Integrand size = 103, antiderivative size = 21 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\log (x) \left (-x+\frac {1}{\log ((-5+x) x)}+x \log \left (x^2\right )\right ) \]
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\log (x) \left (\frac {1}{\log ((-5+x) x)}+x \left (-1+\log \left (x^2\right )\right )\right ) \]
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]
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 |
\(\displaystyle -\int \frac {\log (x)}{(x-5) \log ^2((x-5) x)}dx-\int \frac {\log (x)}{x \log ^2((x-5) x)}dx+\int \frac {1}{x \log ((x-5) x)}dx+x \log \left (x^2\right ) \log (x)-x \log (x)\) |
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]
3.3.58.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Leaf count of result is larger than twice the leaf count of optimal. \(93\) vs. \(2(21)=42\).
Time = 4.72 (sec) , antiderivative size = 94, normalized size of antiderivative = 4.48
method | result | size |
parallelrisch | \(\frac {-240 \ln \left (x \right ) \ln \left (x^{2}-5 x \right ) x +240 \ln \left (x \right )+240 \ln \left (x^{2}\right ) x \ln \left (x \right ) \ln \left (x^{2}-5 x \right )+2400 \ln \left (x^{2}-5 x \right ) \ln \left (x \right ) \ln \left (x^{2}\right )-2400 \ln \left (x \right )^{2} \ln \left (x^{2}-5 x \right )-600 \ln \left (x^{2}\right )^{2} \ln \left (x^{2}-5 x \right )}{240 \ln \left (x^{2}-5 x \right )}\) | \(94\) |
risch | \(2 x \ln \left (x \right )^{2}-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) \ln \left (x \right )}{2}+i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} \ln \left (x \right )-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3} \ln \left (x \right )}{2}-x \ln \left (x \right )+\frac {2 \ln \left (x \right )}{2 \ln \left (x \right )+2 \ln \left (-5+x \right )+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (-5+x \right )\right ) \operatorname {csgn}\left (i \left (-5+x \right )\right )-i \pi \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (-5+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-5+x \right )\right )}\) | \(167\) |
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,method=_RETURNVERBOSE)
1/240*(-240*ln(x)*ln(x^2-5*x)*x+240*ln(x)+240*ln(x^2)*x*ln(x)*ln(x^2-5*x)+ 2400*ln(x^2-5*x)*ln(x)*ln(x^2)-2400*ln(x)^2*ln(x^2-5*x)-600*ln(x^2)^2*ln(x ^2-5*x))/ln(x^2-5*x)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {{\left (2 \, x \log \left (x\right )^{2} - x \log \left (x\right )\right )} \log \left (x^{2} - 5 \, x\right ) + \log \left (x\right )}{\log \left (x^{2} - 5 \, x\right )} \]
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=\
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=2 x \log {\left (x \right )}^{2} - x \log {\left (x \right )} + \frac {\log {\left (x \right )}}{\log {\left (x^{2} - 5 x \right )}} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (21) = 42\).
Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {2 \, x \log \left (x\right )^{3} - x \log \left (x\right )^{2} + {\left (2 \, x \log \left (x\right )^{2} - x \log \left (x\right )\right )} \log \left (x - 5\right ) + \log \left (x\right )}{\log \left (x - 5\right ) + \log \left (x\right )} \]
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=\
(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 = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=2 \, x \log \left (x\right )^{2} - x \log \left (x\right ) + \frac {\log \left (x\right )}{\log \left (x - 5\right ) + \log \left (x\right )} \]
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=\
Time = 10.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.62 \[ \int \frac {(5-2 x) \log (x)+(-5+x) \log \left (-5 x+x^2\right )+\left (5 x-x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log ^2\left (-5 x+x^2\right )+\left (-5 x+x^2+\left (-5 x+x^2\right ) \log (x)\right ) \log \left (x^2\right ) \log ^2\left (-5 x+x^2\right )}{\left (-5 x+x^2\right ) \log ^2\left (-5 x+x^2\right )} \, dx=\frac {\ln \left (x\right )}{\ln \left (x^2-5\,x\right )}-\frac {5}{4\,\left (x-\frac {5}{2}\right )}-\frac {x}{2\,x-5}+\frac {5}{2\,x-5}-x\,\ln \left (x\right )+x\,\ln \left (x^2\right )\,\ln \left (x\right ) \]
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)