Integrand size = 103, antiderivative size = 27 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=4+\frac {x}{2-x}-\log (-4-\log (4)+2 (x+\log (\log (x)))) \]
Time = 0.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=-2 \left (\frac {1}{-2+x}+\frac {1}{2} \log (4-2 x+\log (4)-2 \log (\log (x)))\right ) \]
Integrate[(-8 + 8*x - 2*x^2 + (-16*x + 12*x^2 - 2*x^3 - 2*x*Log[4])*Log[x] + 4*x*Log[x]*Log[Log[x]])/((-16*x + 24*x^2 - 12*x^3 + 2*x^4 + (-4*x + 4*x ^2 - x^3)*Log[4])*Log[x] + (8*x - 8*x^2 + 2*x^3)*Log[x]*Log[Log[x]]),x]
Time = 3.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {7292, 27, 27, 7292, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-2 x^2+\left (-2 x^3+12 x^2-16 x-2 x \log (4)\right ) \log (x)+8 x+4 x \log (x) \log (\log (x))-8}{\left (2 x^3-8 x^2+8 x\right ) \log (\log (x)) \log (x)+\left (2 x^4-12 x^3+24 x^2+\left (-x^3+4 x^2-4 x\right ) \log (4)-16 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (x^3 (-\log (x))-x^2+6 x^2 \log (x)+4 x-8 x \left (1+\frac {\log (2)}{4}\right ) \log (x)+2 x \log (x) \log (\log (x))-4\right )}{(2-x)^2 x \log (x) \left (2 x+2 \log (\log (x))-4 \left (1+\frac {\log (2)}{2}\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\log (x) x^3-6 \log (x) x^2+x^2+2 (4+\log (2)) \log (x) x-2 \log (x) \log (\log (x)) x-4 x+4}{2 (2-x)^2 x \log (x) (-x-\log (\log (x))+\log (2)+2)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^3 \log (x)+x^2-6 x^2 \log (x)-4 x+2 x (4+\log (2)) \log (x)-2 x \log (x) \log (\log (x))+4}{(2-x)^2 x \log (x) (-x-\log (\log (x))+2+\log (2))}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^3 (-\log (x))-x^2+6 x^2 \log (x)+4 x-2 x (4+\log (2)) \log (x)+2 x \log (x) \log (\log (x))-4}{(2-x)^2 x \log (x) \left (x+\log (\log (x))-2 \left (1+\frac {\log (2)}{2}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2}{(x-2)^2}+\frac {-x \log (x)-1}{x \log (x) \left (x+\log \left (\frac {\log (x)}{2}\right )-2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{2-x}-\log \left (-x-\log \left (\frac {\log (x)}{2}\right )+2\right )\) |
Int[(-8 + 8*x - 2*x^2 + (-16*x + 12*x^2 - 2*x^3 - 2*x*Log[4])*Log[x] + 4*x *Log[x]*Log[Log[x]])/((-16*x + 24*x^2 - 12*x^3 + 2*x^4 + (-4*x + 4*x^2 - x ^3)*Log[4])*Log[x] + (8*x - 8*x^2 + 2*x^3)*Log[x]*Log[Log[x]]),x]
3.23.54.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {2}{-2+x}-\ln \left (\ln \left (\ln \left (x \right )\right )-\ln \left (2\right )+x -2\right )\) | \(22\) |
parallelrisch | \(\frac {-\ln \left (\ln \left (\ln \left (x \right )\right )-\ln \left (2\right )+x -2\right ) x -2+2 \ln \left (\ln \left (\ln \left (x \right )\right )-\ln \left (2\right )+x -2\right )}{-2+x}\) | \(36\) |
int((4*x*ln(x)*ln(ln(x))+(-4*x*ln(2)-2*x^3+12*x^2-16*x)*ln(x)-2*x^2+8*x-8) /((2*x^3-8*x^2+8*x)*ln(x)*ln(ln(x))+(2*(-x^3+4*x^2-4*x)*ln(2)+2*x^4-12*x^3 +24*x^2-16*x)*ln(x)),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=-\frac {{\left (x - 2\right )} \log \left (x - \log \left (2\right ) + \log \left (\log \left (x\right )\right ) - 2\right ) + 2}{x - 2} \]
integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2 *x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log( 2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=- \log {\left (x + \log {\left (\log {\left (x \right )} \right )} - 2 - \log {\left (2 \right )} \right )} - \frac {2}{x - 2} \]
integrate((4*x*ln(x)*ln(ln(x))+(-4*x*ln(2)-2*x**3+12*x**2-16*x)*ln(x)-2*x* *2+8*x-8)/((2*x**3-8*x**2+8*x)*ln(x)*ln(ln(x))+(2*(-x**3+4*x**2-4*x)*ln(2) +2*x**4-12*x**3+24*x**2-16*x)*ln(x)),x)
Time = 0.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=-\frac {2}{x - 2} - \log \left (x - \log \left (2\right ) + \log \left (\log \left (x\right )\right ) - 2\right ) \]
integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2 *x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log( 2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=-\frac {2}{x - 2} - \log \left (x - \log \left (2\right ) + \log \left (\log \left (x\right )\right ) - 2\right ) \]
integrate((4*x*log(x)*log(log(x))+(-4*x*log(2)-2*x^3+12*x^2-16*x)*log(x)-2 *x^2+8*x-8)/((2*x^3-8*x^2+8*x)*log(x)*log(log(x))+(2*(-x^3+4*x^2-4*x)*log( 2)+2*x^4-12*x^3+24*x^2-16*x)*log(x)),x, algorithm=\
Timed out. \[ \int \frac {-8+8 x-2 x^2+\left (-16 x+12 x^2-2 x^3-2 x \log (4)\right ) \log (x)+4 x \log (x) \log (\log (x))}{\left (-16 x+24 x^2-12 x^3+2 x^4+\left (-4 x+4 x^2-x^3\right ) \log (4)\right ) \log (x)+\left (8 x-8 x^2+2 x^3\right ) \log (x) \log (\log (x))} \, dx=\int \frac {\ln \left (x\right )\,\left (16\,x+4\,x\,\ln \left (2\right )-12\,x^2+2\,x^3\right )-8\,x+2\,x^2-4\,x\,\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )+8}{\ln \left (x\right )\,\left (16\,x+2\,\ln \left (2\right )\,\left (x^3-4\,x^2+4\,x\right )-24\,x^2+12\,x^3-2\,x^4\right )-\ln \left (\ln \left (x\right )\right )\,\ln \left (x\right )\,\left (2\,x^3-8\,x^2+8\,x\right )} \,d x \]
int((log(x)*(16*x + 4*x*log(2) - 12*x^2 + 2*x^3) - 8*x + 2*x^2 - 4*x*log(l og(x))*log(x) + 8)/(log(x)*(16*x + 2*log(2)*(4*x - 4*x^2 + x^3) - 24*x^2 + 12*x^3 - 2*x^4) - log(log(x))*log(x)*(8*x - 8*x^2 + 2*x^3)),x)