Integrand size = 85, antiderivative size = 21 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x+\log (x)}{\left (x+x^2\right ) \log ((-1+x) x)} \]
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x+\log (x)}{x (1+x) \log ((-1+x) x)} \]
Integrate[(-x + x^2 + 2*x^3 + (-1 + x + 2*x^2)*Log[x] + (1 - 2*x^2 + x^3 + (-1 - x + 2*x^2)*Log[x])*Log[-x + x^2])/((-x^2 - x^3 + x^4 + x^5)*Log[-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.
\(\displaystyle \int \frac {2 x^3+x^2+\left (2 x^2+x-1\right ) \log (x)+\left (x^3-2 x^2+\left (2 x^2-x-1\right ) \log (x)+1\right ) \log \left (x^2-x\right )-x}{\left (x^5+x^4-x^3-x^2\right ) \log ^2\left (x^2-x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2 x^3+x^2+\left (2 x^2+x-1\right ) \log (x)+\left (x^3-2 x^2+\left (2 x^2-x-1\right ) \log (x)+1\right ) \log \left (x^2-x\right )-x}{x^2 \left (x^3+x^2-x-1\right ) \log ^2\left (x^2-x\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {2 x^3+x^2+\left (2 x^2+x-1\right ) \log (x)+\left (x^3-2 x^2+\left (2 x^2-x-1\right ) \log (x)+1\right ) \log \left (x^2-x\right )-x}{2 x^2 \left (x^2-1\right ) \log ^2\left (x^2-x\right )}-\frac {2 x^3+x^2+\left (2 x^2+x-1\right ) \log (x)+\left (x^3-2 x^2+\left (2 x^2-x-1\right ) \log (x)+1\right ) \log \left (x^2-x\right )-x}{2 x^2 (x+1)^2 \log ^2\left (x^2-x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\log (x)}{x^2 \log ^2((x-1) x)}dx-\int \frac {1}{x^2 \log ((x-1) x)}dx+\int \frac {x^2}{(x+1)^2 \log ((x-1) x)}dx+2 \int \frac {x^2}{(x+1) \log ((x-1) x)}dx+\int \frac {\log (x)}{x^2 \log ((x-1) x)}dx-2 \int \frac {1}{\log ^2((x-1) x)}dx+\int \frac {1}{x \log ^2((x-1) x)}dx+\frac {1}{2} \int \frac {x}{(x-1) \log ^2((x-1) x)}dx+\frac {3}{2} \int \frac {x}{(x+1) \log ^2((x-1) x)}dx+\frac {1}{2} \int \frac {\log (x)}{(x-1) \log ^2((x-1) x)}dx-2 \int \frac {\log (x)}{x \log ^2((x-1) x)}dx+\frac {3}{2} \int \frac {\log (x)}{(x+1) \log ^2((x-1) x)}dx+3 \int \frac {1}{\log ((x-1) x)}dx+\int \frac {1}{x \log ((x-1) x)}dx-2 \int \frac {x}{\log ((x-1) x)}dx-\int \frac {1}{(x+1)^2 \log ((x-1) x)}dx-\int \frac {x}{(x+1)^2 \log ((x-1) x)}dx-2 \int \frac {1}{(x+1) \log ((x-1) x)}dx-2 \int \frac {x}{(x+1) \log ((x-1) x)}dx-\int \frac {\log (x)}{(x+1)^2 \log ((x-1) x)}dx\) |
Int[(-x + x^2 + 2*x^3 + (-1 + x + 2*x^2)*Log[x] + (1 - 2*x^2 + x^3 + (-1 - x + 2*x^2)*Log[x])*Log[-x + x^2])/((-x^2 - x^3 + x^4 + x^5)*Log[-x + x^2] ^2),x]
3.4.14.3.1 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])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 262.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(\frac {-x -\ln \left (x \right )}{x \ln \left (x^{2}-x \right ) \left (1+x \right )}\) | \(28\) |
risch | \(-\frac {2 \left (x +\ln \left (x \right )\right )}{\left (1+x \right ) x \left (-i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )\right )-i \pi \,\operatorname {csgn}\left (i x \left (-1+x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-1+x \right )\right )+2 \ln \left (x \right )+2 \ln \left (-1+x \right )\right )}\) | \(106\) |
int((((2*x^2-x-1)*ln(x)+x^3-2*x^2+1)*ln(x^2-x)+(2*x^2+x-1)*ln(x)+2*x^3+x^2 -x)/(x^5+x^4-x^3-x^2)/ln(x^2-x)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x + \log \left (x\right )}{{\left (x^{2} + x\right )} \log \left (x^{2} - x\right )} \]
integrate((((2*x^2-x-1)*log(x)+x^3-2*x^2+1)*log(x^2-x)+(2*x^2+x-1)*log(x)+ 2*x^3+x^2-x)/(x^5+x^4-x^3-x^2)/log(x^2-x)^2,x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=\frac {- x - \log {\left (x \right )}}{\left (x^{2} + x\right ) \log {\left (x^{2} - x \right )}} \]
integrate((((2*x**2-x-1)*ln(x)+x**3-2*x**2+1)*ln(x**2-x)+(2*x**2+x-1)*ln(x )+2*x**3+x**2-x)/(x**5+x**4-x**3-x**2)/ln(x**2-x)**2,x)
Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x + \log \left (x\right )}{{\left (x^{2} + x\right )} \log \left (x - 1\right ) + {\left (x^{2} + x\right )} \log \left (x\right )} \]
integrate((((2*x^2-x-1)*log(x)+x^3-2*x^2+1)*log(x^2-x)+(2*x^2+x-1)*log(x)+ 2*x^3+x^2-x)/(x^5+x^4-x^3-x^2)/log(x^2-x)^2,x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x + \log \left (x\right )}{x^{2} \log \left (x - 1\right ) + x^{2} \log \left (x\right ) + x \log \left (x - 1\right ) + x \log \left (x\right )} \]
integrate((((2*x^2-x-1)*log(x)+x^3-2*x^2+1)*log(x^2-x)+(2*x^2+x-1)*log(x)+ 2*x^3+x^2-x)/(x^5+x^4-x^3-x^2)/log(x^2-x)^2,x, algorithm=\
Time = 9.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-x+x^2+2 x^3+\left (-1+x+2 x^2\right ) \log (x)+\left (1-2 x^2+x^3+\left (-1-x+2 x^2\right ) \log (x)\right ) \log \left (-x+x^2\right )}{\left (-x^2-x^3+x^4+x^5\right ) \log ^2\left (-x+x^2\right )} \, dx=-\frac {x+\ln \left (x\right )}{x\,\ln \left (x^2-x\right )\,\left (x+1\right )} \]