Integrand size = 79, antiderivative size = 24 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{e^3 \left (5-\frac {3 \left (\log (3)+\log \left (3 x^2\right )\right )}{-1+x}\right )} \end {dmath*}
Time = 0.49 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=729^{\frac {e^3}{1-x}} e^{5 e^3} \left (x^2\right )^{-\frac {3 e^3}{-1+x}} \end {dmath*}
Integrate[(E^((E^3*(-5 + 5*x) - 3*E^3*Log[3] - 3*E^3*Log[3*x^2])/(-1 + x)) *(E^3*(6 - 6*x) + 3*E^3*x*Log[3] + 3*E^3*x*Log[3*x^2]))/(x - 2*x^2 + x^3), 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.
\(\displaystyle \int \frac {\left (3 e^3 x \log \left (3 x^2\right )+e^3 (6-6 x)+3 e^3 x \log (3)\right ) \exp \left (\frac {-3 e^3 \log \left (3 x^2\right )+e^3 (5 x-5)-3 e^3 \log (3)}{x-1}\right )}{x^3-2 x^2+x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (3 e^3 x \log \left (3 x^2\right )+e^3 (6-6 x)+3 e^3 x \log (3)\right ) \exp \left (\frac {-3 e^3 \log \left (3 x^2\right )+e^3 (5 x-5)-3 e^3 \log (3)}{x-1}\right )}{x \left (x^2-2 x+1\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {3^{1+\frac {6 e^3}{1-x}} e^{5 e^3} \left (x^2\right )^{\frac {3 e^3}{1-x}} \left (2 e^3 (1-x)+e^3 x \log \left (3 x^2\right )+e^3 x \log (3)\right )}{4 (1-x)^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{5 e^3} \int \frac {3^{1+\frac {6 e^3}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}} \left (2 e^3 (1-x)+e^3 x \log \left (3 x^2\right )+e^3 x \log (3)\right )}{(1-x)^2 x}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle e^{5 e^3} \int \frac {3^{\frac {x-6 e^3-1}{x-1}} e^3 \left (x^2\right )^{\frac {3 e^3}{1-x}} \left (\log \left (x^2\right ) x-2 (1-\log (3)) x+2\right )}{(1-x)^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{3+5 e^3} \int \frac {3^{\frac {-x+6 e^3+1}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}} \left (\log \left (x^2\right ) x-2 (1-\log (3)) x+2\right )}{(1-x)^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^{3+5 e^3} \int \left (\frac {3^{\frac {-x+6 e^3+1}{1-x}} (2-x (2-\log (9))) \left (x^2\right )^{\frac {3 e^3}{1-x}}}{(1-x)^2 x}+\frac {3^{\frac {-x+6 e^3+1}{1-x}} \log \left (x^2\right ) \left (x^2\right )^{\frac {3 e^3}{1-x}}}{(x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{3+5 e^3} \left (-2 \int \frac {3^{\frac {-x+6 e^3+1}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}}}{x-1}dx+2 \int \frac {3^{\frac {-x+6 e^3+1}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}}}{x}dx-2 \int \frac {\int \frac {3^{\frac {x-6 e^3-1}{x-1}} \left (x^2\right )^{-\frac {3 e^3}{x-1}}}{(x-1)^2}dx}{x}dx+\log \left (x^2\right ) \int \frac {3^{\frac {-x+6 e^3+1}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}}}{(x-1)^2}dx+\log (9) \int \frac {3^{\frac {-x+6 e^3+1}{1-x}} \left (x^2\right )^{\frac {3 e^3}{1-x}}}{(x-1)^2}dx\right )\) |
Int[(E^((E^3*(-5 + 5*x) - 3*E^3*Log[3] - 3*E^3*Log[3*x^2])/(-1 + x))*(E^3* (6 - 6*x) + 3*E^3*x*Log[3] + 3*E^3*x*Log[3*x^2]))/(x - 2*x^2 + x^3),x]
3.9.68.3.1 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_)*((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 = 0.94 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
risch | \({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (3 \ln \left (3 x^{2}\right )+3 \ln \left (3\right )-5 x +5\right )}{-1+x}}\) | \(28\) |
parallelrisch | \({\mathrm e}^{-\frac {{\mathrm e}^{3} \left (3 \ln \left (3 x^{2}\right )+3 \ln \left (3\right )-5 x +5\right )}{-1+x}}\) | \(28\) |
default | \(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}}{-1+x}\) | \(76\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}-{\mathrm e}^{\frac {-3 \,{\mathrm e}^{3} \ln \left (3 x^{2}\right )-3 \,{\mathrm e}^{3} \ln \left (3\right )+\left (5 x -5\right ) {\mathrm e}^{3}}{-1+x}}}{-1+x}\) | \(76\) |
int((3*x*exp(3)*ln(3*x^2)+3*x*exp(3)*ln(3)+(6-6*x)*exp(3))*exp((-3*exp(3)* ln(3*x^2)-3*exp(3)*ln(3)+(5*x-5)*exp(3))/(-1+x))/(x^3-2*x^2+x),x,method=_R ETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (\frac {5 \, {\left (x - 1\right )} e^{3} - 3 \, e^{3} \log \left (3\right ) - 3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1}\right )} \end {dmath*}
integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3 *exp(3)*log(3*x^2)-3*exp(3)*log(3)+(5*x-5)*exp(3))/(-1+x))/(x^3-2*x^2+x),x , algorithm=\
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\frac {\left (5 x - 5\right ) e^{3} - 3 e^{3} \log {\left (3 x^{2} \right )} - 3 e^{3} \log {\left (3 \right )}}{x - 1}} \end {dmath*}
integrate((3*x*exp(3)*ln(3*x**2)+3*x*exp(3)*ln(3)+(6-6*x)*exp(3))*exp((-3* exp(3)*ln(3*x**2)-3*exp(3)*ln(3)+(5*x-5)*exp(3))/(-1+x))/(x**3-2*x**2+x),x )
Time = 0.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (-\frac {6 \, e^{3} \log \left (3\right )}{x - 1} - \frac {6 \, e^{3} \log \left (x\right )}{x - 1} + 5 \, e^{3}\right )} \end {dmath*}
integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3 *exp(3)*log(3*x^2)-3*exp(3)*log(3)+(5*x-5)*exp(3))/(-1+x))/(x^3-2*x^2+x),x , algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx=e^{\left (\frac {5 \, x e^{3}}{x - 1} - \frac {3 \, e^{3} \log \left (3\right )}{x - 1} - \frac {3 \, e^{3} \log \left (3 \, x^{2}\right )}{x - 1} - \frac {5 \, e^{3}}{x - 1}\right )} \end {dmath*}
integrate((3*x*exp(3)*log(3*x^2)+3*x*exp(3)*log(3)+(6-6*x)*exp(3))*exp((-3 *exp(3)*log(3*x^2)-3*exp(3)*log(3)+(5*x-5)*exp(3))/(-1+x))/(x^3-2*x^2+x),x , algorithm=\
Time = 18.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \begin {dmath*} \int \frac {e^{\frac {e^3 (-5+5 x)-3 e^3 \log (3)-3 e^3 \log \left (3 x^2\right )}{-1+x}} \left (e^3 (6-6 x)+3 e^3 x \log (3)+3 e^3 x \log \left (3 x^2\right )\right )}{x-2 x^2+x^3} \, dx={\mathrm {e}}^{-\frac {5\,{\mathrm {e}}^3}{x-1}}\,{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^3}{x-1}}\,{\left (\frac {1}{729\,x^6}\right )}^{\frac {{\mathrm {e}}^3}{x-1}} \end {dmath*}