\(\int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+(12 x+6 x^2) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx\) [8683]

Optimal result

Integrand size = 62, antiderivative size = 18 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=2+\frac {1}{3+\frac {3 \log (x) \log (\log (x))}{2+x}} \]

[Out]

4/(12*ln(ln(x))*ln(x)/(2+x)+12)+2

Rubi [F]

\[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=-\frac {-2-x}{3 (2+x+\log (x) \log (\log (x)))} \]

[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]])

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\frac {x + 2}{3 \, {\left (\log \left (x\right ) \log \left (\log \left (x\right )\right ) + x + 2\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\frac {x + 2}{3 x + 3 \log {\left (x \right )} \log {\left (\log {\left (x \right )} \right )} + 6} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\frac {x + 2}{3 \, {\left (\log \left (x\right ) \log \left (\log \left (x\right )\right ) + x + 2\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\frac {x + 2}{3 \, {\left (\log \left (x\right ) \log \left (\log \left (x\right )\right ) + x + 2\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 3.50 \[ \int \frac {-2-x+(-2-x+x \log (x)) \log (\log (x))}{12 x+12 x^2+3 x^3+\left (12 x+6 x^2\right ) \log (x) \log (\log (x))+3 x \log ^2(x) \log ^2(\log (x))} \, dx=\frac {\frac {4\,x}{3}-\ln \left (x\right )\,\left (\frac {x^3}{3}+x^2+\frac {2\,x}{3}\right )+\frac {4\,x^2}{3}+\frac {x^3}{3}}{\left (x+\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )+2\right )\,\left (2\,x-x^2\,\ln \left (x\right )-x\,\ln \left (x\right )+x^2\right )} \]

[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))