Integrand size = 164, antiderivative size = 29 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=\frac {\left (x-\log (3)+x^2 \log ^2\left (\frac {12 \left (e^2+x\right )}{x}\right )\right )^2}{x^2} \]
\[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=\int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx \]
Integrate[((2*E^2*x + 2*x^2)*Log[3] + (-2*E^2 - 2*x)*Log[3]^2 + (-4*E^2*x^ 3 + 4*E^2*x^2*Log[3])*Log[(12*E^2 + 12*x)/x] + (2*E^2*x^3 + 2*x^4)*Log[(12 *E^2 + 12*x)/x]^2 - 4*E^2*x^4*Log[(12*E^2 + 12*x)/x]^3 + (2*E^2*x^4 + 2*x^ 5)*Log[(12*E^2 + 12*x)/x]^4)/(E^2*x^3 + x^4),x]
Integrate[((2*E^2*x + 2*x^2)*Log[3] + (-2*E^2 - 2*x)*Log[3]^2 + (-4*E^2*x^ 3 + 4*E^2*x^2*Log[3])*Log[(12*E^2 + 12*x)/x] + (2*E^2*x^3 + 2*x^4)*Log[(12 *E^2 + 12*x)/x]^2 - 4*E^2*x^4*Log[(12*E^2 + 12*x)/x]^3 + (2*E^2*x^4 + 2*x^ 5)*Log[(12*E^2 + 12*x)/x]^4)/(E^2*x^3 + 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.
| \(\displaystyle \int \frac {-4 e^2 x^4 \log ^3\left (\frac {12 x+12 e^2}{x}\right )+\left (2 x^2+2 e^2 x\right ) \log (3)+\left (2 x^5+2 e^2 x^4\right ) \log ^4\left (\frac {12 x+12 e^2}{x}\right )+\left (2 x^4+2 e^2 x^3\right ) \log ^2\left (\frac {12 x+12 e^2}{x}\right )+\left (4 e^2 x^2 \log (3)-4 e^2 x^3\right ) \log \left (\frac {12 x+12 e^2}{x}\right )+\left (-2 x-2 e^2\right ) \log ^2(3)}{x^4+e^2 x^3} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-4 e^2 x^4 \log ^3\left (\frac {12 x+12 e^2}{x}\right )+\left (2 x^2+2 e^2 x\right ) \log (3)+\left (2 x^5+2 e^2 x^4\right ) \log ^4\left (\frac {12 x+12 e^2}{x}\right )+\left (2 x^4+2 e^2 x^3\right ) \log ^2\left (\frac {12 x+12 e^2}{x}\right )+\left (4 e^2 x^2 \log (3)-4 e^2 x^3\right ) \log \left (\frac {12 x+12 e^2}{x}\right )+\left (-2 x-2 e^2\right ) \log ^2(3)}{x^3 \left (x+e^2\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 \left (x^2 \log ^2\left (\frac {12 \left (x+e^2\right )}{x}\right )+x-\log (3)\right ) \left (\left (x+e^2\right ) x^2 \log ^2\left (\frac {12 \left (x+e^2\right )}{x}\right )-2 e^2 x^2 \log \left (\frac {12 \left (x+e^2\right )}{x}\right )+\left (x+e^2\right ) \log (3)\right )}{x^3 \left (x+e^2\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {\left (x^2 \log ^2\left (\frac {12 \left (x+e^2\right )}{x}\right )+x-\log (3)\right ) \left (\left (x+e^2\right ) \log ^2\left (\frac {12 \left (x+e^2\right )}{x}\right ) x^2-2 e^2 \log \left (\frac {12 \left (x+e^2\right )}{x}\right ) x^2+\left (x+e^2\right ) \log (3)\right )}{x^3 \left (x+e^2\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 2 \int \left (x \log ^4\left (12+\frac {12 e^2}{x}\right )+\frac {2 e^2 x \log ^3\left (12+\frac {12 e^2}{x}\right )}{-x-e^2}+\log ^2\left (12+\frac {12 e^2}{x}\right )+\frac {2 e^2 (\log (3)-x) \log \left (12+\frac {12 e^2}{x}\right )}{x \left (x+e^2\right )}+\frac {(x-\log (3)) \log (3)}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \left (2 e^4 \int \frac {\log ^3\left (12+\frac {12 e^2}{x}\right )}{x+e^2}dx+2 e^2 \operatorname {PolyLog}\left (2,-\frac {e^2}{x}\right )-12 e^4 \operatorname {PolyLog}\left (4,\frac {1}{1+\frac {e^2}{x}}\right )-6 e^4 \operatorname {PolyLog}\left (2,\frac {1}{1+\frac {e^2}{x}}\right ) \log ^2\left (\frac {12 \left (x+e^2\right )}{x}\right )-12 e^4 \operatorname {PolyLog}\left (3,\frac {1}{1+\frac {e^2}{x}}\right ) \log \left (\frac {12 \left (x+e^2\right )}{x}\right )+2 \left (e^2+\log (3)\right ) \operatorname {PolyLog}\left (2,-\frac {x}{e^2}\right )-2 \log (3) \operatorname {PolyLog}\left (2,-\frac {x}{e^2}\right )+\frac {1}{2} x^2 \log ^4\left (\frac {12 \left (x+e^2\right )}{x}\right )+\frac {(x-\log (3))^2}{2 x^2}+2 e^2 \left (\frac {e^2}{x}+1\right ) x \log ^3\left (\frac {12 \left (x+e^2\right )}{x}\right )+2 e^4 \log \left (1-\frac {1}{\frac {e^2}{x}+1}\right ) \log ^3\left (\frac {12 \left (x+e^2\right )}{x}\right )-2 e^2 \left (x+e^2\right ) \log ^3\left (\frac {12 e^2}{x}+12\right )+\left (x+e^2\right ) \log ^2\left (\frac {12 e^2}{x}+12\right )-\log (3) \log ^2(x)-\left (e^2+\log (3)\right ) \log ^2\left (12 \left (x+e^2\right )\right )+2 \log (3) (2+\log (12)) \log (x)+2 e^2 \log (12) \log (x)+2 \log (3) \log (x) \left (\log (x)+\log \left (\frac {12 \left (x+e^2\right )}{x}\right )-\log \left (12 x+12 e^2\right )\right )-2 \left (e^2+\log (3)\right ) \log \left (x+e^2\right ) \left (\log (x)+\log \left (\frac {12 \left (x+e^2\right )}{x}\right )-\log \left (12 x+12 e^2\right )\right )+2 \left (e^2+\log (3)\right ) \log (x) \log \left (\frac {x}{e^2}+1\right )\right )\) |
Int[((2*E^2*x + 2*x^2)*Log[3] + (-2*E^2 - 2*x)*Log[3]^2 + (-4*E^2*x^3 + 4* E^2*x^2*Log[3])*Log[(12*E^2 + 12*x)/x] + (2*E^2*x^3 + 2*x^4)*Log[(12*E^2 + 12*x)/x]^2 - 4*E^2*x^4*Log[(12*E^2 + 12*x)/x]^3 + (2*E^2*x^4 + 2*x^5)*Log [(12*E^2 + 12*x)/x]^4)/(E^2*x^3 + x^4),x]
3.24.100.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(66\) vs. \(2(28)=56\).
