\right ) - 1} + \frac {1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + x/(log(log(x)) - 1) + 1/log(x)

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maple [A]  time = 0.04, size = 22, normalized size = 1.05

method	result	size
risch	\(\frac {x \ln \relax (x )+1}{\ln \relax (x )}+\frac {x}{\ln \left (\ln \relax (x )\right )-1}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)^2-1)*ln(ln(x))^2+(-x*ln(x)^2+2)*ln(ln(x))-x*ln(x)-1)/(x*ln(x)^2*ln(ln(x))^2-2*x*ln(x)^2*ln(ln(x)
)+x*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 26, normalized size = 1.24 \begin {gather*} \frac {{\left (x \log \relax (x) + 1\right )} \log \left (\log \relax (x)\right ) - 1}{\log \relax (x) \log \left (\log \relax (x)\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*log(x) + 1)*log(log(x)) - 1)/(log(x)*log(log(x)) - log(x))

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mupad [B]  time = 3.65, size = 33, normalized size = 1.57 \begin {gather*} x+\frac {1}{\ln \relax (x)}+x\,\ln \relax (x)+\frac {x\,\left (\ln \relax (x)+1\right )-x\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)}{\ln \left (\ln \relax (x)\right )-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 1/log(x) + x*log(x) + (x*(log(x) + 1) - x*log(log(x))*log(x))/(log(log(x)) - 1)

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sympy [A]  time = 0.25, size = 14, normalized size = 0.67 \begin {gather*} x + \frac {x}{\log {\left (\log {\relax (x )} \right )} - 1} + \frac {1}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + x/(log(log(x)) - 1) + 1/log(x)

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