Integrand size = 74, antiderivative size = 21 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (x+\frac {1}{4} (2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right ) \] Output:
ln(ln(x+ln(5/exp(2))*ln(x)*(1/2-1/4*x)))
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (x-\frac {1}{4} (-2+x) (-2+\log (5)) \log (x)\right )\right ) \] Input:
Integrate[(-4*x + (-2 + x)*Log[5/E^2] + x*Log[5/E^2]*Log[x])/((-4*x^2 + (- 2*x + x^2)*Log[5/E^2]*Log[x])*Log[(4*x + (2 - x)*Log[5/E^2]*Log[x])/4]),x]
Output:
Log[Log[x - ((-2 + x)*(-2 + Log[5])*Log[x])/4]]
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.
\(\displaystyle \int \frac {-4 x+(x-2) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (\left (x^2-2 x\right ) \log \left (\frac {5}{e^2}\right ) \log (x)-4 x^2\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(2-\log (5)) \log (x)}{\left (4 x+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (x-2) (\log (5)-2) \log (x)\right )}+\frac {2 (\log (5)-2)}{x \left (4 x+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (x-2) (\log (5)-2) \log (x)\right )}+\frac {4 \left (1+\frac {1}{4} (2-\log (5))\right )}{\left (4 x+2 x \left (1-\frac {\log (5)}{2}\right ) \log (x)-4 \left (1-\frac {\log (5)}{2}\right ) \log (x)\right ) \log \left (x-\frac {1}{4} (x-2) (\log (5)-2) \log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (6-\log (5)) \int \frac {1}{(4 x-(x-2) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (x-2) (-2+\log (5)) \log (x)\right )}dx-2 (2-\log (5)) \int \frac {1}{x (4 x-(x-2) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (x-2) (-2+\log (5)) \log (x)\right )}dx+(2-\log (5)) \int \frac {\log (x)}{(4 x-(x-2) (-2+\log (5)) \log (x)) \log \left (x-\frac {1}{4} (x-2) (-2+\log (5)) \log (x)\right )}dx\) |
Input:
Int[(-4*x + (-2 + x)*Log[5/E^2] + x*Log[5/E^2]*Log[x])/((-4*x^2 + (-2*x + x^2)*Log[5/E^2]*Log[x])*Log[(4*x + (2 - x)*Log[5/E^2]*Log[x])/4]),x]
Output:
$Aborted
Time = 1.98 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \left (\ln \left (5\right )-2\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(18\) |
norman | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(21\) |
parallelrisch | \(\ln \left (\ln \left (\frac {\left (2-x \right ) \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )}{4}+x \right )\right )\) | \(21\) |
default | \(\ln \left (2 \ln \left (2\right )-\ln \left (-x \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )+2 \ln \left (5 \,{\mathrm e}^{-2}\right ) \ln \left (x \right )+4 x \right )\right )\) | \(37\) |
Input:
int((x*ln(5/exp(2))*ln(x)+(-2+x)*ln(5/exp(2))-4*x)/((x^2-2*x)*ln(5/exp(2)) *ln(x)-4*x^2)/ln(1/4*(2-x)*ln(5/exp(2))*ln(x)+x),x,method=_RETURNVERBOSE)
Output:
ln(ln(1/4*(2-x)*(ln(5)-2)*ln(x)+x))
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (\log \left (-\frac {1}{4} \, {\left ({\left (x - 2\right )} \log \left (5\right ) - 2 \, x + 4\right )} \log \left (x\right ) + x\right )\right ) \] Input:
integrate((x*log(5/exp(2))*log(x)+(-2+x)*log(5/exp(2))-4*x)/((x^2-2*x)*log (5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x)*log(5/exp(2))*log(x)+x),x, algorith m="fricas")
Output:
log(log(-1/4*((x - 2)*log(5) - 2*x + 4)*log(x) + x))
Time = 0.61 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log {\left (\log {\left (x + \left (\frac {1}{2} - \frac {x}{4}\right ) \log {\left (x \right )} \log {\left (\frac {5}{e^{2}} \right )} \right )} \right )} \] Input:
integrate((x*ln(5/exp(2))*ln(x)+(-2+x)*ln(5/exp(2))-4*x)/((x**2-2*x)*ln(5/ exp(2))*ln(x)-4*x**2)/ln(1/4*(2-x)*ln(5/exp(2))*ln(x)+x),x)
Output:
log(log(x + (1/2 - x/4)*log(x)*log(5*exp(-2))))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (-2 \, \log \left (2\right ) + \log \left (-{\left (x {\left (\log \left (5\right ) - 2\right )} - 2 \, \log \left (5\right ) + 4\right )} \log \left (x\right ) + 4 \, x\right )\right ) \] Input:
integrate((x*log(5/exp(2))*log(x)+(-2+x)*log(5/exp(2))-4*x)/((x^2-2*x)*log (5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x)*log(5/exp(2))*log(x)+x),x, algorith m="maxima")
Output:
log(-2*log(2) + log(-(x*(log(5) - 2) - 2*log(5) + 4)*log(x) + 4*x))
Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (16) = 32\).