Time = 4.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31
| method | result | size |
| parallelrisch | \(\frac {\ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{4} x^{4}-2 \ln \left (3\right ) x^{2} \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{2}+2 \ln \left (\frac {12 \,{\mathrm e}^{2}+12 x}{x}\right )^{2} x^{3}+\ln \left (3\right )^{2}-2 x \ln \left (3\right )}{x^{2}}\) | \(67\) |
int(((2*x^4*exp(2)+2*x^5)*ln((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*ln((12*exp (2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*ln((12*exp(2)+12*x)/x)^2+(4*x^2*exp(2) *ln(3)-4*x^3*exp(2))*ln((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*ln(3)^2+(2*exp (2)*x+2*x^2)*ln(3))/(x^3*exp(2)+x^4),x,method=_RETURNVERBOSE)
(ln(12*(x+exp(2))/x)^4*x^4-2*ln(3)*x^2*ln(12*(x+exp(2))/x)^2+2*ln(12*(x+ex p(2))/x)^2*x^3+ln(3)^2-2*x*ln(3))/x^2
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=\frac {x^{4} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{4} + 2 \, {\left (x^{3} - x^{2} \log \left (3\right )\right )} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} - 2 \, x \log \left (3\right ) + \log \left (3\right )^{2}}{x^{2}} \]
integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log ((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x ^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*log (3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm=\
(x^4*log(12*(x + e^2)/x)^4 + 2*(x^3 - x^2*log(3))*log(12*(x + e^2)/x)^2 - 2*x*log(3) + log(3)^2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=x^{2} \log {\left (\frac {12 x + 12 e^{2}}{x} \right )}^{4} + \left (2 x - 2 \log {\left (3 \right )}\right ) \log {\left (\frac {12 x + 12 e^{2}}{x} \right )}^{2} + \frac {- 2 x \log {\left (3 \right )} + \log {\left (3 \right )}^{2}}{x^{2}} \]
integrate(((2*x**4*exp(2)+2*x**5)*ln((12*exp(2)+12*x)/x)**4-4*x**4*exp(2)* ln((12*exp(2)+12*x)/x)**3+(2*x**3*exp(2)+2*x**4)*ln((12*exp(2)+12*x)/x)**2 +(4*x**2*exp(2)*ln(3)-4*x**3*exp(2))*ln((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x )*ln(3)**2+(2*exp(2)*x+2*x**2)*ln(3))/(x**3*exp(2)+x**4),x)
x**2*log((12*x + 12*exp(2))/x)**4 + (2*x - 2*log(3))*log((12*x + 12*exp(2) )/x)**2 + (-2*x*log(3) + log(3)**2)/x**2
Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (28) = 56\).
Time = 0.34 (sec) , antiderivative size = 528, normalized size of antiderivative = 18.21 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx =\text {Too large to display} \]
integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log ((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x ^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*log (3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm=\
x^2*log(x + e^2)^4 - 4*x^2*(log(3) + 2*log(2))*log(x)^3 + x^2*log(x)^4 + ( 2*e^(-6)*log(x + e^2) - 2*e^(-6)*log(x) - (2*x - e^2)*e^(-4)/x^2)*e^2*log( 3)^2 + 4*(x^2*(log(3) + 2*log(2)) - x^2*log(x))*log(x + e^2)^3 - 4*(e^(-2) *log(x + e^2) - e^(-2)*log(x))*e^2*log(3)*log(12*e^2/x + 12) + (log(3)^4 + 8*log(3)^3*log(2) + 24*log(3)^2*log(2)^2 + 32*log(3)*log(2)^3 + 16*log(2) ^4)*x^2 + 2*(e^(-4)*log(x + e^2) - e^(-4)*log(x) - e^(-2)/x)*e^2*log(3) - 2*(e^(-4)*log(x + e^2) - e^(-4)*log(x) - e^(-2)/x)*log(3)^2 - 2*(6*x^2*(lo g(3) + 2*log(2))*log(x) - 3*x^2*log(x)^2 - 3*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2)*x^2 - x)*log(x + e^2)^2 + 2*(3*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2)*x^2 + x)*log(x)^2 + 2*(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 )*x - 2*(e^(-2)*log(x + e^2) - e^(-2)*log(x))*log(3) + 2*(log(x + e^2)^2 - 2*log(x + e^2)*log(x) + log(x)^2)*log(3) + 4*(3*x^2*(log(3) + 2*log(2))*l og(x)^2 - x^2*log(x)^3 + (log(3)^3 + 6*log(3)^2*log(2) + 12*log(3)*log(2)^ 2 + 8*log(2)^3)*x^2 + x*(log(3) + 2*log(2)) - (3*(log(3)^2 + 4*log(3)*log( 2) + 4*log(2)^2)*x^2 + x)*log(x))*log(x + e^2) - 4*((log(3)^3 + 6*log(3)^2 *log(2) + 12*log(3)*log(2)^2 + 8*log(2)^3)*x^2 + x*(log(3) + 2*log(2)))*lo g(x)
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (28) = 56\).
