Integrand size = 264, antiderivative size = 27 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\left (1-\frac {e^2}{2}+e^x+x\right )^2\right )\right ) \]
Time = 1.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \log \left (x^2\right ) \log \left (x+\log \left (\frac {1}{4} \left (2-e^2+2 e^x+2 x\right )^2\right )\right ) \]
Integrate[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E ^x*x + 4*x^2 + (4 - 2*E^2 + 4*E^x + 4*x)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(- 4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4 *x^2 - 2*E^2*x^2 + 4*E^x*x^2 + 4*x^3 + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*L og[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]),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.
| \(\displaystyle \int \frac {\left (2 x^2+6 e^x x-e^2 x+6 x\right ) \log \left (x^2\right )+\left (4 x^2+\left (4 x+4 e^x-2 e^2+4\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+4 e^x x-2 e^2 x+4 x\right ) \log \left (\log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+x\right )}{4 x^3+4 e^x x^2-2 e^2 x^2+4 x^2+\left (4 x^2+4 e^x x-2 e^2 x+4 x\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (2 x^2+6 e^x x-e^2 x+6 x\right ) \log \left (x^2\right )+\left (4 x^2+\left (4 x+4 e^x-2 e^2+4\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+4 e^x x-2 e^2 x+4 x\right ) \log \left (\log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )+x\right )}{4 x^3+4 e^x x^2+\left (4-2 e^2\right ) x^2+\left (4 x^2+4 e^x x-2 e^2 x+4 x\right ) \log \left (\frac {1}{4} \left (4 x^2+8 x+4 e^{2 x}+e^2 (-4 x-4)+e^x \left (8 x-4 e^2+8\right )+e^4+4\right )\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \left (\frac {\left (2 x+6 e^x+6 \left (1-\frac {e^2}{6}\right )\right ) \log \left (x^2\right )}{2 \left (2 x+2 e^x+2 \left (1-\frac {e^2}{2}\right )\right ) \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}+\frac {\log \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{2} \int \frac {\log \left (x^2\right )}{x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )}dx+2 \int \frac {x \log \left (x^2\right )}{\left (-2 x-2 e^x-2 \left (1-\frac {e^2}{2}\right )\right ) \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}dx+e^2 \int \frac {\log \left (x^2\right )}{\left (2 x+2 e^x+2 \left (1-\frac {e^2}{2}\right )\right ) \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}dx+\int \frac {\log \left (x+\log \left (\frac {1}{4} \left (2 x+2 e^x-e^2+2\right )^2\right )\right )}{x}dx\) |
Int[((6*x - E^2*x + 6*E^x*x + 2*x^2)*Log[x^2] + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2 + (4 - 2*E^2 + 4*E^x + 4*x)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4* x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4])*Log[x + Log[(4 + E^4 + 4*E^( 2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/4]])/(4*x^2 - 2*E^2*x^2 + 4*E^x*x^2 + 4*x^3 + (4*x - 2*E^2*x + 4*E^x*x + 4*x^2)*Log[(4 + E^4 + 4*E^(2*x) + E^2*(-4 - 4*x) + 8*x + 4*x^2 + E^x*(8 - 4*E^2 + 8*x))/ 4]),x]
3.10.63.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 25.99 (sec) , antiderivative size = 249, normalized size of antiderivative = 9.22
| method | result | size |
| risch | \(\ln \left (x \right ) \ln \left (-2 \ln \left (2\right )+2 \ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )\right )}^{2}}{2}+x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right ) \ln \left (\ln \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )+\frac {i \left (-\pi {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right )}^{2} \operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )+2 \pi \,\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )\right ) {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )}^{2}-\pi {\operatorname {csgn}\left (i \left (-2 \,{\mathrm e}^{x}+{\mathrm e}^{2}-2 x -2\right )^{2}\right )}^{3}+4 i \ln \left (2\right )-2 i x \right )}{4}\right )}{4}\) | \(249\) |
int((((4*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 /4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4*x^2+4*x )*ln(ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*ex p(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*ln(x^2))/((4*exp(x)*x-2 *exp(2)*x+4*x^2+4*x)*ln(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2 +1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3+4*x^2),x,m ethod=_RETURNVERBOSE)
