Integrand size = 106, antiderivative size = 33 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {x}{5}-\frac {x-x^2}{3-e^{\frac {x^2}{e^2}}+2 x} \]
Time = 4.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {1}{5} \left (x-\frac {5 (-1+x) x}{-3+e^{\frac {x^2}{e^2}}-2 x}\right ) \]
Integrate[(E^(2 + (2*x^2)/E^2) + E^2*(-6 + 42*x + 14*x^2) + E^(x^2/E^2)*(E ^2*(-1 - 14*x) - 10*x^2 + 10*x^3))/(5*E^(2 + (2*x^2)/E^2) + E^(2 + x^2/E^2 )*(-30 - 20*x) + E^2*(45 + 60*x + 20*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.
\(\displaystyle \int \frac {e^2 \left (14 x^2+42 x-6\right )+e^{\frac {2 x^2}{e^2}+2}+e^{\frac {x^2}{e^2}} \left (10 x^3-10 x^2+e^2 (-14 x-1)\right )}{e^{\frac {x^2}{e^2}+2} (-20 x-30)+5 e^{\frac {2 x^2}{e^2}+2}+e^2 \left (20 x^2+60 x+45\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^2 \left (14 x^2+42 x-6\right )+e^{\frac {2 x^2}{e^2}+2}+e^{\frac {x^2}{e^2}} \left (10 x^3-10 x^2+e^2 (-14 x-1)\right )}{5 e^2 \left (-e^{\frac {x^2}{e^2}}+2 x+3\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-2 e^2 \left (-7 x^2-21 x+3\right )+e^{\frac {2 x^2}{e^2}+2}-e^{\frac {x^2}{e^2}} \left (-10 x^3+10 x^2+e^2 (14 x+1)\right )}{\left (2 x-e^{\frac {x^2}{e^2}}+3\right )^2}dx}{5 e^2}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {10 (x-1) x \left (2 x^2+3 x-e^2\right )}{\left (-2 x+e^{\frac {x^2}{e^2}}-3\right )^2}+\frac {5 \left (2 x^3-2 x^2-2 e^2 x+e^2\right )}{-2 x+e^{\frac {x^2}{e^2}}-3}+e^2\right )dx}{5 e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 e^2 \int \frac {1}{-2 x+e^{\frac {x^2}{e^2}}-3}dx+10 e^2 \int \frac {x}{\left (-2 x+e^{\frac {x^2}{e^2}}-3\right )^2}dx-10 \left (3+e^2\right ) \int \frac {x^2}{\left (-2 x+e^{\frac {x^2}{e^2}}-3\right )^2}dx+10 e^2 \int \frac {x}{2 x-e^{\frac {x^2}{e^2}}+3}dx+10 \int \frac {x^2}{2 x-e^{\frac {x^2}{e^2}}+3}dx+20 \int \frac {x^4}{\left (-2 x+e^{\frac {x^2}{e^2}}-3\right )^2}dx+10 \int \frac {x^3}{\left (-2 x+e^{\frac {x^2}{e^2}}-3\right )^2}dx+10 \int \frac {x^3}{-2 x+e^{\frac {x^2}{e^2}}-3}dx+e^2 x}{5 e^2}\) |
Int[(E^(2 + (2*x^2)/E^2) + E^2*(-6 + 42*x + 14*x^2) + E^(x^2/E^2)*(E^2*(-1 - 14*x) - 10*x^2 + 10*x^3))/(5*E^(2 + (2*x^2)/E^2) + E^(2 + x^2/E^2)*(-30 - 20*x) + E^2*(45 + 60*x + 20*x^2)),x]
3.27.57.3.1 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]]
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x}{5}+\frac {x \left (-1+x \right )}{2 x -{\mathrm e}^{{\mathrm e}^{-2} x^{2}}+3}\) | \(26\) |
norman | \(\frac {\left (-\frac {2 x \,{\mathrm e}}{5}+\frac {7 x^{2} {\mathrm e}}{5}-\frac {x \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{-2} x^{2}}}{5}\right ) {\mathrm e}^{-1}}{2 x -{\mathrm e}^{{\mathrm e}^{-2} x^{2}}+3}\) | \(51\) |