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\log \left (2 \, \log \left (2\right ) - \log \left (-x \log \left (5\right ) \log \left (x\right ) + 2 \, x \log \left (x\right ) + 2 \, \log \left (5\right ) \log \left (x\right ) + 4 \, x - 4 \, \log \left (x\right )\right )\right ) \] Input:
integrate((x*log(5/exp(2))*log(x)+(-2+x)*log(5/exp(2))-4*x)/((x^2-2*x)*log (5/exp(2))*log(x)-4*x^2)/log(1/4*(2-x)*log(5/exp(2))*log(x)+x),x, algorith m="giac")
Output:
log(2*log(2) - log(-x*log(5)*log(x) + 2*x*log(x) + 2*log(5)*log(x) + 4*x - 4*log(x)))
Timed out. \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=-\int \frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\left (x-2\right )-4\,x+x\,\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )}{\ln \left (x-\frac {\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )\,\left (x-2\right )}{4}\right )\,\left (4\,x^2+\ln \left (5\,{\mathrm {e}}^{-2}\right )\,\ln \left (x\right )\,\left (2\,x-x^2\right )\right )} \,d x \] Input:
int(-(log(5*exp(-2))*(x - 2) - 4*x + x*log(5*exp(-2))*log(x))/(log(x - (lo g(5*exp(-2))*log(x)*(x - 2))/4)*(4*x^2 + log(5*exp(-2))*log(x)*(2*x - x^2) )),x)
Output:
-int((log(5*exp(-2))*(x - 2) - 4*x + x*log(5*exp(-2))*log(x))/(log(x - (lo g(5*exp(-2))*log(x)*(x - 2))/4)*(4*x^2 + log(5*exp(-2))*log(x)*(2*x - x^2) )), x)
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-4 x+(-2+x) \log \left (\frac {5}{e^2}\right )+x \log \left (\frac {5}{e^2}\right ) \log (x)}{\left (-4 x^2+\left (-2 x+x^2\right ) \log \left (\frac {5}{e^2}\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 x+(2-x) \log \left (\frac {5}{e^2}\right ) \log (x)\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (-\frac {\mathrm {log}\left (\frac {5}{e^{2}}\right ) \mathrm {log}\left (x \right ) x}{4}+\frac {\mathrm {log}\left (\frac {5}{e^{2}}\right ) \mathrm {log}\left (x \right )}{2}+x \right )\right ) \] Input:
int((x*log(5/exp(2))*log(x)+(-2+x)*log(5/exp(2))-4*x)/((x^2-2*x)*log(5/exp (2))*log(x)-4*x^2)/log(1/4*(2-x)*log(5/exp(2))*log(x)+x),x)
Output:
log(log(( - log(5/e**2)*log(x)*x + 2*log(5/e**2)*log(x) + 4*x)/4))