Integrand size = 233, antiderivative size = 30 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=\frac {1+e^{e^{12}}-x}{-\frac {2}{e^4}+3 x+\log (-3+x)-\log (x)} \]
Time = 0.13 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=-\frac {e^4 \left (1+e^{e^{12}}-x\right )}{2-3 e^4 x-e^4 \log (-3+x)+e^4 \log (x)} \]
Integrate[(E^(8 + E^12)*(-3 + 9*x - 3*x^2) + E^8*(-3 + 12*x - 3*x^2) + E^4 *(-6*x + 2*x^2) + E^8*(3*x - x^2)*Log[-3 + x] + E^8*(-3*x + x^2)*Log[x])/( -12*x + 4*x^2 + E^4*(36*x^2 - 12*x^3) + E^8*(-27*x^3 + 9*x^4) + (E^4*(12*x - 4*x^2) + E^8*(-18*x^2 + 6*x^3))*Log[-3 + x] + E^8*(-3*x + x^2)*Log[-3 + x]^2 + (E^4*(-12*x + 4*x^2) + E^8*(18*x^2 - 6*x^3) + E^8*(6*x - 2*x^2)*Lo g[-3 + x])*Log[x] + E^8*(-3*x + x^2)*Log[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.
| \(\displaystyle \int \frac {e^{8+e^{12}} \left (-3 x^2+9 x-3\right )+e^8 \left (-3 x^2+12 x-3\right )+e^4 \left (2 x^2-6 x\right )+e^8 \left (3 x-x^2\right ) \log (x-3)+e^8 \left (x^2-3 x\right ) \log (x)}{4 x^2+e^8 \left (x^2-3 x\right ) \log ^2(x-3)+e^8 \left (x^2-3 x\right ) \log ^2(x)+e^8 \left (9 x^4-27 x^3\right )+e^4 \left (36 x^2-12 x^3\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (6 x^3-18 x^2\right )\right ) \log (x-3)+\left (e^4 \left (4 x^2-12 x\right )+e^8 \left (6 x-2 x^2\right ) \log (x-3)+e^8 \left (18 x^2-6 x^3\right )\right ) \log (x)-12 x} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^4 \left (-2 \left (1-\frac {3}{2} e^4 \left (1+e^{e^{12}}\right )\right ) x^2+6 \left (1-\frac {1}{2} e^4 \left (4+3 e^{e^{12}}\right )\right ) x+e^4 (x-3) x \log (x-3)-e^4 (x-3) x \log (x)+3 e^4 \left (1+e^{e^{12}}\right )\right )}{(3-x) x \left (-3 e^4 x-e^4 \log (x-3)+e^4 \log (x)+2\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^4 \int \frac {-\left (\left (2-3 e^4 \left (1+e^{e^{12}}\right )\right ) x^2\right )-e^4 (3-x) \log (x-3) x+e^4 (3-x) \log (x) x+3 \left (2-e^4 \left (4+3 e^{e^{12}}\right )\right ) x+3 e^4 \left (1+e^{e^{12}}\right )}{(3-x) x \left (-3 e^4 x-e^4 \log (x-3)+e^4 \log (x)+2\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle e^4 \int \left (\frac {3 e^4 \left (x-e^{e^{12}}-1\right ) \left (x^2-3 x+1\right )}{(x-3) x \left (3 e^4 x+e^4 \log (x-3)-e^4 \log (x)-2\right )^2}+\frac {1}{-3 e^4 x-e^4 \log (x-3)+e^4 \log (x)+2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^4 \left (-3 e^4 \left (1+e^{e^{12}}\right ) \int \frac {1}{\left (3 e^4 x+e^4 \log (x-3)-e^4 \log (x)-2\right )^2}dx+e^4 \left (2-e^{e^{12}}\right ) \int \frac {1}{(x-3) \left (3 e^4 x+e^4 \log (x-3)-e^4 \log (x)-2\right )^2}dx+e^4 \left (1+e^{e^{12}}\right ) \int \frac {1}{x \left (3 e^4 x+e^4 \log (x-3)-e^4 \log (x)-2\right )^2}dx+3 e^4 \int \frac {x}{\left (3 e^4 x+e^4 \log (x-3)-e^4 \log (x)-2\right )^2}dx+\int \frac {1}{-3 e^4 x-e^4 \log (x-3)+e^4 \log (x)+2}dx\right )\) |
