3.87.83 \(\int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+(12 x+6 x^2) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=18 \[ 2+\frac {1}{3+\frac {3 \log (x) \log (\log (x))}{2+x}} \]

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Rubi [F]  time = 1.57, 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 {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - x + (-2 - x + x*Log[x])*Log[Log[x]])/(12*x + 12*x^2 + 3*x^3 + (12*x + 6*x^2)*Log[x]*Log[Log[x]] + 3*
x*Log[x]^2*Log[Log[x]]^2),x]

[Out]

-Defer[Int][(2 + x + Log[x]*Log[Log[x]])^(-2), x] - (2*Defer[Int][1/(x*(2 + x + Log[x]*Log[Log[x]])^2), x])/3
- Defer[Int][x/(2 + x + Log[x]*Log[Log[x]])^2, x]/3 + (4*Defer[Int][1/(Log[x]*(2 + x + Log[x]*Log[Log[x]])^2),
x])/3 + (4*Defer[Int][1/(x*Log[x]*(2 + x + Log[x]*Log[Log[x]])^2), x])/3 + Defer[Int][x/(Log[x]*(2 + x + Log[
x]*Log[Log[x]])^2), x]/3 + Defer[Int][(2 + x + Log[x]*Log[Log[x]])^(-1), x]/3 - Defer[Int][1/(Log[x]*(2 + x +
Log[x]*Log[Log[x]])), x]/3 - (2*Defer[Int][1/(x*Log[x]*(2 + x + Log[x]*Log[Log[x]])), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-x-(2+x-x \log (x)) \log (\log (x))}{3 x (2+x+\log (x) \log (\log (x)))^2} \, dx\\ &=\frac {1}{3} \int \frac {-2-x-(2+x-x \log (x)) \log (\log (x))}{x (2+x+\log (x) \log (\log (x)))^2} \, dx\\ &=\frac {1}{3} \int \left (-\frac {(2+x) (-2-x+\log (x)+x \log (x))}{x \log (x) (2+x+\log (x) \log (\log (x)))^2}+\frac {-2-x+x \log (x)}{x \log (x) (2+x+\log (x) \log (\log (x)))}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {(2+x) (-2-x+\log (x)+x \log (x))}{x \log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx\right )+\frac {1}{3} \int \frac {-2-x+x \log (x)}{x \log (x) (2+x+\log (x) \log (\log (x)))} \, dx\\ &=-\left (\frac {1}{3} \int \left (\frac {-2-x+\log (x)+x \log (x)}{\log (x) (2+x+\log (x) \log (\log (x)))^2}+\frac {2 (-2-x+\log (x)+x \log (x))}{x \log (x) (2+x+\log (x) \log (\log (x)))^2}\right ) \, dx\right )+\frac {1}{3} \int \left (\frac {1}{2+x+\log (x) \log (\log (x))}-\frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))}-\frac {2}{x \log (x) (2+x+\log (x) \log (\log (x)))}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {-2-x+\log (x)+x \log (x)}{\log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx\right )+\frac {1}{3} \int \frac {1}{2+x+\log (x) \log (\log (x))} \, dx-\frac {1}{3} \int \frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))} \, dx-\frac {2}{3} \int \frac {-2-x+\log (x)+x \log (x)}{x \log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx-\frac {2}{3} \int \frac {1}{x \log (x) (2+x+\log (x) \log (\log (x)))} \, dx\\ &=\frac {1}{3} \int \frac {1}{2+x+\log (x) \log (\log (x))} \, dx-\frac {1}{3} \int \frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))} \, dx-\frac {1}{3} \int \left (\frac {1}{(2+x+\log (x) \log (\log (x)))^2}+\frac {x}{(2+x+\log (x) \log (\log (x)))^2}-\frac {2}{\log (x) (2+x+\log (x) \log (\log (x)))^2}-\frac {x}{\log (x) (2+x+\log (x) \log (\log (x)))^2}\right ) \, dx-\frac {2}{3} \int \frac {1}{x \log (x) (2+x+\log (x) \log (\log (x)))} \, dx-\frac {2}{3} \int \left (\frac {1}{(2+x+\log (x) \log (\log (x)))^2}+\frac {1}{x (2+x+\log (x) \log (\log (x)))^2}-\frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))^2}-\frac {2}{x \log (x) (2+x+\log (x) \log (\log (x)))^2}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{(2+x+\log (x) \log (\log (x)))^2} \, dx\right )-\frac {1}{3} \int \frac {x}{(2+x+\log (x) \log (\log (x)))^2} \, dx+\frac {1}{3} \int \frac {x}{\log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx+\frac {1}{3} \int \frac {1}{2+x+\log (x) \log (\log (x))} \, dx-\frac {1}{3} \int \frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))} \, dx-\frac {2}{3} \int \frac {1}{(2+x+\log (x) \log (\log (x)))^2} \, dx-\frac {2}{3} \int \frac {1}{x (2+x+\log (x) \log (\log (x)))^2} \, dx+2 \left (\frac {2}{3} \int \frac {1}{\log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx\right )-\frac {2}{3} \int \frac {1}{x \log (x) (2+x+\log (x) \log (\log (x)))} \, dx+\frac {4}{3} \int \frac {1}{x \log (x) (2+x+\log (x) \log (\log (x)))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 20, normalized size = 1.11 \begin {gather*} -\frac {-2-x}{3 (2+x+\log (x) \log (\log (x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - x + (-2 - x + x*Log[x])*Log[Log[x]])/(12*x + 12*x^2 + 3*x^3 + (12*x + 6*x^2)*Log[x]*Log[Log[x]
] + 3*x*Log[x]^2*Log[Log[x]]^2),x]