ln(x)*ln(-2*ln(2)+2*ln(-2*exp(x)+exp(2)-2*x-2)-1/2*I*Pi*csgn(I*(-2*exp(x)+ exp(2)-2*x-2)^2)*(-csgn(I*(-2*exp(x)+exp(2)-2*x-2)^2)+csgn(I*(-2*exp(x)+ex p(2)-2*x-2)))^2+x)-1/4*I*Pi*csgn(I*x^2)*(csgn(I*x)^2-2*csgn(I*x^2)*csgn(I* x)+csgn(I*x^2)^2)*ln(ln(-2*exp(x)+exp(2)-2*x-2)+1/4*I*(-Pi*csgn(I*(-2*exp( x)+exp(2)-2*x-2))^2*csgn(I*(-2*exp(x)+exp(2)-2*x-2)^2)+2*Pi*csgn(I*(-2*exp (x)+exp(2)-2*x-2))*csgn(I*(-2*exp(x)+exp(2)-2*x-2)^2)^2-Pi*csgn(I*(-2*exp( x)+exp(2)-2*x-2)^2)^3+4*I*ln(2)-2*I*x))
Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right ) \]
integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 /4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 +4*x^2),x, algorithm=\
1/2*log(x^2)*log(x + log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1 /4*e^4 + e^(2*x) + 1))
Timed out. \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\text {Timed out} \]
integrate((((4*exp(x)-2*exp(2)+4*x+4)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*e xp(x)+1/4*exp(2)**2+1/4*(-4-4*x)*exp(2)+x**2+2*x+1)+4*exp(x)*x-2*exp(2)*x+ 4*x**2+4*x)*ln(ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)**2+1/4 *(-4-4*x)*exp(2)+x**2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x**2+6*x)*ln(x**2)) /((4*exp(x)*x-2*exp(2)*x+4*x**2+4*x)*ln(exp(x)**2+1/4*(-4*exp(2)+8*x+8)*ex p(x)+1/4*exp(2)**2+1/4*(-4-4*x)*exp(2)+x**2+2*x+1)+4*exp(x)*x**2-2*x**2*ex p(2)+4*x**3+4*x**2),x)
Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\log \left (x - 2 \, \log \left (2\right ) + 2 \, \log \left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )\right ) \log \left (x\right ) \]
integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 /4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 +4*x^2),x, algorithm=\
\[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\int { \frac {{\left (2 \, x^{2} - x e^{2} + 6 \, x e^{x} + 6 \, x\right )} \log \left (x^{2}\right ) + 2 \, {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + {\left (2 \, x - e^{2} + 2 \, e^{x} + 2\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right ) + 2 \, x\right )} \log \left (x + \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}{2 \, {\left (2 \, x^{3} - x^{2} e^{2} + 2 \, x^{2} e^{x} + 2 \, x^{2} + {\left (2 \, x^{2} - x e^{2} + 2 \, x e^{x} + 2 \, x\right )} \log \left (x^{2} - {\left (x + 1\right )} e^{2} + {\left (2 \, x - e^{2} + 2\right )} e^{x} + 2 \, x + \frac {1}{4} \, e^{4} + e^{\left (2 \, x\right )} + 1\right )\right )}} \,d x } \]
integrate((((4*exp(x)-2*exp(2)+4*x+4)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*e xp(x)+1/4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x-2*exp(2)*x+4* x^2+4*x)*log(log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1/4*exp(2)^2+1/4*(- 4-4*x)*exp(2)+x^2+2*x+1)+x)+(6*exp(x)*x-exp(2)*x+2*x^2+6*x)*log(x^2))/((4* exp(x)*x-2*exp(2)*x+4*x^2+4*x)*log(exp(x)^2+1/4*(-4*exp(2)+8*x+8)*exp(x)+1 /4*exp(2)^2+1/4*(-4-4*x)*exp(2)+x^2+2*x+1)+4*exp(x)*x^2-2*x^2*exp(2)+4*x^3 +4*x^2),x, algorithm=\
integrate(1/2*((2*x^2 - x*e^2 + 6*x*e^x + 6*x)*log(x^2) + 2*(2*x^2 - x*e^2 + 2*x*e^x + (2*x - e^2 + 2*e^x + 2)*log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1) + 2*x)*log(x + log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1)))/(2*x^3 - x^2*e^2 + 2*x^2*e^x + 2*x^2 + (2*x^2 - x*e^2 + 2*x*e^x + 2*x)*log(x^2 - (x + 1)*e^2 + (2*x - e^2 + 2)*e^x + 2*x + 1/4*e^4 + e^(2*x) + 1)), x)
Time = 13.95 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {\left (6 x-e^2 x+6 e^x x+2 x^2\right ) \log \left (x^2\right )+\left (4 x-2 e^2 x+4 e^x x+4 x^2+\left (4-2 e^2+4 e^x+4 x\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right ) \log \left (x+\log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )\right )}{4 x^2-2 e^2 x^2+4 e^x x^2+4 x^3+\left (4 x-2 e^2 x+4 e^x x+4 x^2\right ) \log \left (\frac {1}{4} \left (4+e^4+4 e^{2 x}+e^2 (-4-4 x)+8 x+4 x^2+e^x \left (8-4 e^2+8 x\right )\right )\right )} \, dx=\frac {\ln \left (x^2\right )\,\ln \left (x+\ln \left (2\,x+{\mathrm {e}}^{2\,x}+\frac {{\mathrm {e}}^4}{4}+\frac {{\mathrm {e}}^x\,\left (8\,x-4\,{\mathrm {e}}^2+8\right )}{4}+x^2-\frac {{\mathrm {e}}^2\,\left (4\,x+4\right )}{4}+1\right )\right )}{2} \]
int((log(x^2)*(6*x - x*exp(2) + 6*x*exp(x) + 2*x^2) + log(x + log(2*x + ex p(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1))*(4*x - 2*x*exp(2) + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*( 8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1)*(4*x - 2*exp(2) + 4*exp(x) + 4) + 4*x*exp(x) + 4*x^2))/(4*x^2*exp(x) + log(2*x + exp(2*x) + exp(4)/4 + (exp(x)*(8*x - 4*exp(2) + 8))/4 + x^2 - (exp(2)*(4*x + 4))/4 + 1)*(4*x - 2*x*exp(2) + 4*x*exp(x) + 4*x^2) - 2*x^2*exp(2) + 4*x^2 + 4*x^3 ),x)