Integrand size = 126, antiderivative size = 26 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=x+x^2-\frac {x \log (x)}{-1+\log ^2\left (\frac {x}{5-x}\right )} \]
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=x+x^2-\frac {x \log (x)}{-1+\log ^2\left (-\frac {x}{-5+x}\right )} \]
Integrate[(-10 - 8*x + 2*x^2 + (15 + 17*x - 4*x^2)*Log[-(x/(-5 + x))]^2 + (-5 - 9*x + 2*x^2)*Log[-(x/(-5 + x))]^4 + Log[x]*(-5 + x - 10*Log[-(x/(-5 + x))] + (5 - x)*Log[-(x/(-5 + x))]^2))/(-5 + x + (10 - 2*x)*Log[-(x/(-5 + x))]^2 + (-5 + x)*Log[-(x/(-5 + x))]^4),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.
| \(\displaystyle \int \frac {2 x^2+\left (2 x^2-9 x-5\right ) \log ^4\left (-\frac {x}{x-5}\right )+\left (-4 x^2+17 x+15\right ) \log ^2\left (-\frac {x}{x-5}\right )-8 x+\log (x) \left (x+(5-x) \log ^2\left (-\frac {x}{x-5}\right )-10 \log \left (-\frac {x}{x-5}\right )-5\right )-10}{x+(x-5) \log ^4\left (-\frac {x}{x-5}\right )+(10-2 x) \log ^2\left (-\frac {x}{x-5}\right )-5} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-2 x^2-\left (\left (2 x^2-9 x-5\right ) \log ^4\left (-\frac {x}{x-5}\right )\right )-\left (-4 x^2+17 x+15\right ) \log ^2\left (-\frac {x}{x-5}\right )+8 x-\log (x) \left (x+(5-x) \log ^2\left (-\frac {x}{x-5}\right )-10 \log \left (-\frac {x}{x-5}\right )-5\right )+10}{(5-x) \left (1-\log ^2\left (-\frac {x}{x-5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (2 x+\frac {-\log (x)-1}{2 \left (\log \left (-\frac {x}{x-5}\right )-1\right )}+\frac {\log (x)+1}{2 \left (\log \left (-\frac {x}{x-5}\right )+1\right )}-\frac {5 \log (x)}{2 (x-5) \left (\log \left (-\frac {x}{x-5}\right )-1\right )^2}+\frac {5 \log (x)}{2 (x-5) \left (\log \left (-\frac {x}{x-5}\right )+1\right )^2}+1\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{1-\log \left (-\frac {x}{x-5}\right )}dx-\frac {5}{2} \int \frac {\log (x)}{(x-5) \left (\log \left (-\frac {x}{x-5}\right )-1\right )^2}dx-\frac {1}{2} \int \frac {\log (x)}{\log \left (-\frac {x}{x-5}\right )-1}dx+\frac {5}{2} \int \frac {\log (x)}{(x-5) \left (\log \left (-\frac {x}{x-5}\right )+1\right )^2}dx+\frac {1}{2} \int \frac {1}{\log \left (-\frac {x}{x-5}\right )+1}dx+\frac {1}{2} \int \frac {\log (x)}{\log \left (-\frac {x}{x-5}\right )+1}dx+x^2+x\) |
Int[(-10 - 8*x + 2*x^2 + (15 + 17*x - 4*x^2)*Log[-(x/(-5 + x))]^2 + (-5 - 9*x + 2*x^2)*Log[-(x/(-5 + x))]^4 + Log[x]*(-5 + x - 10*Log[-(x/(-5 + x))] + (5 - x)*Log[-(x/(-5 + x))]^2))/(-5 + x + (10 - 2*x)*Log[-(x/(-5 + x))]^ 2 + (-5 + x)*Log[-(x/(-5 + x))]^4),x]
3.19.74.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(75\) vs. \(2(26)=52\).