parallelrisch | \(\frac {\left (7 x^{2} {\mathrm e}^{2}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{-2} x^{2}} x -2 \,{\mathrm e}^{2} x \right ) {\mathrm e}^{-2}}{10 x -5 \,{\mathrm e}^{{\mathrm e}^{-2} x^{2}}+15}\) | \(58\) |
int((exp(1)^2*exp(x^2/exp(1)^2)^2+((-14*x-1)*exp(1)^2+10*x^3-10*x^2)*exp(x ^2/exp(1)^2)+(14*x^2+42*x-6)*exp(1)^2)/(5*exp(1)^2*exp(x^2/exp(1)^2)^2+(-2 0*x-30)*exp(1)^2*exp(x^2/exp(1)^2)+(20*x^2+60*x+45)*exp(1)^2),x,method=_RE TURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {{\left (7 \, x^{2} - 2 \, x\right )} e^{2} - x e^{\left ({\left (x^{2} + 2 \, e^{2}\right )} e^{\left (-2\right )}\right )}}{5 \, {\left ({\left (2 \, x + 3\right )} e^{2} - e^{\left ({\left (x^{2} + 2 \, e^{2}\right )} e^{\left (-2\right )}\right )}\right )}} \]
integrate((exp(1)^2*exp(x^2/exp(1)^2)^2+((-14*x-1)*exp(1)^2+10*x^3-10*x^2) *exp(x^2/exp(1)^2)+(14*x^2+42*x-6)*exp(1)^2)/(5*exp(1)^2*exp(x^2/exp(1)^2) ^2+(-20*x-30)*exp(1)^2*exp(x^2/exp(1)^2)+(20*x^2+60*x+45)*exp(1)^2),x, alg orithm=\
1/5*((7*x^2 - 2*x)*e^2 - x*e^((x^2 + 2*e^2)*e^(-2)))/((2*x + 3)*e^2 - e^(( x^2 + 2*e^2)*e^(-2)))
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {x}{5} + \frac {- x^{2} + x}{- 2 x + e^{\frac {x^{2}}{e^{2}}} - 3} \]
integrate((exp(1)**2*exp(x**2/exp(1)**2)**2+((-14*x-1)*exp(1)**2+10*x**3-1 0*x**2)*exp(x**2/exp(1)**2)+(14*x**2+42*x-6)*exp(1)**2)/(5*exp(1)**2*exp(x **2/exp(1)**2)**2+(-20*x-30)*exp(1)**2*exp(x**2/exp(1)**2)+(20*x**2+60*x+4 5)*exp(1)**2),x)
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {7 \, x^{2} - x e^{\left (x^{2} e^{\left (-2\right )}\right )} - 2 \, x}{5 \, {\left (2 \, x - e^{\left (x^{2} e^{\left (-2\right )}\right )} + 3\right )}} \]
integrate((exp(1)^2*exp(x^2/exp(1)^2)^2+((-14*x-1)*exp(1)^2+10*x^3-10*x^2) *exp(x^2/exp(1)^2)+(14*x^2+42*x-6)*exp(1)^2)/(5*exp(1)^2*exp(x^2/exp(1)^2) ^2+(-20*x-30)*exp(1)^2*exp(x^2/exp(1)^2)+(20*x^2+60*x+45)*exp(1)^2),x, alg orithm=\
Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=\frac {7 \, x^{2} - x e^{\left (x^{2} e^{\left (-2\right )}\right )} - 2 \, x}{5 \, {\left (2 \, x - e^{\left (x^{2} e^{\left (-2\right )}\right )} + 3\right )}} \]
integrate((exp(1)^2*exp(x^2/exp(1)^2)^2+((-14*x-1)*exp(1)^2+10*x^3-10*x^2) *exp(x^2/exp(1)^2)+(14*x^2+42*x-6)*exp(1)^2)/(5*exp(1)^2*exp(x^2/exp(1)^2) ^2+(-20*x-30)*exp(1)^2*exp(x^2/exp(1)^2)+(20*x^2+60*x+45)*exp(1)^2),x, alg orithm=\
Time = 8.93 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{2+\frac {2 x^2}{e^2}}+e^2 \left (-6+42 x+14 x^2\right )+e^{\frac {x^2}{e^2}} \left (e^2 (-1-14 x)-10 x^2+10 x^3\right )}{5 e^{2+\frac {2 x^2}{e^2}}+e^{2+\frac {x^2}{e^2}} (-30-20 x)+e^2 \left (45+60 x+20 x^2\right )} \, dx=-\frac {x\,\left ({\mathrm {e}}^{x^2\,{\mathrm {e}}^{-2}}-7\,x+2\right )}{5\,\left (2\,x-{\mathrm {e}}^{x^2\,{\mathrm {e}}^{-2}}+3\right )} \]