Integrand size = 92, antiderivative size = 37 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=\frac {1}{3} \left (-x+\frac {\log ((1-x) x)}{10 \left (-x+\frac {x}{-1+x}\right ) \log (x)}\right ) \]
Time = 1.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=\frac {1}{30} \left (-10 x-\frac {(-1+x) \log (-((-1+x) x))}{(-2+x) x \log (x)}\right ) \]
Integrate[((-40*x^2 + 40*x^3 - 10*x^4)*Log[x]^2 + (2 - 3*x + x^2)*Log[x - x^2] + Log[x]*(-2 + 5*x - 2*x^2 + (2 - 2*x + x^2)*Log[x - x^2]))/((120*x^2 - 120*x^3 + 30*x^4)*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.
\(\displaystyle \int \frac {\left (-2 x^2+\left (x^2-2 x+2\right ) \log \left (x-x^2\right )+5 x-2\right ) \log (x)+\left (x^2-3 x+2\right ) \log \left (x-x^2\right )+\left (-10 x^4+40 x^3-40 x^2\right ) \log ^2(x)}{\left (30 x^4-120 x^3+120 x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-2 x^2+\left (x^2-2 x+2\right ) \log \left (x-x^2\right )+5 x-2\right ) \log (x)+\left (x^2-3 x+2\right ) \log \left (x-x^2\right )+\left (-10 x^4+40 x^3-40 x^2\right ) \log ^2(x)}{x^2 \left (30 x^2-120 x+120\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 120 \int -\frac {10 \left (x^4-4 x^3+4 x^2\right ) \log ^2(x)+\left (2 x^2-5 x-\left (x^2-2 x+2\right ) \log \left (x-x^2\right )+2\right ) \log (x)-\left (x^2-3 x+2\right ) \log \left (x-x^2\right )}{3600 (2-x)^2 x^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{30} \int \frac {10 \left (x^4-4 x^3+4 x^2\right ) \log ^2(x)+\left (2 x^2-5 x-\left (x^2-2 x+2\right ) \log \left (x-x^2\right )+2\right ) \log (x)-\left (x^2-3 x+2\right ) \log \left (x-x^2\right )}{(2-x)^2 x^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{30} \int \left (\frac {10 \log (x) x^3-20 \log (x) x^2+2 x-1}{(x-2) x^2 \log (x)}+\frac {\left (-\log (x) x^2-x^2+2 \log (x) x+3 x-2 \log (x)-2\right ) \log ((1-x) x)}{(2-x)^2 x^2 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{30} \left (\frac {1}{2} \int \frac {\log ((1-x) x)}{x^2 \log ^2(x)}dx-\int \frac {2 x-1}{(x-2) x^2 \log (x)}dx+\frac {1}{2} \int \frac {\log ((1-x) x)}{x^2 \log (x)}dx-\frac {3}{2} \int \frac {\log ((1-x) x)}{(2-x) \log ^2(x)}dx-\frac {5}{4} \int \frac {\log ((1-x) x)}{(x-2) \log ^2(x)}dx-\frac {1}{4} \int \frac {\log ((1-x) x)}{x \log ^2(x)}dx+\frac {1}{2} \int \frac {\log ((1-x) x)}{(2-x)^2 \log (x)}dx-\int \frac {\log ((1-x) x)}{(2-x) \log (x)}dx-\int \frac {\log ((1-x) x)}{(x-2) \log (x)}dx-10 x\right )\) |
Int[((-40*x^2 + 40*x^3 - 10*x^4)*Log[x]^2 + (2 - 3*x + x^2)*Log[x - x^2] + Log[x]*(-2 + 5*x - 2*x^2 + (2 - 2*x + x^2)*Log[x - x^2]))/((120*x^2 - 120 *x^3 + 30*x^4)*Log[x]^2),x]
3.30.7.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 5.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51
method | result | size |
parallelrisch | \(\frac {-60 x^{3} \ln \left (x \right )+80 x^{2} \ln \left (x \right )+80 x \ln \left (x \right )-6 \ln \left (-x^{2}+x \right ) x +6 \ln \left (-x^{2}+x \right )}{180 x \ln \left (x \right ) \left (-2+x \right )}\) | \(56\) |