Time = 87.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.92
| method | result | size |
| parallelrisch | \(\frac {-25+10 \ln \left (-\frac {x}{-5+x}\right )^{2} x^{2}+10 \ln \left (-\frac {x}{-5+x}\right )^{2} x -10 x^{2}-10 x \ln \left (x \right )+25 \ln \left (-\frac {x}{-5+x}\right )^{2}-10 x}{10 \ln \left (-\frac {x}{-5+x}\right )^{2}-10}\) | \(76\) |
| risch | \(x^{2}+x -\frac {4 x \ln \left (x \right )}{-4-4 \pi ^{2}+4 \ln \left (-5+x \right )^{2}+4 \ln \left (x \right )^{2}+8 \pi ^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-4 \pi ^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}+4 \pi ^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{5}-4 \pi ^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{4}-\pi ^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{6}+8 i \pi \ln \left (x \right )-8 i \pi \ln \left (-5+x \right )+4 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{4}+4 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{4}-2 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{5}-2 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{5}-4 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-4 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-\pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right )^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{4}-\pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{4}-8 \ln \left (x \right ) \ln \left (-5+x \right )-4 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )+4 i \ln \left (-5+x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )-8 i \ln \left (x \right ) \pi \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}+8 i \ln \left (-5+x \right ) \pi \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-4 i \ln \left (-5+x \right ) \pi \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}+4 i \ln \left (x \right ) \pi \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}+2 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right )^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}+2 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}-4 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{3}+4 \pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )-\pi ^{2} \operatorname {csgn}\left (\frac {i}{-5+x}\right )^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}+4 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}+4 i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-4 i \ln \left (-5+x \right ) \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}-4 i \ln \left (-5+x \right ) \pi \,\operatorname {csgn}\left (\frac {i}{-5+x}\right ) \operatorname {csgn}\left (\frac {i x}{-5+x}\right )^{2}}\) | \(740\) |
int((((5-x)*ln(-x/(-5+x))^2-10*ln(-x/(-5+x))+x-5)*ln(x)+(2*x^2-9*x-5)*ln(- x/(-5+x))^4+(-4*x^2+17*x+15)*ln(-x/(-5+x))^2+2*x^2-8*x-10)/((-5+x)*ln(-x/( -5+x))^4+(-2*x+10)*ln(-x/(-5+x))^2+x-5),x,method=_RETURNVERBOSE)
1/10*(-25+10*ln(-x/(-5+x))^2*x^2+10*ln(-x/(-5+x))^2*x-10*x^2-10*x*ln(x)+25 *ln(-x/(-5+x))^2-10*x)/(ln(-x/(-5+x))^2-1)
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=\frac {{\left (x^{2} + x\right )} \log \left (-\frac {x}{x - 5}\right )^{2} - x^{2} - x \log \left (x\right ) - x}{\log \left (-\frac {x}{x - 5}\right )^{2} - 1} \]
integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9* x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+15)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5 +x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=x^{2} + x - \frac {x \log {\left (x \right )}}{\log {\left (- \frac {x}{x - 5} \right )}^{2} - 1} \]
integrate((((5-x)*ln(-x/(-5+x))**2-10*ln(-x/(-5+x))+x-5)*ln(x)+(2*x**2-9*x -5)*ln(-x/(-5+x))**4+(-4*x**2+17*x+15)*ln(-x/(-5+x))**2+2*x**2-8*x-10)/((- 5+x)*ln(-x/(-5+x))**4+(-2*x+10)*ln(-x/(-5+x))**2+x-5),x)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (25) = 50\).
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.08 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=\frac {{\left (x^{2} + x\right )} \log \left (x\right )^{2} - 2 \, {\left (x^{2} + x\right )} \log \left (x\right ) \log \left (-x + 5\right ) + {\left (x^{2} + x\right )} \log \left (-x + 5\right )^{2} - x^{2} - x \log \left (x\right ) - x}{\log \left (x\right )^{2} - 2 \, \log \left (x\right ) \log \left (-x + 5\right ) + \log \left (-x + 5\right )^{2} - 1} \]
integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9* x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+15)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5 +x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm=\
((x^2 + x)*log(x)^2 - 2*(x^2 + x)*log(x)*log(-x + 5) + (x^2 + x)*log(-x + 5)^2 - x^2 - x*log(x) - x)/(log(x)^2 - 2*log(x)*log(-x + 5) + log(-x + 5)^ 2 - 1)
Result contains complex when optimal does not.
Time = 0.37 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=x^{2} + x + \frac {x \log \left (x\right )}{\pi ^{2} + 2 i \, \pi \log \left (x - 5\right ) - \log \left (x - 5\right )^{2} - 2 i \, \pi \log \left (x\right ) + 2 \, \log \left (x - 5\right ) \log \left (x\right ) - \log \left (x\right )^{2} + 1} \]
integrate((((5-x)*log(-x/(-5+x))^2-10*log(-x/(-5+x))+x-5)*log(x)+(2*x^2-9* x-5)*log(-x/(-5+x))^4+(-4*x^2+17*x+15)*log(-x/(-5+x))^2+2*x^2-8*x-10)/((-5 +x)*log(-x/(-5+x))^4+(-2*x+10)*log(-x/(-5+x))^2+x-5),x, algorithm=\
x^2 + x + x*log(x)/(pi^2 + 2*I*pi*log(x - 5) - log(x - 5)^2 - 2*I*pi*log(x ) + 2*log(x - 5)*log(x) - log(x)^2 + 1)
Time = 9.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {-10-8 x+2 x^2+\left (15+17 x-4 x^2\right ) \log ^2\left (-\frac {x}{-5+x}\right )+\left (-5-9 x+2 x^2\right ) \log ^4\left (-\frac {x}{-5+x}\right )+\log (x) \left (-5+x-10 \log \left (-\frac {x}{-5+x}\right )+(5-x) \log ^2\left (-\frac {x}{-5+x}\right )\right )}{-5+x+(10-2 x) \log ^2\left (-\frac {x}{-5+x}\right )+(-5+x) \log ^4\left (-\frac {x}{-5+x}\right )} \, dx=x+x^2+\frac {\ln \left (x\right )\,\left (5\,x-x^2\right )}{\left ({\ln \left (-\frac {x}{x-5}\right )}^2-1\right )\,\left (x-5\right )} \]