Int[(E^(8 + E^12)*(-3 + 9*x - 3*x^2) + E^8*(-3 + 12*x - 3*x^2) + E^4*(-6*x + 2*x^2) + E^8*(3*x - x^2)*Log[-3 + x] + E^8*(-3*x + x^2)*Log[x])/(-12*x + 4*x^2 + E^4*(36*x^2 - 12*x^3) + E^8*(-27*x^3 + 9*x^4) + (E^4*(12*x - 4*x ^2) + E^8*(-18*x^2 + 6*x^3))*Log[-3 + x] + E^8*(-3*x + x^2)*Log[-3 + x]^2 + (E^4*(-12*x + 4*x^2) + E^8*(18*x^2 - 6*x^3) + E^8*(6*x - 2*x^2)*Log[-3 + x])*Log[x] + E^8*(-3*x + x^2)*Log[x]^2),x]
3.1.90.3.1 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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
| method | result | size |
| risch | \(\frac {\left ({\mathrm e}^{{\mathrm e}^{12}}-x +1\right ) {\mathrm e}^{4}}{{\mathrm e}^{4} \ln \left (-3+x \right )-{\mathrm e}^{4} \ln \left (x \right )+3 x \,{\mathrm e}^{4}-2}\) | \(34\) |
| default | \(-\frac {{\mathrm e}^{4} \left ({\mathrm e}^{{\mathrm e}^{12}}-x +1\right )}{{\mathrm e}^{4} \ln \left (x \right )-3 \,{\mathrm e}^{\ln \left (x \right )+4}-{\mathrm e}^{4} \ln \left (-3+x \right )+2}\) | \(37\) |
| parallelrisch | \(\frac {15 \,{\mathrm e}^{8} \ln \left (-3+x \right )+18 \,{\mathrm e}^{{\mathrm e}^{12}} {\mathrm e}^{8} x -6 \,{\mathrm e}^{{\mathrm e}^{12}} {\mathrm e}^{8} \ln \left (x \right )+6 \,{\mathrm e}^{{\mathrm e}^{12}} {\mathrm e}^{8} \ln \left (-3+x \right )+18 x \,{\mathrm e}^{8}-15 \,{\mathrm e}^{8} \ln \left (x \right )-12 x \,{\mathrm e}^{4}-18 \,{\mathrm e}^{4}+27 \,{\mathrm e}^{8}+27 \,{\mathrm e}^{8} {\mathrm e}^{{\mathrm e}^{12}}}{3 \left ({\mathrm e}^{4} \ln \left (-3+x \right )-{\mathrm e}^{4} \ln \left (x \right )+3 x \,{\mathrm e}^{4}-2\right ) \left (9 \,{\mathrm e}^{4}+4\right )}\) | \(117\) |
int(((x^2-3*x)*exp(4)^2*ln(x)+(-x^2+3*x)*exp(4)^2*ln(-3+x)+(-3*x^2+9*x-3)* exp(4)^2*exp(exp(12))+(-3*x^2+12*x-3)*exp(4)^2+(2*x^2-6*x)*exp(4))/((x^2-3 *x)*exp(4)^2*ln(x)^2+((-2*x^2+6*x)*exp(4)^2*ln(-3+x)+(-6*x^3+18*x^2)*exp(4 )^2+(4*x^2-12*x)*exp(4))*ln(x)+(x^2-3*x)*exp(4)^2*ln(-3+x)^2+((6*x^3-18*x^ 2)*exp(4)^2+(-4*x^2+12*x)*exp(4))*ln(-3+x)+(9*x^4-27*x^3)*exp(4)^2+(-12*x^ 3+36*x^2)*exp(4)+4*x^2-12*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=-\frac {{\left (x - 1\right )} e^{8} - e^{\left (e^{12} + 8\right )}}{3 \, x e^{8} + e^{8} \log \left (x - 3\right ) - e^{8} \log \left (x\right ) - 2 \, e^{4}} \]
integrate(((x^2-3*x)*exp(4)^2*log(x)+(-x^2+3*x)*exp(4)^2*log(-3+x)+(-3*x^2 +9*x-3)*exp(4)^2*exp(exp(12))+(-3*x^2+12*x-3)*exp(4)^2+(2*x^2-6*x)*exp(4)) /((x^2-3*x)*exp(4)^2*log(x)^2+((-2*x^2+6*x)*exp(4)^2*log(-3+x)+(-6*x^3+18* x^2)*exp(4)^2+(4*x^2-12*x)*exp(4))*log(x)+(x^2-3*x)*exp(4)^2*log(-3+x)^2+( (6*x^3-18*x^2)*exp(4)^2+(-4*x^2+12*x)*exp(4))*log(-3+x)+(9*x^4-27*x^3)*exp (4)^2+(-12*x^3+36*x^2)*exp(4)+4*x^2-12*x),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=\frac {- x e^{4} + e^{4} + e^{4} e^{e^{12}}}{3 x e^{4} - e^{4} \log {\left (x \right )} + e^{4} \log {\left (x - 3 \right )} - 2} \]
integrate(((x**2-3*x)*exp(4)**2*ln(x)+(-x**2+3*x)*exp(4)**2*ln(-3+x)+(-3*x **2+9*x-3)*exp(4)**2*exp(exp(12))+(-3*x**2+12*x-3)*exp(4)**2+(2*x**2-6*x)* exp(4))/((x**2-3*x)*exp(4)**2*ln(x)**2+((-2*x**2+6*x)*exp(4)**2*ln(-3+x)+( -6*x**3+18*x**2)*exp(4)**2+(4*x**2-12*x)*exp(4))*ln(x)+(x**2-3*x)*exp(4)** 2*ln(-3+x)**2+((6*x**3-18*x**2)*exp(4)**2+(-4*x**2+12*x)*exp(4))*ln(-3+x)+ (9*x**4-27*x**3)*exp(4)**2+(-12*x**3+36*x**2)*exp(4)+4*x**2-12*x),x)
