Integrand size = 67, antiderivative size = 26 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \left (x^2+\log ^2(x)-\log \left (5 \log \left (\frac {-1+x}{2 x}\right )\right )\right ) \end {dmath*}
Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \left (x^2+\log ^2(x)-\log \left (\log \left (\frac {-1+x}{2 x}\right )\right )\right ) \end {dmath*}
Integrate[(-5 + (-10*x^2 + 10*x^3)*Log[(-1 + x)/(2*x)] + (-10 + 10*x)*Log[ (-1 + x)/(2*x)]*Log[x])/((-x + x^2)*Log[(-1 + x)/(2*x)]),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.
\(\displaystyle \int \frac {\left (10 x^3-10 x^2\right ) \log \left (\frac {x-1}{2 x}\right )+(10 x-10) \log (x) \log \left (\frac {x-1}{2 x}\right )-5}{\left (x^2-x\right ) \log \left (\frac {x-1}{2 x}\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (10 x^3-10 x^2\right ) \log \left (\frac {x-1}{2 x}\right )+(10 x-10) \log (x) \log \left (\frac {x-1}{2 x}\right )-5}{(x-1) x \log \left (\frac {x-1}{2 x}\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (2 \left (x^2+\log (x)\right )-\frac {1}{(x-1) \log \left (\frac {x-1}{2 x}\right )}\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int \frac {2 \left (x^2+\log (x)\right )+\frac {1}{(1-x) \log \left (-\frac {1-x}{2 x}\right )}}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle 5 \int \left (\frac {-2 \log \left (\frac {x-1}{2 x}\right ) x^3+2 \log \left (\frac {x-1}{2 x}\right ) x^2+1}{(1-x) x \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}+\frac {2 \log (x)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \left (-\int \frac {1}{(x-1) \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}dx+\int \frac {1}{x \log \left (\frac {1}{2}-\frac {1}{2 x}\right )}dx+x^2+\log ^2(x)\right )\) |
Int[(-5 + (-10*x^2 + 10*x^3)*Log[(-1 + x)/(2*x)] + (-10 + 10*x)*Log[(-1 + x)/(2*x)]*Log[x])/((-x + x^2)*Log[(-1 + x)/(2*x)]),x]
3.3.60.3.1 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[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (\frac {-1+x}{2 x}\right )\right )\) | \(25\) |
parts | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (\frac {-1+x}{2 x}\right )\right )\) | \(25\) |
default | \(5 \ln \left (x \right )^{2}+5 x^{2}-5 \ln \left (\ln \left (2\right )-\ln \left (1-\frac {1}{x}\right )\right )\) | \(29\) |
risch | \(5 x^{2}+5 \ln \left (x \right )^{2}-5 \ln \left (\ln \left (-1+x \right )-\frac {i \left (\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}-\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )-\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2} \operatorname {csgn}\left (i \left (-1+x \right )\right )+\pi \,\operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right )-2 i \ln \left (2\right )-2 i \ln \left (x \right )\right )}{2}\right )\) | \(119\) |
int(((10*x-10)*ln(1/2*(-1+x)/x)*ln(x)+(10*x^3-10*x^2)*ln(1/2*(-1+x)/x)-5)/ (x^2-x)/ln(1/2*(-1+x)/x),x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (\log \left (\frac {x - 1}{2 \, x}\right )\right ) \end {dmath*}
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 x^{2} + 5 \log {\left (x \right )}^{2} - 5 \log {\left (\log {\left (\frac {\frac {x}{2} - \frac {1}{2}}{x} \right )} \right )} \end {dmath*}
integrate(((10*x-10)*ln(1/2*(-1+x)/x)*ln(x)+(10*x**3-10*x**2)*ln(1/2*(-1+x )/x)-5)/(x**2-x)/ln(1/2*(-1+x)/x),x)
Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (-\log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right )\right ) \end {dmath*}
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5 \, x^{2} + 5 \, \log \left (x\right )^{2} - 5 \, \log \left (-\log \left (2\right ) + \log \left (x - 1\right ) - \log \left (x\right )\right ) \end {dmath*}
integrate(((10*x-10)*log(1/2*(-1+x)/x)*log(x)+(10*x^3-10*x^2)*log(1/2*(-1+ x)/x)-5)/(x^2-x)/log(1/2*(-1+x)/x),x, algorithm=\
Time = 14.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {-5+\left (-10 x^2+10 x^3\right ) \log \left (\frac {-1+x}{2 x}\right )+(-10+10 x) \log \left (\frac {-1+x}{2 x}\right ) \log (x)}{\left (-x+x^2\right ) \log \left (\frac {-1+x}{2 x}\right )} \, dx=5\,{\ln \left (x\right )}^2-5\,\ln \left (\ln \left (\frac {\frac {x}{2}-\frac {1}{2}}{x}\right )\right )+5\,x^2 \end {dmath*}