Integrand size = 75, antiderivative size = 30 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=\frac {1}{\log \left (\frac {1-4 x}{5 \left (1-x+\frac {5 (-x+\log (x))}{x}\right )}\right )} \]
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=\frac {1}{\log \left (\frac {x (-1+4 x)}{5 (x (4+x)-5 \log (x))}\right )} \]
Integrate[(-5 + 20*x + 17*x^2 + (5 - 40*x)*Log[x])/((4*x^2 - 15*x^3 - 4*x^ 4 + (-5*x + 20*x^2)*Log[x])*Log[(x - 4*x^2)/(-20*x - 5*x^2 + 25*Log[x])]^2 ),x]
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.
\(\displaystyle \int \frac {17 x^2+20 x+(5-40 x) \log (x)-5}{\left (-4 x^4-15 x^3+4 x^2+\left (20 x^2-5 x\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-5 x^2-20 x+25 \log (x)}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {17 x^2+20 x+(5-40 x) \log (x)-5}{(1-4 x) x \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (-\frac {(1-4 x) x}{5 \left (x^2+4 x-5 \log (x)\right )}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {17 x^2+20 x-40 x \log (x)+5 \log (x)-5}{x \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}-\frac {4 \left (17 x^2+20 x-40 x \log (x)+5 \log (x)-5\right )}{(4 x-1) \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {17}{4} \int \frac {1}{\left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}dx-5 \int \frac {1}{x \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}dx-\frac {17}{4} \int \frac {1}{(4 x-1) \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}dx+5 \int \frac {\log (x)}{x \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}dx+20 \int \frac {\log (x)}{(4 x-1) \left (x^2+4 x-5 \log (x)\right ) \log ^2\left (\frac {x (4 x-1)}{5 (x (x+4)-5 \log (x))}\right )}dx\) |
Int[(-5 + 20*x + 17*x^2 + (5 - 40*x)*Log[x])/((4*x^2 - 15*x^3 - 4*x^4 + (- 5*x + 20*x^2)*Log[x])*Log[(x - 4*x^2)/(-20*x - 5*x^2 + 25*Log[x])]^2),x]
3.28.99.3.1 Defintions of rubi rules used
Time = 2.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(\frac {1}{\ln \left (-\frac {x \left (-1+4 x \right )}{5 \left (-x^{2}+5 \ln \left (x \right )-4 x \right )}\right )}\) | \(27\) |
risch | \(\frac {2 i}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}+\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (\frac {i}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )-\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}+\pi \,\operatorname {csgn}\left (\frac {i}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{3}+\pi \,\operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right ) \operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{2}-\pi \operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-x^{2}+5 \ln \left (x \right )-4 x}\right )^{3}-2 i \ln \left (x^{2}-5 \ln \left (x \right )+4 x \right )+4 i \ln \left (2\right )+2 i \ln \left (x \right )-2 i \ln \left (5\right )+2 i \ln \left (x -\frac {1}{4}\right )}\) | \(362\) |
default | \(\frac {2 i}{2 \pi {\operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) \operatorname {csgn}\left (\frac {i}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )-\pi \,\operatorname {csgn}\left (i \left (x -\frac {1}{4}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{2}-\pi \,\operatorname {csgn}\left (\frac {i}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) {\operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) \operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )-\pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{3}-\pi \,\operatorname {csgn}\left (\frac {i \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right ) {\operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i x \left (x -\frac {1}{4}\right )}{-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}}\right )}^{3}-2 \pi +2 i \ln \left (x \right )+4 i \ln \left (2\right )+2 i \ln \left (x -\frac {1}{4}\right )-4 i \ln \left (5\right )-2 i \ln \left (-\frac {x^{2}}{5}+\ln \left (x \right )-\frac {4 x}{5}\right )}\) | \(368\) |
int(((-40*x+5)*ln(x)+17*x^2+20*x-5)/((20*x^2-5*x)*ln(x)-4*x^4-15*x^3+4*x^2 )/ln((-4*x^2+x)/(25*ln(x)-5*x^2-20*x))^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=\frac {1}{\log \left (\frac {4 \, x^{2} - x}{5 \, {\left (x^{2} + 4 \, x - 5 \, \log \left (x\right )\right )}}\right )} \]
integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x ^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-5*x^2-20*x))^2,x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=\frac {1}{\log {\left (\frac {- 4 x^{2} + x}{- 5 x^{2} - 20 x + 25 \log {\left (x \right )}} \right )}} \]
integrate(((-40*x+5)*ln(x)+17*x**2+20*x-5)/((20*x**2-5*x)*ln(x)-4*x**4-15* x**3+4*x**2)/ln((-4*x**2+x)/(25*ln(x)-5*x**2-20*x))**2,x)
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=-\frac {1}{\log \left (5\right ) + \log \left (x^{2} + 4 \, x - 5 \, \log \left (x\right )\right ) - \log \left (4 \, x - 1\right ) - \log \left (x\right )} \]
integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x ^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-5*x^2-20*x))^2,x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=-\frac {1}{\log \left (5 \, x^{2} + 20 \, x - 25 \, \log \left (x\right )\right ) - \log \left (4 \, x - 1\right ) - \log \left (x\right )} \]
integrate(((-40*x+5)*log(x)+17*x^2+20*x-5)/((20*x^2-5*x)*log(x)-4*x^4-15*x ^3+4*x^2)/log((-4*x^2+x)/(25*log(x)-5*x^2-20*x))^2,x, algorithm=\
Time = 13.74 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-5+20 x+17 x^2+(5-40 x) \log (x)}{\left (4 x^2-15 x^3-4 x^4+\left (-5 x+20 x^2\right ) \log (x)\right ) \log ^2\left (\frac {x-4 x^2}{-20 x-5 x^2+25 \log (x)}\right )} \, dx=\frac {1}{\ln \left (-\frac {x-4\,x^2}{20\,x-25\,\ln \left (x\right )+5\,x^2}\right )} \]