Integrand size = 284, antiderivative size = 29 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \]
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \]
Integrate[(-10*x - 3*x^2 + x^3 + E^4*(5*x - x^2) + (-10 + E^4*(5 - x) - 3* x + x^2)*Log[3] + (15*x + 14*x^2 - 2*x^3 + E^4*(-10*x + x^2) + (20 + E^4*( -10 + x) + 13*x - 2*x^2)*Log[3])*Log[x] + (5*x - x^2 + (5 - x)*Log[3] + (- 10*x + x^2 + (-10 + x)*Log[3])*Log[x])*Log[x + Log[3]])/(16*x^4 + 4*E^8*x^ 4 + 16*x^5 + 4*x^6 + E^4*(-16*x^4 - 8*x^5) + (16*x^3 + 4*E^8*x^3 + 16*x^4 + 4*x^5 + E^4*(-16*x^3 - 8*x^4))*Log[3] + (-16*x^4 + 8*E^4*x^4 - 8*x^5 + ( -16*x^3 + 8*E^4*x^3 - 8*x^4)*Log[3])*Log[x + Log[3]] + (4*x^4 + 4*x^3*Log[ 3])*Log[x + Log[3]]^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.
| \(\displaystyle \int \frac {x^3-3 x^2+e^4 \left (5 x-x^2\right )+\left (-x^2+\left (x^2-10 x+(x-10) \log (3)\right ) \log (x)+5 x+(5-x) \log (3)\right ) \log (x+\log (3))+\left (x^2-3 x+e^4 (5-x)-10\right ) \log (3)+\left (-2 x^3+14 x^2+e^4 \left (x^2-10 x\right )+\left (-2 x^2+13 x+e^4 (x-10)+20\right ) \log (3)+15 x\right ) \log (x)-10 x}{4 x^6+16 x^5+4 e^8 x^4+16 x^4+e^4 \left (-8 x^5-16 x^4\right )+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))+\left (-8 x^5+8 e^4 x^4-16 x^4+\left (-8 x^4+8 e^4 x^3-16 x^3\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^5+16 x^4+4 e^8 x^3+16 x^3+e^4 \left (-8 x^4-16 x^3\right )\right ) \log (3)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x^3-3 x^2+e^4 \left (5 x-x^2\right )+\left (-x^2+\left (x^2-10 x+(x-10) \log (3)\right ) \log (x)+5 x+(5-x) \log (3)\right ) \log (x+\log (3))+\left (x^2-3 x+e^4 (5-x)-10\right ) \log (3)+\left (-2 x^3+14 x^2+e^4 \left (x^2-10 x\right )+\left (-2 x^2+13 x+e^4 (x-10)+20\right ) \log (3)+15 x\right ) \log (x)-10 x}{4 x^6+16 x^5+\left (16+4 e^8\right ) x^4+e^4 \left (-8 x^5-16 x^4\right )+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))+\left (-8 x^5+8 e^4 x^4-16 x^4+\left (-8 x^4+8 e^4 x^3-16 x^3\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^5+16 x^4+4 e^8 x^3+16 x^3+e^4 \left (-8 x^4-16 x^3\right )\right ) \log (3)}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\log (x) \left (-2 x^3+x^2 \left (14+e^4-\log (9)\right )+x \left (15+e^4 (\log (3)-10)+13 \log (3)\right )+(x-10) (x+\log (3)) \log (x+\log (3))-10 \left (e^4-2\right ) \log (3)\right )+(x-5) (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )}{4 x^3 (x+\log (3)) \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int -\frac {(5-x) (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )+\log (x) \left (2 x^3-\left (14+e^4-\log (9)\right ) x^2-\left (15-e^4 (10-\log (3))+13 \log (3)\right ) x+(10-x) (x+\log (3)) \log (x+\log (3))-10 \left (2-e^4\right ) \log (3)\right )}{x^3 (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{4} \int \frac {(5-x) (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )+\log (x) \left (2 x^3-\left (14+e^4-\log (9)\right ) x^2-\left (15-e^4 (10-\log (3))+13 \log (3)\right ) x+(10-x) (x+\log (3)) \log (x+\log (3))-10 \left (2-e^4\right ) \log (3)\right )}{x^3 (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {1}{4} \int \frac {(5-x) (x+\log (3)) \left (x-\log (x+\log (3))-e^4+2\right )+\log (x) \left (2 x^3-\left (14+e^4-\log (9)\right ) x^2-\left (15-e^4 (10-\log (3))+13 \log (3)\right ) x+(10-x) (x+\log (3)) \log (x+\log (3))-10 \left (2-e^4\right ) \log (3)\right )}{x^3 (x+\log (3)) \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{4} \int \left (\frac {(5-x) (-x-\log (3)+1) \log (x)}{x^2 (x+\log (3)) \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}+\frac {\log (x) x-x-10 \log (x)+5}{x^3 \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (-5 \int \frac {1}{x^3 \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )}dx-10 \int \frac {\log (x)}{x^3 \left (-x+\log (x+\log (3))-2 \left (1-\frac {e^4}{2}\right )\right )}dx-\frac {5 (1-\log (3)) \int \frac {\log (x)}{x^2 \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}dx}{\log (3)}-\int \frac {\log (x)}{x^2 \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )}dx-\int \frac {1}{x^2 \left (-x+\log (x+\log (3))-2 \left (1-\frac {e^4}{2}\right )\right )}dx+\frac {\left (5-\log ^2(3)+\log (3)\right ) \int \frac {\log (x)}{x \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}dx}{\log ^2(3)}-\frac {(5+\log (3)) \int \frac {\log (x)}{(x+\log (3)) \left (x-\log (x+\log (3))+2 \left (1-\frac {e^4}{2}\right )\right )^2}dx}{\log ^2(3)}\right )\) |
Int[(-10*x - 3*x^2 + x^3 + E^4*(5*x - x^2) + (-10 + E^4*(5 - x) - 3*x + x^ 2)*Log[3] + (15*x + 14*x^2 - 2*x^3 + E^4*(-10*x + x^2) + (20 + E^4*(-10 + x) + 13*x - 2*x^2)*Log[3])*Log[x] + (5*x - x^2 + (5 - x)*Log[3] + (-10*x + x^2 + (-10 + x)*Log[3])*Log[x])*Log[x + Log[3]])/(16*x^4 + 4*E^8*x^4 + 16 *x^5 + 4*x^6 + E^4*(-16*x^4 - 8*x^5) + (16*x^3 + 4*E^8*x^3 + 16*x^4 + 4*x^ 5 + E^4*(-16*x^3 - 8*x^4))*Log[3] + (-16*x^4 + 8*E^4*x^4 - 8*x^5 + (-16*x^ 3 + 8*E^4*x^3 - 8*x^4)*Log[3])*Log[x + Log[3]] + (4*x^4 + 4*x^3*Log[3])*Lo g[x + Log[3]]^2),x]
3.27.3.3.1 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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 5.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
| method | result | size |
| risch | \(-\frac {\left (-5+x \right ) \ln \left (x \right )}{4 x^{2} \left ({\mathrm e}^{4}+\ln \left (\ln \left (3\right )+x \right )-x -2\right )}\) | \(25\) |
| parallelrisch | \(\frac {-x \ln \left (x \right )+5 \ln \left (x \right )}{4 x^{2} \left ({\mathrm e}^{4}+\ln \left (\ln \left (3\right )+x \right )-x -2\right )}\) | \(30\) |
int(((((x-10)*ln(3)+x^2-10*x)*ln(x)+(5-x)*ln(3)-x^2+5*x)*ln(ln(3)+x)+(((x- 10)*exp(4)-2*x^2+13*x+20)*ln(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2+15*x)*ln(x) +((5-x)*exp(4)+x^2-3*x-10)*ln(3)+(-x^2+5*x)*exp(4)+x^3-3*x^2-10*x)/((4*x^3 *ln(3)+4*x^4)*ln(ln(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)*ln(3)+8*x^4*exp(4 )-8*x^5-16*x^4)*ln(ln(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16*x^3)*exp(4)+4*x^5+1 6*x^4+16*x^3)*ln(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4)*exp(4)+4*x^6+16*x^5+16* x^4),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {{\left (x - 5\right )} \log \left (x\right )}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \left (3\right )\right ) + 2 \, x^{2}\right )}} \]
integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log( 3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2 +15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x^2 -10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)* log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16 *x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4)*ex p(4)+4*x^6+16*x^5+16*x^4),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {- x \log {\left (x \right )} + 5 \log {\left (x \right )}}{- 4 x^{3} + 4 x^{2} \log {\left (x + \log {\left (3 \right )} \right )} - 8 x^{2} + 4 x^{2} e^{4}} \]
integrate(((((x-10)*ln(3)+x**2-10*x)*ln(x)+(5-x)*ln(3)-x**2+5*x)*ln(ln(3)+ x)+(((x-10)*exp(4)-2*x**2+13*x+20)*ln(3)+(x**2-10*x)*exp(4)-2*x**3+14*x**2 +15*x)*ln(x)+((5-x)*exp(4)+x**2-3*x-10)*ln(3)+(-x**2+5*x)*exp(4)+x**3-3*x* *2-10*x)/((4*x**3*ln(3)+4*x**4)*ln(ln(3)+x)**2+((8*x**3*exp(4)-8*x**4-16*x **3)*ln(3)+8*x**4*exp(4)-8*x**5-16*x**4)*ln(ln(3)+x)+(4*x**3*exp(4)**2+(-8 *x**4-16*x**3)*exp(4)+4*x**5+16*x**4+16*x**3)*ln(3)+4*x**4*exp(4)**2+(-8*x **5-16*x**4)*exp(4)+4*x**6+16*x**5+16*x**4),x)
Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {{\left (x - 5\right )} \log \left (x\right )}{4 \, {\left (x^{3} - x^{2} {\left (e^{4} - 2\right )} - x^{2} \log \left (x + \log \left (3\right )\right )\right )}} \]
integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log( 3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2 +15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x^2 -10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)* log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16 *x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4)*ex p(4)+4*x^6+16*x^5+16*x^4),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\frac {x \log \left (x\right ) - 5 \, \log \left (x\right )}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \left (3\right )\right ) + 2 \, x^{2}\right )}} \]
integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log( 3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2 +15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x^2 -10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)* log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16 *x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4)*ex p(4)+4*x^6+16*x^5+16*x^4),x, algorithm=\
Timed out. \[ \int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx=\int \frac {\ln \left (x+\ln \left (3\right )\right )\,\left (5\,x-\ln \left (3\right )\,\left (x-5\right )-x^2+\ln \left (x\right )\,\left (\ln \left (3\right )\,\left (x-10\right )-10\,x+x^2\right )\right )-10\,x+{\mathrm {e}}^4\,\left (5\,x-x^2\right )-\ln \left (3\right )\,\left (3\,x+{\mathrm {e}}^4\,\left (x-5\right )-x^2+10\right )-3\,x^2+x^3+\ln \left (x\right )\,\left (15\,x-{\mathrm {e}}^4\,\left (10\,x-x^2\right )+\ln \left (3\right )\,\left (13\,x+{\mathrm {e}}^4\,\left (x-10\right )-2\,x^2+20\right )+14\,x^2-2\,x^3\right )}{4\,x^4\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^5+16\,x^4\right )+{\ln \left (x+\ln \left (3\right )\right )}^2\,\left (4\,x^4+4\,\ln \left (3\right )\,x^3\right )-\ln \left (x+\ln \left (3\right )\right )\,\left (\ln \left (3\right )\,\left (16\,x^3-8\,x^3\,{\mathrm {e}}^4+8\,x^4\right )-8\,x^4\,{\mathrm {e}}^4+16\,x^4+8\,x^5\right )+16\,x^4+16\,x^5+4\,x^6+\ln \left (3\right )\,\left (4\,x^3\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^4+16\,x^3\right )+16\,x^3+16\,x^4+4\,x^5\right )} \,d x \]
int((log(x + log(3))*(5*x - log(3)*(x - 5) - x^2 + log(x)*(log(3)*(x - 10) - 10*x + x^2)) - 10*x + exp(4)*(5*x - x^2) - log(3)*(3*x + exp(4)*(x - 5) - x^2 + 10) - 3*x^2 + x^3 + log(x)*(15*x - exp(4)*(10*x - x^2) + log(3)*( 13*x + exp(4)*(x - 10) - 2*x^2 + 20) + 14*x^2 - 2*x^3))/(4*x^4*exp(8) - ex p(4)*(16*x^4 + 8*x^5) + log(x + log(3))^2*(4*x^3*log(3) + 4*x^4) - log(x + log(3))*(log(3)*(16*x^3 - 8*x^3*exp(4) + 8*x^4) - 8*x^4*exp(4) + 16*x^4 + 8*x^5) + 16*x^4 + 16*x^5 + 4*x^6 + log(3)*(4*x^3*exp(8) - exp(4)*(16*x^3 + 8*x^4) + 16*x^3 + 16*x^4 + 4*x^5)),x)
int((log(x + log(3))*(5*x - log(3)*(x - 5) - x^2 + log(x)*(log(3)*(x - 10) - 10*x + x^2)) - 10*x + exp(4)*(5*x - x^2) - log(3)*(3*x + exp(4)*(x - 5) - x^2 + 10) - 3*x^2 + x^3 + log(x)*(15*x - exp(4)*(10*x - x^2) + log(3)*( 13*x + exp(4)*(x - 10) - 2*x^2 + 20) + 14*x^2 - 2*x^3))/(4*x^4*exp(8) - ex p(4)*(16*x^4 + 8*x^5) + log(x + log(3))^2*(4*x^3*log(3) + 4*x^4) - log(x + log(3))*(log(3)*(16*x^3 - 8*x^3*exp(4) + 8*x^4) - 8*x^4*exp(4) + 16*x^4 + 8*x^5) + 16*x^4 + 16*x^5 + 4*x^6 + log(3)*(4*x^3*exp(8) - exp(4)*(16*x^3 + 8*x^4) + 16*x^3 + 16*x^4 + 4*x^5)), x)