Integrand size = 106, antiderivative size = 34 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=\frac {-e+x-\frac {x}{x-5 \left (x+\frac {x}{1-e^{x^2}+x}\right )}}{x} \end {dmath*}
Time = 5.49 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=\frac {1-4 e+\frac {5}{-9+4 e^{x^2}-4 x}}{4 x} \end {dmath*}
Integrate[(-9 + E^(2*x^2)*(-4 + 16*E) - 8*x - 4*x^2 + E^x^2*(13 + E*(-72 - 32*x) + 8*x - 10*x^2) + E*(81 + 72*x + 16*x^2))/(81*x^2 + 16*E^(2*x^2)*x^ 2 + 72*x^3 + 16*x^4 + E^x^2*(-72*x^2 - 32*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.
\(\displaystyle \int \frac {-4 x^2+e^{x^2} \left (-10 x^2+8 x+e (-32 x-72)+13\right )+e \left (16 x^2+72 x+81\right )+(16 e-4) e^{2 x^2}-8 x-9}{16 x^4+72 x^3+16 e^{2 x^2} x^2+81 x^2+e^{x^2} \left (-32 x^3-72 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-4 x^2+e^{x^2} \left (-10 x^2+8 x+e (-32 x-72)+13\right )+e \left (16 x^2+72 x+81\right )+(16 e-4) e^{2 x^2}-8 x-9}{x^2 \left (-4 e^{x^2}+4 x+9\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 \left (2 x^2+1\right )}{4 \left (4 e^{x^2}-4 x-9\right ) x^2}-\frac {5 \left (4 x^2+9 x-2\right )}{2 x \left (-4 e^{x^2}+4 x+9\right )^2}+\frac {4 e-1}{4 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {45}{2} \int \frac {1}{\left (-4 x+4 e^{x^2}-9\right )^2}dx-\frac {5}{2} \int \frac {1}{-4 x+4 e^{x^2}-9}dx-\frac {5}{4} \int \frac {1}{\left (-4 x+4 e^{x^2}-9\right ) x^2}dx-10 \int \frac {x}{\left (-4 x+4 e^{x^2}-9\right )^2}dx+5 \int \frac {1}{x \left (4 x-4 e^{x^2}+9\right )^2}dx+\frac {1-4 e}{4 x}\) |
Int[(-9 + E^(2*x^2)*(-4 + 16*E) - 8*x - 4*x^2 + E^x^2*(13 + E*(-72 - 32*x) + 8*x - 10*x^2) + E*(81 + 72*x + 16*x^2))/(81*x^2 + 16*E^(2*x^2)*x^2 + 72 *x^3 + 16*x^4 + E^x^2*(-72*x^2 - 32*x^3)),x]
3.11.19.3.1 Defintions of rubi rules used
Time = 0.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {{\mathrm e}}{x}+\frac {1}{4 x}-\frac {5}{4 x \left (4 x -4 \,{\mathrm e}^{x^{2}}+9\right )}\) | \(32\) |
norman | \(\frac {\left (4 \,{\mathrm e}-1\right ) {\mathrm e}^{x^{2}}+\left (-4 \,{\mathrm e}+1\right ) x +1-9 \,{\mathrm e}}{x \left (4 x -4 \,{\mathrm e}^{x^{2}}+9\right )}\) | \(43\) |
parallelrisch | \(-\frac {-4+16 x \,{\mathrm e}-16 \,{\mathrm e} \,{\mathrm e}^{x^{2}}+36 \,{\mathrm e}-4 x +4 \,{\mathrm e}^{x^{2}}}{4 x \left (4 x -4 \,{\mathrm e}^{x^{2}}+9\right )}\) | \(47\) |