Time = 0.39 (sec) , antiderivative size = 256, normalized size of antiderivative = 8.83 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=\frac {{\left (e^{8} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{4} - \frac {2 \, {\left (x + e^{2}\right )}^{2} e^{4} \log \left (3\right ) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x^{2}} + \frac {4 \, {\left (x + e^{2}\right )} e^{4} \log \left (3\right ) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x} - 2 \, e^{4} \log \left (3\right ) \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} + \frac {2 \, {\left (x + e^{2}\right )} e^{6} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2}}{x} - 2 \, e^{6} \log \left (\frac {12 \, {\left (x + e^{2}\right )}}{x}\right )^{2} - \frac {2 \, {\left (x + e^{2}\right )}^{3} e^{2} \log \left (3\right )}{x^{3}} + \frac {4 \, {\left (x + e^{2}\right )}^{2} e^{2} \log \left (3\right )}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} e^{2} \log \left (3\right )}{x} + \frac {{\left (x + e^{2}\right )}^{4} \log \left (3\right )^{2}}{x^{4}} - \frac {4 \, {\left (x + e^{2}\right )}^{3} \log \left (3\right )^{2}}{x^{3}} + \frac {5 \, {\left (x + e^{2}\right )}^{2} \log \left (3\right )^{2}}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} \log \left (3\right )^{2}}{x}\right )} e^{\left (-2\right )}}{\frac {{\left (x + e^{2}\right )}^{2} e^{2}}{x^{2}} - \frac {2 \, {\left (x + e^{2}\right )} e^{2}}{x} + e^{2}} \]
integrate(((2*x^4*exp(2)+2*x^5)*log((12*exp(2)+12*x)/x)^4-4*x^4*exp(2)*log ((12*exp(2)+12*x)/x)^3+(2*x^3*exp(2)+2*x^4)*log((12*exp(2)+12*x)/x)^2+(4*x ^2*exp(2)*log(3)-4*x^3*exp(2))*log((12*exp(2)+12*x)/x)+(-2*exp(2)-2*x)*log (3)^2+(2*exp(2)*x+2*x^2)*log(3))/(x^3*exp(2)+x^4),x, algorithm=\
(e^8*log(12*(x + e^2)/x)^4 - 2*(x + e^2)^2*e^4*log(3)*log(12*(x + e^2)/x)^ 2/x^2 + 4*(x + e^2)*e^4*log(3)*log(12*(x + e^2)/x)^2/x - 2*e^4*log(3)*log( 12*(x + e^2)/x)^2 + 2*(x + e^2)*e^6*log(12*(x + e^2)/x)^2/x - 2*e^6*log(12 *(x + e^2)/x)^2 - 2*(x + e^2)^3*e^2*log(3)/x^3 + 4*(x + e^2)^2*e^2*log(3)/ x^2 - 2*(x + e^2)*e^2*log(3)/x + (x + e^2)^4*log(3)^2/x^4 - 4*(x + e^2)^3* log(3)^2/x^3 + 5*(x + e^2)^2*log(3)^2/x^2 - 2*(x + e^2)*log(3)^2/x)*e^(-2) /((x + e^2)^2*e^2/x^2 - 2*(x + e^2)*e^2/x + e^2)
Time = 14.82 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {\left (2 e^2 x+2 x^2\right ) \log (3)+\left (-2 e^2-2 x\right ) \log ^2(3)+\left (-4 e^2 x^3+4 e^2 x^2 \log (3)\right ) \log \left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^3+2 x^4\right ) \log ^2\left (\frac {12 e^2+12 x}{x}\right )-4 e^2 x^4 \log ^3\left (\frac {12 e^2+12 x}{x}\right )+\left (2 e^2 x^4+2 x^5\right ) \log ^4\left (\frac {12 e^2+12 x}{x}\right )}{e^2 x^3+x^4} \, dx=-\frac {\left (\ln \left (3\right )-x^2\,{\ln \left (\frac {12\,x+12\,{\mathrm {e}}^2}{x}\right )}^2\right )\,\left (x^2\,{\ln \left (\frac {12\,x+12\,{\mathrm {e}}^2}{x}\right )}^2+2\,x-\ln \left (3\right )\right )}{x^2} \]
int(-(log((12*x + 12*exp(2))/x)*(4*x^3*exp(2) - 4*x^2*exp(2)*log(3)) + log (3)^2*(2*x + 2*exp(2)) - log((12*x + 12*exp(2))/x)^2*(2*x^3*exp(2) + 2*x^4 ) - log((12*x + 12*exp(2))/x)^4*(2*x^4*exp(2) + 2*x^5) - log(3)*(2*x*exp(2 ) + 2*x^2) + 4*x^4*exp(2)*log((12*x + 12*exp(2))/x)^3)/(x^3*exp(2) + x^4), x)