Integrand size = 112, antiderivative size = 26 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 x-\log \left (\frac {x \log \left (\frac {3+x}{3}\right )}{\log ^2(\log (2-x))}\right ) \]
Time = 0.80 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 x-\log (x)-\log \left (\log \left (\frac {3+x}{3}\right )\right )+2 \log (\log (\log (2-x))) \]
Integrate[((6*x + 2*x^2)*Log[(3 + x)/3] + ((2*x - x^2)*Log[2 - x] + (6 - 1 3*x + x^2 + 2*x^3)*Log[2 - x]*Log[(3 + x)/3])*Log[Log[2 - x]])/((-6*x + x^ 2 + x^3)*Log[2 - x]*Log[(3 + x)/3]*Log[Log[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.
\(\displaystyle \int \frac {\left (2 x^2+6 x\right ) \log \left (\frac {x+3}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (2 x^3+x^2-13 x+6\right ) \log \left (\frac {x+3}{3}\right ) \log (2-x)\right ) \log (\log (2-x))}{\left (x^3+x^2-6 x\right ) \log (2-x) \log \left (\frac {x+3}{3}\right ) \log (\log (2-x))} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (2 x^2+6 x\right ) \log \left (\frac {x+3}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (2 x^3+x^2-13 x+6\right ) \log \left (\frac {x+3}{3}\right ) \log (2-x)\right ) \log (\log (2-x))}{x \left (x^2+x-6\right ) \log (2-x) \log \left (\frac {x+3}{3}\right ) \log (\log (2-x))}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2 x^2 \log \left (\frac {x+3}{3}\right )-x+5 x \log \left (\frac {x+3}{3}\right )-3 \log (x+3)+\log (27)}{x (x+3) \log \left (\frac {x}{3}+1\right )}+\frac {2}{(x-2) \log (2-x) \log (\log (2-x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \log (27) \int \frac {1}{x \log \left (\frac {x}{3}+1\right )}dx-\int \frac {\log (x+3)}{x \log \left (\frac {x}{3}+1\right )}dx+2 x-\log \left (\frac {x+3}{3}\right ) \log \left (\log \left (\frac {x+3}{3}\right )\right )+\log (x+3) \log \left (\log \left (\frac {x+3}{3}\right )\right )-\frac {1}{3} (3+\log (27)) \log \left (\log \left (\frac {x+3}{3}\right )\right )+2 \log (\log (\log (2-x)))\) |
Int[((6*x + 2*x^2)*Log[(3 + x)/3] + ((2*x - x^2)*Log[2 - x] + (6 - 13*x + x^2 + 2*x^3)*Log[2 - x]*Log[(3 + x)/3])*Log[Log[2 - x]])/((-6*x + x^2 + x^ 3)*Log[2 - x]*Log[(3 + x)/3]*Log[Log[2 - x]]),x]
3.21.68.3.1 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_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 3.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(2 x -\ln \left (x \right )-\ln \left (\ln \left (1+\frac {x}{3}\right )\right )+2 \ln \left (\ln \left (\ln \left (2-x \right )\right )\right )\) | \(28\) |
parts | \(2 x -\ln \left (x \right )-\ln \left (\ln \left (1+\frac {x}{3}\right )\right )+2 \ln \left (\ln \left (\ln \left (2-x \right )\right )\right )\) | \(28\) |
parallelrisch | \(11-\ln \left (x \right )-\ln \left (\ln \left (1+\frac {x}{3}\right )\right )+2 \ln \left (\ln \left (\ln \left (2-x \right )\right )\right )+2 x\) | \(29\) |
default | \(2 x +6-\ln \left (x \right )-\ln \left (\ln \left (3\right )-\ln \left (3+x \right )\right )+2 \ln \left (\ln \left (\ln \left (2-x \right )\right )\right )\) | \(32\) |
int((((2*x^3+x^2-13*x+6)*ln(2-x)*ln(1+1/3*x)+(-x^2+2*x)*ln(2-x))*ln(ln(2-x ))+(2*x^2+6*x)*ln(1+1/3*x))/(x^3+x^2-6*x)/ln(2-x)/ln(1+1/3*x)/ln(ln(2-x)), x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 \, x - \log \left (x\right ) - \log \left (\log \left (\frac {1}{3} \, x + 1\right )\right ) + 2 \, \log \left (\log \left (\log \left (-x + 2\right )\right )\right ) \]
integrate((((2*x^3+x^2-13*x+6)*log(2-x)*log(1+1/3*x)+(-x^2+2*x)*log(2-x))* log(log(2-x))+(2*x^2+6*x)*log(1+1/3*x))/(x^3+x^2-6*x)/log(2-x)/log(1+1/3*x )/log(log(2-x)),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 x - \log {\left (x \right )} - \log {\left (\log {\left (\frac {x}{3} + 1 \right )} \right )} + 2 \log {\left (\log {\left (\log {\left (2 - x \right )} \right )} \right )} \]
integrate((((2*x**3+x**2-13*x+6)*ln(2-x)*ln(1+1/3*x)+(-x**2+2*x)*ln(2-x))* ln(ln(2-x))+(2*x**2+6*x)*ln(1+1/3*x))/(x**3+x**2-6*x)/ln(2-x)/ln(1+1/3*x)/ ln(ln(2-x)),x)
Time = 0.41 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 \, x - \log \left (x\right ) - \log \left (-\log \left (3\right ) + \log \left (x + 3\right )\right ) + 2 \, \log \left (\log \left (\log \left (-x + 2\right )\right )\right ) \]
integrate((((2*x^3+x^2-13*x+6)*log(2-x)*log(1+1/3*x)+(-x^2+2*x)*log(2-x))* log(log(2-x))+(2*x^2+6*x)*log(1+1/3*x))/(x^3+x^2-6*x)/log(2-x)/log(1+1/3*x )/log(log(2-x)),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2 \, x - \log \left (-x\right ) - \log \left (-\log \left (3\right ) + \log \left (x + 3\right )\right ) + 2 \, \log \left (\log \left (\log \left (-x + 2\right )\right )\right ) - 4 \]
integrate((((2*x^3+x^2-13*x+6)*log(2-x)*log(1+1/3*x)+(-x^2+2*x)*log(2-x))* log(log(2-x))+(2*x^2+6*x)*log(1+1/3*x))/(x^3+x^2-6*x)/log(2-x)/log(1+1/3*x )/log(log(2-x)),x, algorithm=\
Time = 11.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (6 x+2 x^2\right ) \log \left (\frac {3+x}{3}\right )+\left (\left (2 x-x^2\right ) \log (2-x)+\left (6-13 x+x^2+2 x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right )\right ) \log (\log (2-x))}{\left (-6 x+x^2+x^3\right ) \log (2-x) \log \left (\frac {3+x}{3}\right ) \log (\log (2-x))} \, dx=2\,x-\ln \left (\ln \left (\frac {x}{3}+1\right )\right )+2\,\ln \left (\ln \left (\ln \left (2-x\right )\right )\right )-\ln \left (x\right ) \]