[Out]

-1/3*(-2 - x)/(2 + x + Log[x]*Log[Log[x]])

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fricas [A]  time = 0.82, size = 16, normalized size = 0.89 \begin {gather*} \frac {x + 2}{3 \, {\left (\log \relax (x) \log \left (\log \relax (x)\right ) + x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x-2)*log(log(x))-x-2)/(3*x*log(x)^2*log(log(x))^2+(6*x^2+12*x)*log(x)*log(log(x))+3*x^3+1
2*x^2+12*x),x, algorithm="fricas")

[Out]

1/3*(x + 2)/(log(x)*log(log(x)) + x + 2)

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giac [A]  time = 0.97, size = 16, normalized size = 0.89 \begin {gather*} \frac {x + 2}{3 \, {\left (\log \relax (x) \log \left (\log \relax (x)\right ) + x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x-2)*log(log(x))-x-2)/(3*x*log(x)^2*log(log(x))^2+(6*x^2+12*x)*log(x)*log(log(x))+3*x^3+1
2*x^2+12*x),x, algorithm="giac")

[Out]

1/3*(x + 2)/(log(x)*log(log(x)) + x + 2)

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maple [A]  time = 0.03, size = 17, normalized size = 0.94

method	result	size
risch	\(\frac {2+x}{3 \ln \relax (x ) \ln \left (\ln \relax (x )\right )+3 x +6}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)-x-2)*ln(ln(x))-x-2)/(3*x*ln(x)^2*ln(ln(x))^2+(6*x^2+12*x)*ln(x)*ln(ln(x))+3*x^3+12*x^2+12*x),x,m
ethod=_RETURNVERBOSE)

[Out]

1/3*(2+x)/(ln(x)*ln(ln(x))+x+2)

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maxima [A]  time = 0.40, size = 16, normalized size = 0.89 \begin {gather*} \frac {x + 2}{3 \, {\left (\log \relax (x) \log \left (\log \relax (x)\right ) + x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x-2)*log(log(x))-x-2)/(3*x*log(x)^2*log(log(x))^2+(6*x^2+12*x)*log(x)*log(log(x))+3*x^3+1
2*x^2+12*x),x, algorithm="maxima")

[Out]

1/3*(x + 2)/(log(x)*log(log(x)) + x + 2)

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mupad [B]  time = 6.12, size = 63, normalized size = 3.50 \begin {gather*} \frac {\frac {4\,x}{3}-\ln \relax (x)\,\left (\frac {x^3}{3}+x^2+\frac {2\,x}{3}\right )+\frac {4\,x^2}{3}+\frac {x^3}{3}}{\left (x+\ln \left (\ln \relax (x)\right )\,\ln \relax (x)+2\right )\,\left (2\,x-x^2\,\ln \relax (x)-x\,\ln \relax (x)+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(log(x))*(x - x*log(x) + 2) + 2)/(12*x + 12*x^2 + 3*x^3 + log(log(x))*log(x)*(12*x + 6*x^2) + 3*x
*log(log(x))^2*log(x)^2),x)

[Out]

((4*x)/3 - log(x)*((2*x)/3 + x^2 + x^3/3) + (4*x^2)/3 + x^3/3)/((x + log(log(x))*log(x) + 2)*(2*x - x^2*log(x)
- x*log(x) + x^2))

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sympy [A]  time = 0.29, size = 17, normalized size = 0.94 \begin {gather*} \frac {x + 2}{3 x + 3 \log {\relax (x )} \log {\left (\log {\relax (x )} \right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)-x-2)*ln(ln(x))-x-2)/(3*x*ln(x)**2*ln(ln(x))**2+(6*x**2+12*x)*ln(x)*ln(ln(x))+3*x**3+12*x**
2+12*x),x)

[Out]

(x + 2)/(3*x + 3*log(x)*log(log(x)) + 6)

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