Integrand size = 88, antiderivative size = 32 \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=1+\frac {e^x x}{3+e^{-2 x+2 \left (1-e+e^4\right ) x} x^2} \]
Time = 4.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\frac {e^{(1+2 e) x} x}{3 e^{2 e x}+e^{2 e^4 x} x^2} \]
Integrate[(E^x*(3 + 3*x) + E^(x - 2*E*x + 2*E^4*x)*(-x^2 + x^3 + 2*E*x^3 - 2*E^4*x^3))/(9 + 6*E^(-2*E*x + 2*E^4*x)*x^2 + E^(-4*E*x + 4*E^4*x)*x^4),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.
\(\displaystyle \int \frac {e^{2 e^4 x-2 e x+x} \left (-2 e^4 x^3+2 e x^3+x^3-x^2\right )+e^x (3 x+3)}{e^{4 e^4 x-4 e x} x^4+6 e^{2 e^4 x-2 e x} x^2+9} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{4 e x} \left (e^{2 e^4 x-2 e x+x} \left (-2 e^4 x^3+2 e x^3+x^3-x^2\right )+e^x (3 x+3)\right )}{\left (e^{2 e^4 x} x^2+3 e^{2 e x}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {6 e^{4 e x+x} \left (1-e \left (1-e^3\right ) x\right )}{\left (e^{2 e^4 x} x^2+3 e^{2 e x}\right )^2}+\frac {e^{4 e x+(1-2 e) x} \left (\left (1+2 e-2 e^4\right ) x-1\right )}{e^{2 e^4 x} x^2+3 e^{2 e x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 6 \int \frac {e^{(1+4 e) x}}{\left (e^{2 e^4 x} x^2+3 e^{2 e x}\right )^2}dx-6 \left (1-e^3\right ) \int \frac {e^{(1+4 e) x+1} x}{\left (e^{2 e^4 x} x^2+3 e^{2 e x}\right )^2}dx-\int \frac {e^{(1+2 e) x}}{e^{2 e^4 x} x^2+3 e^{2 e x}}dx+\left (1+2 e-2 e^4\right ) \int \frac {e^{(1+2 e) x} x}{e^{2 e^4 x} x^2+3 e^{2 e x}}dx\) |
Int[(E^x*(3 + 3*x) + E^(x - 2*E*x + 2*E^4*x)*(-x^2 + x^3 + 2*E*x^3 - 2*E^4 *x^3))/(9 + 6*E^(-2*E*x + 2*E^4*x)*x^2 + E^(-4*E*x + 4*E^4*x)*x^4),x]
3.8.64.3.1 Defintions of rubi rules used
Time = 1.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{x}}{x^{2} {\mathrm e}^{-2 x \left ({\mathrm e}-{\mathrm e}^{4}\right )}+3}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{x} x}{x^{2} {\mathrm e}^{2 x \left (-{\mathrm e}+{\mathrm e}^{4}\right )}+3}\) | \(25\) |
int(((-2*x^3*exp(4)+2*x^3*exp(1)+x^3-x^2)*exp(x)*exp(x*exp(4)-x*exp(1))^2+ (3*x+3)*exp(x))/(x^4*exp(x*exp(4)-x*exp(1))^4+6*x^2*exp(x*exp(4)-x*exp(1)) ^2+9),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\frac {x e^{\left (2 \, x e^{4} - 2 \, x e + x\right )}}{x^{2} e^{\left (4 \, x e^{4} - 4 \, x e\right )} + 3 \, e^{\left (2 \, x e^{4} - 2 \, x e\right )}} \]
integrate(((-2*x^3*exp(4)+2*x^3*exp(1)+x^3-x^2)*exp(x)*exp(x*exp(4)-x*exp( 1))^2+(3*x+3)*exp(x))/(x^4*exp(x*exp(4)-x*exp(1))^4+6*x^2*exp(x*exp(4)-x*e xp(1))^2+9),x, algorithm=\
Timed out. \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\text {Timed out} \]
integrate(((-2*x**3*exp(4)+2*x**3*exp(1)+x**3-x**2)*exp(x)*exp(x*exp(4)-x* exp(1))**2+(3*x+3)*exp(x))/(x**4*exp(x*exp(4)-x*exp(1))**4+6*x**2*exp(x*ex p(4)-x*exp(1))**2+9),x)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\frac {x e^{\left (2 \, x e + x\right )}}{x^{2} e^{\left (2 \, x e^{4}\right )} + 3 \, e^{\left (2 \, x e\right )}} \]
integrate(((-2*x^3*exp(4)+2*x^3*exp(1)+x^3-x^2)*exp(x)*exp(x*exp(4)-x*exp( 1))^2+(3*x+3)*exp(x))/(x^4*exp(x*exp(4)-x*exp(1))^4+6*x^2*exp(x*exp(4)-x*e xp(1))^2+9),x, algorithm=\
\[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\int { -\frac {{\left (2 \, x^{3} e^{4} - 2 \, x^{3} e - x^{3} + x^{2}\right )} e^{\left (2 \, x e^{4} - 2 \, x e + x\right )} - 3 \, {\left (x + 1\right )} e^{x}}{x^{4} e^{\left (4 \, x e^{4} - 4 \, x e\right )} + 6 \, x^{2} e^{\left (2 \, x e^{4} - 2 \, x e\right )} + 9} \,d x } \]
integrate(((-2*x^3*exp(4)+2*x^3*exp(1)+x^3-x^2)*exp(x)*exp(x*exp(4)-x*exp( 1))^2+(3*x+3)*exp(x))/(x^4*exp(x*exp(4)-x*exp(1))^4+6*x^2*exp(x*exp(4)-x*e xp(1))^2+9),x, algorithm=\
integrate(-((2*x^3*e^4 - 2*x^3*e - x^3 + x^2)*e^(2*x*e^4 - 2*x*e + x) - 3* (x + 1)*e^x)/(x^4*e^(4*x*e^4 - 4*x*e) + 6*x^2*e^(2*x*e^4 - 2*x*e) + 9), x)
Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {e^x (3+3 x)+e^{x-2 e x+2 e^4 x} \left (-x^2+x^3+2 e x^3-2 e^4 x^3\right )}{9+6 e^{-2 e x+2 e^4 x} x^2+e^{-4 e x+4 e^4 x} x^4} \, dx=\frac {x^2\,{\mathrm {e}}^x-x^3\,\left ({\mathrm {e}}^{x+1}-{\mathrm {e}}^{x+4}\right )}{\left (x^2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^4-2\,x\,\mathrm {e}}+3\right )\,\left (x-x^2\,\mathrm {e}+x^2\,{\mathrm {e}}^4\right )} \]
int((exp(x)*(3*x + 3) + exp(2*x*exp(4) - 2*x*exp(1))*exp(x)*(2*x^3*exp(1) - 2*x^3*exp(4) - x^2 + x^3))/(6*x^2*exp(2*x*exp(4) - 2*x*exp(1)) + x^4*exp (4*x*exp(4) - 4*x*exp(1)) + 9),x)