(-x*exp(4) + exp(4) + exp(4)*exp(exp(12)))/(3*x*exp(4) - exp(4)*log(x) + e xp(4)*log(x - 3) - 2)
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=-\frac {x e^{4} - {\left (e^{\left (e^{12}\right )} + 1\right )} e^{4}}{3 \, x e^{4} + e^{4} \log \left (x - 3\right ) - e^{4} \log \left (x\right ) - 2} \]
integrate(((x^2-3*x)*exp(4)^2*log(x)+(-x^2+3*x)*exp(4)^2*log(-3+x)+(-3*x^2 +9*x-3)*exp(4)^2*exp(exp(12))+(-3*x^2+12*x-3)*exp(4)^2+(2*x^2-6*x)*exp(4)) /((x^2-3*x)*exp(4)^2*log(x)^2+((-2*x^2+6*x)*exp(4)^2*log(-3+x)+(-6*x^3+18* x^2)*exp(4)^2+(4*x^2-12*x)*exp(4))*log(x)+(x^2-3*x)*exp(4)^2*log(-3+x)^2+( (6*x^3-18*x^2)*exp(4)^2+(-4*x^2+12*x)*exp(4))*log(-3+x)+(9*x^4-27*x^3)*exp (4)^2+(-12*x^3+36*x^2)*exp(4)+4*x^2-12*x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=-\frac {x e^{8} - e^{8} - e^{\left (e^{12} + 8\right )}}{3 \, x e^{8} + e^{8} \log \left (x - 3\right ) - e^{8} \log \left (x\right ) - 2 \, e^{4}} \]
integrate(((x^2-3*x)*exp(4)^2*log(x)+(-x^2+3*x)*exp(4)^2*log(-3+x)+(-3*x^2 +9*x-3)*exp(4)^2*exp(exp(12))+(-3*x^2+12*x-3)*exp(4)^2+(2*x^2-6*x)*exp(4)) /((x^2-3*x)*exp(4)^2*log(x)^2+((-2*x^2+6*x)*exp(4)^2*log(-3+x)+(-6*x^3+18* x^2)*exp(4)^2+(4*x^2-12*x)*exp(4))*log(x)+(x^2-3*x)*exp(4)^2*log(-3+x)^2+( (6*x^3-18*x^2)*exp(4)^2+(-4*x^2+12*x)*exp(4))*log(-3+x)+(9*x^4-27*x^3)*exp (4)^2+(-12*x^3+36*x^2)*exp(4)+4*x^2-12*x),x, algorithm=\
Time = 13.67 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {e^{8+e^{12}} \left (-3+9 x-3 x^2\right )+e^8 \left (-3+12 x-3 x^2\right )+e^4 \left (-6 x+2 x^2\right )+e^8 \left (3 x-x^2\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log (x)}{-12 x+4 x^2+e^4 \left (36 x^2-12 x^3\right )+e^8 \left (-27 x^3+9 x^4\right )+\left (e^4 \left (12 x-4 x^2\right )+e^8 \left (-18 x^2+6 x^3\right )\right ) \log (-3+x)+e^8 \left (-3 x+x^2\right ) \log ^2(-3+x)+\left (e^4 \left (-12 x+4 x^2\right )+e^8 \left (18 x^2-6 x^3\right )+e^8 \left (6 x-2 x^2\right ) \log (-3+x)\right ) \log (x)+e^8 \left (-3 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-4}\,\left (3\,{\mathrm {e}}^{{\mathrm {e}}^{12}+8}-2\,{\mathrm {e}}^4+3\,{\mathrm {e}}^8+\ln \left (x-3\right )\,{\mathrm {e}}^8-{\mathrm {e}}^8\,\ln \left (x\right )\right )}{3\,\left (\ln \left (x-3\right )\,{\mathrm {e}}^4+3\,x\,{\mathrm {e}}^4-{\mathrm {e}}^4\,\ln \left (x\right )-2\right )} \]
int((exp(4)*(6*x - 2*x^2) + exp(8)*(3*x^2 - 12*x + 3) + exp(8)*log(x)*(3*x - x^2) - log(x - 3)*exp(8)*(3*x - x^2) + exp(8)*exp(exp(12))*(3*x^2 - 9*x + 3))/(12*x + exp(8)*(27*x^3 - 9*x^4) - exp(4)*(36*x^2 - 12*x^3) - log(x - 3)*(exp(4)*(12*x - 4*x^2) - exp(8)*(18*x^2 - 6*x^3)) - 4*x^2 - log(x)*(e xp(8)*(18*x^2 - 6*x^3) - exp(4)*(12*x - 4*x^2) + log(x - 3)*exp(8)*(6*x - 2*x^2)) + exp(8)*log(x)^2*(3*x - x^2) + log(x - 3)^2*exp(8)*(3*x - x^2)),x )