Integrand size = 103, antiderivative size = 21 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\log (x) \left (-x+\frac {1}{\log ((-5+x) x)}+x \log \left (x^2\right )\right ) \] Output:
ln(x)*(1/ln((-5+x)*x)+x*ln(x^2)-x)
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\log (x) \left (\frac {1}{\log ((-5+x) x)}+x \left (-1+\log \left (x^2\right )\right )\right ) \] Input:
Integrate[((5 - 2*x)*Log[x] + (-5 + x)*Log[-5*x + x^2] + (5*x - x^2 + (-5* x + x^2)*Log[x])*Log[-5*x + x^2]^2 + (-5*x + x^2 + (-5*x + x^2)*Log[x])*Lo g[x^2]*Log[-5*x + x^2]^2)/((-5*x + x^2)*Log[-5*x + x^2]^2),x]
Output:
Log[x]*(Log[(-5 + x)*x]^(-1) + x*(-1 + Log[x^2]))
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.
\(\displaystyle \int \frac {\left (-x^2+\left (x^2-5 x\right ) \log (x)+5 x\right ) \log ^2\left (x^2-5 x\right )+\left (x^2+\left (x^2-5 x\right ) \log (x)-5 x\right ) \log \left (x^2\right ) \log ^2\left (x^2-5 x\right )+(x-5) \log \left (x^2-5 x\right )+(5-2 x) \log (x)}{\left (x^2-5 x\right ) \log ^2\left (x^2-5 x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-x^2+\left (x^2-5 x\right ) \log (x)+5 x\right ) \log ^2\left (x^2-5 x\right )+\left (x^2+\left (x^2-5 x\right ) \log (x)-5 x\right ) \log \left (x^2\right ) \log ^2\left (x^2-5 x\right )+(x-5) \log \left (x^2-5 x\right )+(5-2 x) \log (x)}{(x-5) x \log ^2\left (x^2-5 x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (-x^2+\left (x^2-5 x\right ) \log (x)+5 x\right ) \log ^2\left (x^2-5 x\right )\right )-\left (x^2+\left (x^2-5 x\right ) \log (x)-5 x\right ) \log \left (x^2\right ) \log ^2\left (x^2-5 x\right )-(x-5) \log \left (x^2-5 x\right )-(5-2 x) \log (x)}{(5-x) x \log ^2((x-5) x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\log \left (x^2\right ) \log (x)+\log \left (x^2\right )-\frac {(2 x-5) \log (x)}{(x-5) x \log ^2((x-5) x)}+\log (x)+\frac {1}{x \log ((x-5) x)}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log (x)}{(x-5) \log ^2((x-5) x)}dx-\int \frac {\log (x)}{x \log ^2((x-5) x)}dx+\int \frac {1}{x \log ((x-5) x)}dx+x \log \left (x^2\right ) \log (x)-x \log (x)\) |
Input:
Int[((5 - 2*x)*Log[x] + (-5 + x)*Log[-5*x + x^2] + (5*x - x^2 + (-5*x + x^ 2)*Log[x])*Log[-5*x + x^2]^2 + (-5*x + x^2 + (-5*x + x^2)*Log[x])*Log[x^2] *Log[-5*x + x^2]^2)/((-5*x + x^2)*Log[-5*x + x^2]^2),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.13 (sec) , antiderivative size = 167, normalized size of antiderivative = 7.95
\[2 x \ln \left (x \right )^{2}-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) \ln \left (x \right )}{2}+i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} \ln \left (x \right )-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3} \ln \left (x \right )}{2}-x \ln \left (x \right )+\frac {2 \ln \left (x \right )}{2 \ln \left (x \right )+2 \ln \left (-5+x \right )+i \pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \right )-i \pi \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )}\]
Input:
int((((x^2-5*x)*ln(x)+x^2-5*x)*ln(x^2-5*x)^2*ln(x^2)+((x^2-5*x)*ln(x)-x^2+ 5*x)*ln(x^2-5*x)^2+(-5+x)*ln(x^2-5*x)+(5-2*x)*ln(x))/(x^2-5*x)/ln(x^2-5*x) ^2,x)
Output:
