Integrand size = 109, antiderivative size = 27 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {(-4+\log (x)) \log (4+2 x)}{\log \left (e^{2 e^3} (3-x)\right )} \]
Time = 3.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {(-4+\log (x)) \log (2 (2+x))}{2 e^3+\log (3-x)} \]
Integrate[((8*x + 4*x^2 + (-2*x - x^2)*Log[x])*Log[4 + 2*x] + Log[E^(2*E^3 )*(3 - x)]*(12*x - 4*x^2 + (-3*x + x^2)*Log[x] + (-6 - x + x^2)*Log[4 + 2* x]))/((-6*x - x^2 + x^3)*Log[E^(2*E^3)*(3 - 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.
\(\displaystyle \int \frac {\left (4 x^2+\left (-x^2-2 x\right ) \log (x)+8 x\right ) \log (2 x+4)+\log \left (e^{2 e^3} (3-x)\right ) \left (-4 x^2+\left (x^2-3 x\right ) \log (x)+\left (x^2-x-6\right ) \log (2 x+4)+12 x\right )}{\left (x^3-x^2-6 x\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (4 x^2+\left (-x^2-2 x\right ) \log (x)+8 x\right ) \log (2 x+4)+\log \left (e^{2 e^3} (3-x)\right ) \left (-4 x^2+\left (x^2-3 x\right ) \log (x)+\left (x^2-x-6\right ) \log (2 x+4)+12 x\right )}{x \left (x^2-x-6\right ) \log ^2\left (e^{2 e^3} (3-x)\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {\log (x)-4}{(x+2) \left (\log (3-x)+2 e^3\right )}+\frac {\left (-4 \left (1+\frac {e^3}{2}\right ) x+x (-\log (3-x))+x \log (x)+3 \log (3-x)+6 e^3\right ) \log (2 x+4)}{(3-x) x \left (\log (3-x)+2 e^3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (2+e^3\right ) \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (\log (x)+2 e^3\right )^2}dx,x,3-x\right )-2 e^3 \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (\log (x)+2 e^3\right )^2}dx,x,3-x\right )+\int \frac {\log (x)-4}{(x+2) \left (\log (3-x)+2 e^3\right )}dx+2 e^3 \int \frac {\log (2 x+4)}{x \left (\log (3-x)+2 e^3\right )^2}dx+\int \frac {\log (3-x) \log (2 x+4)}{(3-x) \left (\log (3-x)+2 e^3\right )^2}dx+\int \frac {\log (3-x) \log (2 x+4)}{(x-3) \left (\log (3-x)+2 e^3\right )^2}dx+\int \frac {\log (3-x) \log (2 x+4)}{x \left (\log (3-x)+2 e^3\right )^2}dx+\int \frac {\log (x) \log (2 x+4)}{(3-x) \left (\log (3-x)+2 e^3\right )^2}dx\) |
Int[((8*x + 4*x^2 + (-2*x - x^2)*Log[x])*Log[4 + 2*x] + Log[E^(2*E^3)*(3 - x)]*(12*x - 4*x^2 + (-3*x + x^2)*Log[x] + (-6 - x + x^2)*Log[4 + 2*x]))/( (-6*x - x^2 + x^3)*Log[E^(2*E^3)*(3 - x)]^2),x]
3.27.18.3.1 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 = 5.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {\left (\ln \left (x \right )-4\right ) \ln \left (4+2 x \right )}{\ln \left (\left (-x +3\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}\right )}\) | \(26\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) \ln \left (4+2 x \right )-8 \ln \left (4+2 x \right )}{2 \ln \left (\left (-x +3\right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}\right )}\) | \(36\) |
int((((x^2-x-6)*ln(4+2*x)+(x^2-3*x)*ln(x)-4*x^2+12*x)*ln((-x+3)*exp(exp(3) )^2)+((-x^2-2*x)*ln(x)+4*x^2+8*x)*ln(4+2*x))/(x^3-x^2-6*x)/ln((-x+3)*exp(e xp(3))^2)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {{\left (\log \left (x\right ) - 4\right )} \log \left (2 \, x + 4\right )}{\log \left (-{\left (x - 3\right )} e^{\left (2 \, e^{3}\right )}\right )} \]
integrate((((x^2-x-6)*log(4+2*x)+(x^2-3*x)*log(x)-4*x^2+12*x)*log((-x+3)*e xp(exp(3))^2)+((-x^2-2*x)*log(x)+4*x^2+8*x)*log(4+2*x))/(x^3-x^2-6*x)/log( (-x+3)*exp(exp(3))^2)^2,x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {\log {\left (x \right )} \log {\left (2 x + 4 \right )} - 4 \log {\left (2 x + 4 \right )}}{\log {\left (\left (3 - x\right ) e^{2 e^{3}} \right )}} \]
integrate((((x**2-x-6)*ln(4+2*x)+(x**2-3*x)*ln(x)-4*x**2+12*x)*ln((-x+3)*e xp(exp(3))**2)+((-x**2-2*x)*ln(x)+4*x**2+8*x)*ln(4+2*x))/(x**3-x**2-6*x)/l n((-x+3)*exp(exp(3))**2)**2,x)
Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {{\left (\log \left (x\right ) - 4\right )} \log \left (x + 2\right ) + \log \left (2\right ) \log \left (x\right ) - 4 \, \log \left (2\right )}{2 \, e^{3} + \log \left (-x + 3\right )} \]
integrate((((x^2-x-6)*log(4+2*x)+(x^2-3*x)*log(x)-4*x^2+12*x)*log((-x+3)*e xp(exp(3))^2)+((-x^2-2*x)*log(x)+4*x^2+8*x)*log(4+2*x))/(x^3-x^2-6*x)/log( (-x+3)*exp(exp(3))^2)^2,x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\frac {\log \left (2 \, x + 4\right ) \log \left (x\right ) - 4 \, \log \left (2 \, x + 4\right )}{2 \, e^{3} + \log \left (-x + 3\right )} \]
integrate((((x^2-x-6)*log(4+2*x)+(x^2-3*x)*log(x)-4*x^2+12*x)*log((-x+3)*e xp(exp(3))^2)+((-x^2-2*x)*log(x)+4*x^2+8*x)*log(4+2*x))/(x^3-x^2-6*x)/log( (-x+3)*exp(exp(3))^2)^2,x, algorithm=\
Time = 11.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 4.00 \[ \int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx=\ln \left (x\,\left (x+2\right )\right )+\frac {20}{x+2}-\frac {5\,\ln \left (x\right )}{x+2}+\frac {\ln \left (2\,x+4\right )\,\left (\ln \left (x\right )-4\right )-\frac {\ln \left (-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (x-3\right )\right )\,\left (x-3\right )\,\left (2\,\ln \left (2\,x+4\right )-4\,x+x\,\ln \left (2\,x+4\right )+x\,\ln \left (x\right )\right )}{x\,\left (x+2\right )}}{\ln \left (-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (x-3\right )\right )}-\frac {3\,\ln \left (2\,x+4\right )}{x} \]
int((log(-exp(2*exp(3))*(x - 3))*(log(2*x + 4)*(x - x^2 + 6) - 12*x + log( x)*(3*x - x^2) + 4*x^2) - log(2*x + 4)*(8*x - log(x)*(2*x + x^2) + 4*x^2)) /(log(-exp(2*exp(3))*(x - 3))^2*(6*x + x^2 - x^3)),x)