int(((16*exp(1)-4)*exp(x^2)^2+((-32*x-72)*exp(1)-10*x^2+8*x+13)*exp(x^2)+( 16*x^2+72*x+81)*exp(1)-4*x^2-8*x-9)/(16*x^2*exp(x^2)^2+(-32*x^3-72*x^2)*ex p(x^2)+16*x^4+72*x^3+81*x^2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=-\frac {{\left (4 \, x + 9\right )} e - {\left (4 \, e - 1\right )} e^{\left (x^{2}\right )} - x - 1}{4 \, x^{2} - 4 \, x e^{\left (x^{2}\right )} + 9 \, x} \end {dmath*}
integrate(((16*exp(1)-4)*exp(x^2)^2+((-32*x-72)*exp(1)-10*x^2+8*x+13)*exp( x^2)+(16*x^2+72*x+81)*exp(1)-4*x^2-8*x-9)/(16*x^2*exp(x^2)^2+(-32*x^3-72*x ^2)*exp(x^2)+16*x^4+72*x^3+81*x^2),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=\frac {5}{- 16 x^{2} + 16 x e^{x^{2}} - 36 x} - \frac {- \frac {1}{4} + e}{x} \end {dmath*}
integrate(((16*exp(1)-4)*exp(x**2)**2+((-32*x-72)*exp(1)-10*x**2+8*x+13)*e xp(x**2)+(16*x**2+72*x+81)*exp(1)-4*x**2-8*x-9)/(16*x**2*exp(x**2)**2+(-32 *x**3-72*x**2)*exp(x**2)+16*x**4+72*x**3+81*x**2),x)
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=-\frac {x {\left (4 \, e - 1\right )} - {\left (4 \, e - 1\right )} e^{\left (x^{2}\right )} + 9 \, e - 1}{4 \, x^{2} - 4 \, x e^{\left (x^{2}\right )} + 9 \, x} \end {dmath*}
integrate(((16*exp(1)-4)*exp(x^2)^2+((-32*x-72)*exp(1)-10*x^2+8*x+13)*exp( x^2)+(16*x^2+72*x+81)*exp(1)-4*x^2-8*x-9)/(16*x^2*exp(x^2)^2+(-32*x^3-72*x ^2)*exp(x^2)+16*x^4+72*x^3+81*x^2),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=-\frac {4 \, x e - x + 9 \, e - 4 \, e^{\left (x^{2} + 1\right )} + e^{\left (x^{2}\right )} - 1}{4 \, x^{2} - 4 \, x e^{\left (x^{2}\right )} + 9 \, x} \end {dmath*}
integrate(((16*exp(1)-4)*exp(x^2)^2+((-32*x-72)*exp(1)-10*x^2+8*x+13)*exp( x^2)+(16*x^2+72*x+81)*exp(1)-4*x^2-8*x-9)/(16*x^2*exp(x^2)^2+(-32*x^3-72*x ^2)*exp(x^2)+16*x^4+72*x^3+81*x^2),x, algorithm=\
Time = 15.69 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.74 \begin {dmath*} \int \frac {-9+e^{2 x^2} (-4+16 e)-8 x-4 x^2+e^{x^2} \left (13+e (-72-32 x)+8 x-10 x^2\right )+e \left (81+72 x+16 x^2\right )}{81 x^2+16 e^{2 x^2} x^2+72 x^3+16 x^4+e^{x^2} \left (-72 x^2-32 x^3\right )} \, dx=\frac {x^2\,\left (\frac {16\,\mathrm {e}}{9}-\frac {4}{9}\right )-9\,\mathrm {e}+{\mathrm {e}}^{x^2}\,\left (4\,\mathrm {e}-1\right )-x\,{\mathrm {e}}^{x^2}\,\left (\frac {16\,\mathrm {e}}{9}-\frac {4}{9}\right )+1}{9\,x-4\,x\,{\mathrm {e}}^{x^2}+4\,x^2} \end {dmath*}
int(-(8*x - exp(x^2)*(8*x - 10*x^2 - exp(1)*(32*x + 72) + 13) - exp(2*x^2) *(16*exp(1) - 4) - exp(1)*(72*x + 16*x^2 + 81) + 4*x^2 + 9)/(16*x^2*exp(2* x^2) - exp(x^2)*(72*x^2 + 32*x^3) + 81*x^2 + 72*x^3 + 16*x^4),x)