Integrand size = 104, antiderivative size = 33 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} e^{\frac {x^2 \left (3-e^{x^2}+x-\log (x)\right )}{3-\frac {3}{x}}} \]
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} e^{\frac {x^3 \left (3-e^{x^2}+x\right )}{3 (-1+x)}} x^{\frac {x^3}{3-3 x}} \]
Integrate[(E^((3*x^3 - E^x^2*x^3 + x^4 - x^3*Log[x])/(-3 + 3*x))*(-8*x^2 + x^3 + 3*x^4 + E^x^2*(3*x^2 - 2*x^3 + 2*x^4 - 2*x^5) + (3*x^2 - 2*x^3)*Log [x]))/(15 - 30*x + 15*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.
\(\displaystyle \int \frac {e^{\frac {x^4+3 x^3-x^3 \log (x)-e^{x^2} x^3}{3 x-3}} \left (3 x^4+x^3-8 x^2+\left (3 x^2-2 x^3\right ) \log (x)+e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )\right )}{15 x^2-30 x+15} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 60 \int -\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3-3 x}} \left (-3 x^4-x^3+8 x^2-e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )-\left (3 x^2-2 x^3\right ) \log (x)\right )}{900 (1-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{15} \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}} \left (-3 x^4-x^3+8 x^2-e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )-\left (3 x^2-2 x^3\right ) \log (x)\right )}{(1-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{15} \int \left (\frac {e^{x^2-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} \left (2 x^3-2 x^2+2 x-3\right ) x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}+\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} (2 x-3) \log (x) x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}+\frac {8 e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}-\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+3}}{(x-1)^2}-\frac {3 e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+4}}{(x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{15} \left (-2 \int e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}dx+\int \frac {e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx-4 \int \frac {e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}}{x-1}dx-2 \int e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+3}dx-\int \frac {\int \frac {e^{\frac {x^3 \left (x-e^{x^2}+3\right )}{3 (x-1)}} x^{\frac {x^3}{3-3 x}+2}}{(x-1)^2}dx}{x}dx+2 \int \frac {\int \frac {e^{\frac {x^3 \left (x-e^{x^2}+3\right )}{3 (x-1)}} x^{\frac {x^3}{3-3 x}+2}}{x-1}dx}{x}dx-8 \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx+\int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+3}}{(x-1)^2}dx+3 \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+4}}{(x-1)^2}dx+\log (x) \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx-2 \log (x) \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{x-1}dx\right )\) |
Int[(E^((3*x^3 - E^x^2*x^3 + x^4 - x^3*Log[x])/(-3 + 3*x))*(-8*x^2 + x^3 + 3*x^4 + E^x^2*(3*x^2 - 2*x^3 + 2*x^4 - 2*x^5) + (3*x^2 - 2*x^3)*Log[x]))/ (15 - 30*x + 15*x^2),x]
3.15.11.3.1 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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 6.65 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {{\mathrm e}^{-\frac {x^{3} \left (\ln \left (x \right )+{\mathrm e}^{x^{2}}-x -3\right )}{3 \left (-1+x \right )}}}{5}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {x^{3} \left (\ln \left (x \right )+{\mathrm e}^{x^{2}}-x -3\right )}{3 \left (-1+x \right )}}}{5}\) | \(25\) |
int(((-2*x^3+3*x^2)*ln(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4+x^3-8* x^2)*exp((-x^3*ln(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30*x+15),x, method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4} - x^{3} e^{\left (x^{2}\right )} - x^{3} \log \left (x\right ) + 3 \, x^{3}}{3 \, {\left (x - 1\right )}}\right )} \]
integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 +x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* x+15),x, algorithm=\
Time = 0.44 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {e^{\frac {x^{4} - x^{3} e^{x^{2}} - x^{3} \log {\left (x \right )} + 3 x^{3}}{3 x - 3}}}{5} \]
integrate(((-2*x**3+3*x**2)*ln(x)+(-2*x**5+2*x**4-2*x**3+3*x**2)*exp(x**2) +3*x**4+x**3-8*x**2)*exp((-x**3*ln(x)-x**3*exp(x**2)+x**4+3*x**3)/(-3+3*x) )/(15*x**2-30*x+15),x)
\[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\int { \frac {{\left (3 \, x^{4} + x^{3} - 8 \, x^{2} - {\left (2 \, x^{5} - 2 \, x^{4} + 2 \, x^{3} - 3 \, x^{2}\right )} e^{\left (x^{2}\right )} - {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \left (x\right )\right )} e^{\left (\frac {x^{4} - x^{3} e^{\left (x^{2}\right )} - x^{3} \log \left (x\right ) + 3 \, x^{3}}{3 \, {\left (x - 1\right )}}\right )}}{15 \, {\left (x^{2} - 2 \, x + 1\right )}} \,d x } \]
integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 +x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* x+15),x, algorithm=\
1/15*integrate((3*x^4 + x^3 - 8*x^2 - (2*x^5 - 2*x^4 + 2*x^3 - 3*x^2)*e^(x ^2) - (2*x^3 - 3*x^2)*log(x))*e^(1/3*(x^4 - x^3*e^(x^2) - x^3*log(x) + 3*x ^3)/(x - 1))/(x^2 - 2*x + 1), x)
Time = 0.82 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4}}{3 \, {\left (x - 1\right )}} - \frac {x^{3} e^{\left (x^{2}\right )}}{3 \, {\left (x - 1\right )}} - \frac {x^{3} \log \left (x\right )}{3 \, {\left (x - 1\right )}} + \frac {x^{3}}{x - 1}\right )} \]
integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 +x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* x+15),x, algorithm=\
Time = 11.64 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {x^4}{3\,x-3}}\,{\mathrm {e}}^{-\frac {x^3\,{\mathrm {e}}^{x^2}}{3\,x-3}}\,{\mathrm {e}}^{\frac {x^3}{x-1}}}{5\,x^{\frac {x^3}{3\,x-3}}} \]