Integrand size = 89, antiderivative size = 26 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {3 x}{e^{3+x}+\frac {e^{x^2} x}{3}-\log (\log (3))} \end {dmath*}
Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {9 x}{3 e^{3+x}+e^{x^2} x-3 \log (\log (3))} \end {dmath*}
Integrate[(E^(3 + x)*(27 - 27*x) - 18*E^x^2*x^3 - 27*Log[Log[3]])/(9*E^(6 + 2*x) + 6*E^(3 + x + x^2)*x + E^(2*x^2)*x^2 + (-18*E^(3 + x) - 6*E^x^2*x) *Log[Log[3]] + 9*Log[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.
\(\displaystyle \int \frac {-18 e^{x^2} x^3+e^{x+3} (27-27 x)-27 \log (\log (3))}{e^{2 x^2} x^2+6 e^{x^2+x+3} x+\left (-6 e^{x^2} x-18 e^{x+3}\right ) \log (\log (3))+9 e^{2 x+6}+9 \log ^2(\log (3))} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {9 \left (-2 e^{x^2} x^3-3 e^{x+3} (x-1)-3 \log (\log (3))\right )}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 9 \int \frac {-2 e^{x^2} x^3+3 e^{x+3} (1-x)-3 \log (\log (3))}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 9 \int \left (\frac {3 \left (2 e^{x+3} x^2-2 \log (\log (3)) x^2-e^{x+3} x+e^{x+3}-\log (\log (3))\right )}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}-\frac {2 x^2}{e^{x^2} x+3 e^{x+3}-3 \log (\log (3))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 \left (-3 \log (\log (3)) \int \frac {1}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx+3 \int \frac {e^{x+3}}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx-3 \int \frac {e^{x+3} x}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx-6 \log (\log (3)) \int \frac {x^2}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx+6 \int \frac {e^{x+3} x^2}{\left (e^{x^2} x+3 e^{x+3}-3 \log (\log (3))\right )^2}dx-2 \int \frac {x^2}{e^{x^2} x+3 e^{x+3}-3 \log (\log (3))}dx\right )\) |
Int[(E^(3 + x)*(27 - 27*x) - 18*E^x^2*x^3 - 27*Log[Log[3]])/(9*E^(6 + 2*x) + 6*E^(3 + x + x^2)*x + E^(2*x^2)*x^2 + (-18*E^(3 + x) - 6*E^x^2*x)*Log[L og[3]] + 9*Log[Log[3]]^2),x]
3.1.46.3.1 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 = 0.62 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {9 x}{-{\mathrm e}^{x^{2}} x +3 \ln \left (\ln \left (3\right )\right )-3 \,{\mathrm e}^{3+x}}\) | \(25\) |
parallelrisch | \(-\frac {9 x}{-{\mathrm e}^{x^{2}} x +3 \ln \left (\ln \left (3\right )\right )-3 \,{\mathrm e}^{3+x}}\) | \(25\) |
int((-27*ln(ln(3))-18*x^3*exp(x^2)+(-27*x+27)*exp(3+x))/(9*ln(ln(3))^2+(-6 *exp(x^2)*x-18*exp(3+x))*ln(ln(3))+x^2*exp(x^2)^2+6*x*exp(3+x)*exp(x^2)+9* exp(3+x)^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {9 \, x e^{\left (x^{2}\right )}}{x e^{\left (2 \, x^{2}\right )} - 3 \, e^{\left (x^{2}\right )} \log \left (\log \left (3\right )\right ) + 3 \, e^{\left (x^{2} + x + 3\right )}} \end {dmath*}
integrate((-27*log(log(3))-18*x^3*exp(x^2)+(-27*x+27)*exp(3+x))/(9*log(log (3))^2+(-6*exp(x^2)*x-18*exp(3+x))*log(log(3))+x^2*exp(x^2)^2+6*x*exp(3+x) *exp(x^2)+9*exp(3+x)^2),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {3 x}{\frac {x e^{x^{2}}}{3} + e^{x + 3} - \log {\left (\log {\left (3 \right )} \right )}} \end {dmath*}
integrate((-27*ln(ln(3))-18*x**3*exp(x**2)+(-27*x+27)*exp(3+x))/(9*ln(ln(3 ))**2+(-6*exp(x**2)*x-18*exp(3+x))*ln(ln(3))+x**2*exp(x**2)**2+6*x*exp(3+x )*exp(x**2)+9*exp(3+x)**2),x)
Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {9 \, x}{x e^{\left (x^{2}\right )} + 3 \, e^{\left (x + 3\right )} - 3 \, \log \left (\log \left (3\right )\right )} \end {dmath*}
integrate((-27*log(log(3))-18*x^3*exp(x^2)+(-27*x+27)*exp(3+x))/(9*log(log (3))^2+(-6*exp(x^2)*x-18*exp(3+x))*log(log(3))+x^2*exp(x^2)^2+6*x*exp(3+x) *exp(x^2)+9*exp(3+x)^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\frac {9 \, x}{x e^{\left (x^{2}\right )} + 3 \, e^{\left (x + 3\right )} - 3 \, \log \left (\log \left (3\right )\right )} \end {dmath*}
integrate((-27*log(log(3))-18*x^3*exp(x^2)+(-27*x+27)*exp(3+x))/(9*log(log (3))^2+(-6*exp(x^2)*x-18*exp(3+x))*log(log(3))+x^2*exp(x^2)^2+6*x*exp(3+x) *exp(x^2)+9*exp(3+x)^2),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {e^{3+x} (27-27 x)-18 e^{x^2} x^3-27 \log (\log (3))}{9 e^{6+2 x}+6 e^{3+x+x^2} x+e^{2 x^2} x^2+\left (-18 e^{3+x}-6 e^{x^2} x\right ) \log (\log (3))+9 \log ^2(\log (3))} \, dx=\int -\frac {27\,\ln \left (\ln \left (3\right )\right )+18\,x^3\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{x+3}\,\left (27\,x-27\right )}{9\,{\mathrm {e}}^{2\,x+6}+9\,{\ln \left (\ln \left (3\right )\right )}^2-\ln \left (\ln \left (3\right )\right )\,\left (18\,{\mathrm {e}}^{x+3}+6\,x\,{\mathrm {e}}^{x^2}\right )+x^2\,{\mathrm {e}}^{2\,x^2}+6\,x\,{\mathrm {e}}^{x+3}\,{\mathrm {e}}^{x^2}} \,d x \end {dmath*}
int(-(27*log(log(3)) + 18*x^3*exp(x^2) + exp(x + 3)*(27*x - 27))/(9*exp(2* x + 6) + 9*log(log(3))^2 - log(log(3))*(18*exp(x + 3) + 6*x*exp(x^2)) + x^ 2*exp(2*x^2) + 6*x*exp(x + 3)*exp(x^2)),x)