2*x*ln(x)^2-1/2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)*ln(x)+I*Pi*x*csgn(I*x)*csgn (I*x^2)^2*ln(x)-1/2*I*Pi*x*csgn(I*x^2)^3*ln(x)-x*ln(x)+2*ln(x)/(2*ln(x)+2* ln(-5+x)+I*Pi*csgn(I*(-5+x))*csgn(I*x*(-5+x))^2-I*Pi*csgn(I*(-5+x))*csgn(I *x*(-5+x))*csgn(I*x)-I*Pi*csgn(I*x*(-5+x))^3+I*Pi*csgn(I*x*(-5+x))^2*csgn( I*x))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {{\left (2 \, x \log \left (x\right )^{2} - x \log \left (x\right )\right )} \log \left (x^{2} - 5 \, x\right ) + \log \left (x\right )}{\log \left (x^{2} - 5 \, x\right )} \] Input:
integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*l og(x)-x^2+5*x)*log(x^2-5*x)^2+(-5+x)*log(x^2-5*x)+(5-2*x)*log(x))/(x^2-5*x )/log(x^2-5*x)^2,x, algorithm="fricas")
Output:
((2*x*log(x)^2 - x*log(x))*log(x^2 - 5*x) + log(x))/log(x^2 - 5*x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=2 x \log {\left (x \right )}^{2} - x \log {\left (x \right )} + \frac {\log {\left (x \right )}}{\log {\left (x^{2} - 5 x \right )}} \] Input:
integrate((((x**2-5*x)*ln(x)+x**2-5*x)*ln(x**2-5*x)**2*ln(x**2)+((x**2-5*x )*ln(x)-x**2+5*x)*ln(x**2-5*x)**2+(-5+x)*ln(x**2-5*x)+(5-2*x)*ln(x))/(x**2 -5*x)/ln(x**2-5*x)**2,x)
Output:
2*x*log(x)**2 - x*log(x) + log(x)/log(x**2 - 5*x)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (21) = 42\).
Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {2 \, x \log \left (x\right )^{3} - x \log \left (x\right )^{2} + {\left (2 \, x \log \left (x\right )^{2} - x \log \left (x\right )\right )} \log \left (x - 5\right ) + \log \left (x\right )}{\log \left (x - 5\right ) + \log \left (x\right )} \] Input:
integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*l og(x)-x^2+5*x)*log(x^2-5*x)^2+(-5+x)*log(x^2-5*x)+(5-2*x)*log(x))/(x^2-5*x )/log(x^2-5*x)^2,x, algorithm="maxima")
Output:
(2*x*log(x)^3 - x*log(x)^2 + (2*x*log(x)^2 - x*log(x))*log(x - 5) + log(x) )/(log(x - 5) + log(x))
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=2 \, x \log \left (x\right )^{2} - x \log \left (x\right ) + \frac {\log \left (x\right )}{\log \left (x - 5\right ) + \log \left (x\right )} \] Input:
integrate((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*l og(x)-x^2+5*x)*log(x^2-5*x)^2+(-5+x)*log(x^2-5*x)+(5-2*x)*log(x))/(x^2-5*x )/log(x^2-5*x)^2,x, algorithm="giac")
Output:
2*x*log(x)^2 - x*log(x) + log(x)/(log(x - 5) + log(x))
Time = 2.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {\ln \left (x\right )}{\ln \left (x^2-5\,x\right )}-\frac {5}{4\,\left (x-\frac {5}{2}\right )}-\frac {x}{2\,x-5}+\frac {5}{2\,x-5}-x\,\ln \left (x\right )+x\,\ln \left (x^2\right )\,\ln \left (x\right ) \] Input:
int((log(x)*(2*x - 5) - log(x^2 - 5*x)*(x - 5) + log(x^2 - 5*x)^2*(log(x)* (5*x - x^2) - 5*x + x^2) + log(x^2)*log(x^2 - 5*x)^2*(5*x + log(x)*(5*x - x^2) - x^2))/(log(x^2 - 5*x)^2*(5*x - x^2)),x)
Output:
log(x)/log(x^2 - 5*x) - 5/(4*(x - 5/2)) - x/(2*x - 5) + 5/(2*x - 5) - x*lo g(x) + x*log(x^2)*log(x)
Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {\mathrm {log}\left (x \right ) \left (\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (x^{2}-5 x \right ) x -\mathrm {log}\left (x^{2}-5 x \right ) x +1\right )}{\mathrm {log}\left (x^{2}-5 x \right )} \] Input:
int((((x^2-5*x)*log(x)+x^2-5*x)*log(x^2-5*x)^2*log(x^2)+((x^2-5*x)*log(x)- x^2+5*x)*log(x^2-5*x)^2+(-5+x)*log(x^2-5*x)+(5-2*x)*log(x))/(x^2-5*x)/log( x^2-5*x)^2,x)
Output:
(log(x)*(log(x**2)*log(x**2 - 5*x)*x - log(x**2 - 5*x)*x + 1))/log(x**2 - 5*x)