risch | \(-\frac {\left (-1+x \right ) \ln \left (-1+x \right )}{30 x \left (-2+x \right ) \ln \left (x \right )}-\frac {-i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{3}-i \pi x \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i x \left (-1+x \right )\right ) \operatorname {csgn}\left (i x \right )+2 i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2}+i \pi x \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{3}-2 i \pi x \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2}-i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )\right )-2 i \pi +i \pi \,\operatorname {csgn}\left (i x \left (-1+x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-1+x \right )\right )+2 i \pi x -i \pi \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi x \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi x \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i x \left (-1+x \right )\right )^{2}+20 x^{3} \ln \left (x \right )-40 x^{2} \ln \left (x \right )+2 x \ln \left (x \right )-2 \ln \left (x \right )}{60 x \left (-2+x \right ) \ln \left (x \right )}\) | \(259\) |
int(((-10*x^4+40*x^3-40*x^2)*ln(x)^2+((x^2-2*x+2)*ln(-x^2+x)-2*x^2+5*x-2)* ln(x)+(x^2-3*x+2)*ln(-x^2+x))/(30*x^4-120*x^3+120*x^2)/ln(x)^2,x,method=_R ETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=-\frac {{\left (x - 1\right )} \log \left (-x^{2} + x\right ) + 10 \, {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x\right )}{30 \, {\left (x^{2} - 2 \, x\right )} \log \left (x\right )} \]
integrate(((-10*x^4+40*x^3-40*x^2)*log(x)^2+((x^2-2*x+2)*log(-x^2+x)-2*x^2 +5*x-2)*log(x)+(x^2-3*x+2)*log(-x^2+x))/(30*x^4-120*x^3+120*x^2)/log(x)^2, x, algorithm=\
Exception generated. \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \]
integrate(((-10*x**4+40*x**3-40*x**2)*ln(x)**2+((x**2-2*x+2)*ln(-x**2+x)-2 *x**2+5*x-2)*ln(x)+(x**2-3*x+2)*ln(-x**2+x))/(30*x**4-120*x**3+120*x**2)/l n(x)**2,x)
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=-\frac {{\left (10 \, x^{3} - 20 \, x^{2} + x - 1\right )} \log \left (x\right ) + {\left (x - 1\right )} \log \left (-x + 1\right )}{30 \, {\left (x^{2} - 2 \, x\right )} \log \left (x\right )} \]
integrate(((-10*x^4+40*x^3-40*x^2)*log(x)^2+((x^2-2*x+2)*log(-x^2+x)-2*x^2 +5*x-2)*log(x)+(x^2-3*x+2)*log(-x^2+x))/(30*x^4-120*x^3+120*x^2)/log(x)^2, x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=-\frac {1}{3} \, x - \frac {{\left (x - 1\right )} \log \left (-x + 1\right )}{30 \, {\left (x^{2} \log \left (x\right ) - 2 \, x \log \left (x\right )\right )}} - \frac {1}{60 \, {\left (x - 2\right )}} - \frac {1}{60 \, x} \]
integrate(((-10*x^4+40*x^3-40*x^2)*log(x)^2+((x^2-2*x+2)*log(-x^2+x)-2*x^2 +5*x-2)*log(x)+(x^2-3*x+2)*log(-x^2+x))/(30*x^4-120*x^3+120*x^2)/log(x)^2, x, algorithm=\
Time = 9.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-40 x^2+40 x^3-10 x^4\right ) \log ^2(x)+\left (2-3 x+x^2\right ) \log \left (x-x^2\right )+\log (x) \left (-2+5 x-2 x^2+\left (2-2 x+x^2\right ) \log \left (x-x^2\right )\right )}{\left (120 x^2-120 x^3+30 x^4\right ) \log ^2(x)} \, dx=\frac {\frac {\ln \left (x-x^2\right )}{30}-\frac {x\,\ln \left (x-x^2\right )}{30}}{x\,\ln \left (x\right )\,\left (x-2\right )}-\frac {x}{3} \]