Integrand size = 48, antiderivative size = 21 \[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\frac {1}{5} e^{e+\frac {e^2 (2+x)}{4+x}} x \]
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\frac {1}{5} e^{e+\frac {e^2 (2+x)}{4+x}} x \]
Integrate[(E^((E^2*(2 + x) + E*(4 + x))/(4 + x))*(16 + 8*x + 2*E^2*x + x^2 ))/(80 + 40*x + 5*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.
\(\displaystyle \int \frac {e^{\frac {e^2 (x+2)+e (x+4)}{x+4}} \left (x^2+2 e^2 x+8 x+16\right )}{5 x^2+40 x+80} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{\frac {e^2 (x+2)+e (x+4)}{x+4}} \left (x^2+\left (8+2 e^2\right ) x+16\right )}{5 x^2+40 x+80}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {e^2 (x+2)+e (x+4)}{x+4}} \left (x^2+\left (8+2 e^2\right ) x+16\right )}{\left (\sqrt {5} x+4 \sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {e (1+e) x+2 e (2+e)}{x+4}} \left (x^2+2 \left (4+e^2\right ) x+16\right )}{\left (\sqrt {5} x+4 \sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{5} e^{\frac {e (1+e) x+2 e (2+e)}{x+4}}+\frac {2 e^{\frac {e (1+e) x+2 e (2+e)}{x+4}+2}}{5 (x+4)}-\frac {8 e^{\frac {e (1+e) x+2 e (2+e)}{x+4}+2}}{5 (x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \int e^{\frac {e (1+e) x+2 e (2+e)}{x+4}}dx-\frac {2}{5} e^{2+e+e^2} \operatorname {ExpIntegralEi}\left (-\frac {2 e^2}{x+4}\right )-\frac {4}{5} e^{e (1+e)-\frac {2 e^2}{x+4}}\) |
3.23.20.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {x \,{\mathrm e}^{\frac {\left (2+x \right ) {\mathrm e}^{2}+\left (4+x \right ) {\mathrm e}}{4+x}}}{5}\) | \(24\) |
gosper | \(\frac {{\mathrm e}^{\frac {{\mathrm e}^{2} x +x \,{\mathrm e}+2 \,{\mathrm e}^{2}+4 \,{\mathrm e}}{4+x}} x}{5}\) | \(28\) |
risch | \(\frac {{\mathrm e}^{\frac {{\mathrm e}^{2} x +x \,{\mathrm e}+2 \,{\mathrm e}^{2}+4 \,{\mathrm e}}{4+x}} x}{5}\) | \(28\) |
derivativedivides | \(\frac {2 \,{\mathrm e}^{2} \left (\frac {{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2} \left (4+x \right )}{2}-2 \,{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2}\right )}{5}\) | \(51\) |
default | \(\frac {2 \,{\mathrm e}^{2} \left (\frac {{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2} \left (4+x \right )}{2}-2 \,{\mathrm e}^{{\mathrm e}^{2}+{\mathrm e}-\frac {2 \,{\mathrm e}^{2}}{4+x}} {\mathrm e}^{-2}\right )}{5}\) | \(51\) |
norman | \(\frac {\frac {4 x \,{\mathrm e}^{\frac {\left (2+x \right ) {\mathrm e}^{2}+\left (4+x \right ) {\mathrm e}}{4+x}}}{5}+\frac {x^{2} {\mathrm e}^{\frac {\left (2+x \right ) {\mathrm e}^{2}+\left (4+x \right ) {\mathrm e}}{4+x}}}{5}}{4+x}\) | \(56\) |
int((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/(5*x^2+ 40*x+80),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\frac {1}{5} \, x e^{\left (\frac {{\left (x + 2\right )} e^{2} + {\left (x + 4\right )} e}{x + 4}\right )} \]
integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/( 5*x^2+40*x+80),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\frac {x e^{\frac {\left (x + 2\right ) e^{2} + e \left (x + 4\right )}{x + 4}}}{5} \]
integrate((2*exp(2)*x+x**2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/ (5*x**2+40*x+80),x)
\[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\int { \frac {{\left (x^{2} + 2 \, x e^{2} + 8 \, x + 16\right )} e^{\left (\frac {{\left (x + 2\right )} e^{2} + {\left (x + 4\right )} e}{x + 4}\right )}}{5 \, {\left (x^{2} + 8 \, x + 16\right )}} \,d x } \]
integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/( 5*x^2+40*x+80),x, algorithm=\
1/5*x*e^(-2*e^2/(x + 4) + e^2 + e) + 8/5*e^(-2*e^2/(x + 4) + e^2 + e - 2) - 16/5*integrate(e^(-2*e^2/(x + 4) + e^2 + e)/(x^2 + 8*x + 16), x)
\[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\int { \frac {{\left (x^{2} + 2 \, x e^{2} + 8 \, x + 16\right )} e^{\left (\frac {{\left (x + 2\right )} e^{2} + {\left (x + 4\right )} e}{x + 4}\right )}}{5 \, {\left (x^{2} + 8 \, x + 16\right )}} \,d x } \]
integrate((2*exp(2)*x+x^2+8*x+16)*exp(((2+x)*exp(2)+(4+x)*exp(1))/(4+x))/( 5*x^2+40*x+80),x, algorithm=\
integrate(1/5*(x^2 + 2*x*e^2 + 8*x + 16)*e^(((x + 2)*e^2 + (x + 4)*e)/(x + 4))/(x^2 + 8*x + 16), x)
Timed out. \[ \int \frac {e^{\frac {e^2 (2+x)+e (4+x)}{4+x}} \left (16+8 x+2 e^2 x+x^2\right )}{80+40 x+5 x^2} \, dx=\int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,\left (x+2\right )+\mathrm {e}\,\left (x+4\right )}{x+4}}\,\left (8\,x+2\,x\,{\mathrm {e}}^2+x^2+16\right )}{5\,x^2+40\,x+80} \,d x \]
int((exp((exp(2)*(x + 2) + exp(1)*(x + 4))/(x + 4))*(8*x + 2*x*exp(2) + x^ 2 + 16))/(40*x + 5*x^2 + 80),x)