Integrand size = 89, antiderivative size = 23 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 e^{-\frac {8}{\left (4+e^{\frac {x^2}{5}}\right ) x}} x \end {dmath*}
Time = 1.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} x \end {dmath*}
Integrate[(640 + 320*x + 20*E^((2*x^2)/5)*x + E^(x^2/5)*(160 + 160*x + 64* x^2))/(E^(8/(4*x + E^(x^2/5)*x))*(80*x + 40*E^(x^2/5)*x + 5*E^((2*x^2)/5)* x)),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.
\(\displaystyle \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} \left (20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (64 x^2+160 x+160\right )+320 x+640\right )}{40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x+80 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} \left (20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (64 x^2+160 x+160\right )+320 x+640\right )}{5 \left (e^{\frac {x^2}{5}}+4\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {4 e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} \left (5 e^{\frac {2 x^2}{5}} x+80 x+8 e^{\frac {x^2}{5}} \left (2 x^2+5 x+5\right )+160\right )}{\left (4+e^{\frac {x^2}{5}}\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{5} \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} \left (5 e^{\frac {2 x^2}{5}} x+80 x+8 e^{\frac {x^2}{5}} \left (2 x^2+5 x+5\right )+160\right )}{\left (4+e^{\frac {x^2}{5}}\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4}{5} \int \left (-\frac {64 e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} x}{\left (4+e^{\frac {x^2}{5}}\right )^2}+5 e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}}+\frac {8 e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} \left (2 x^2+5\right )}{\left (4+e^{\frac {x^2}{5}}\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{5} \left (5 \int e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}}dx+40 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}}}{\left (4+e^{\frac {x^2}{5}}\right ) x}dx-64 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} x}{\left (4+e^{\frac {x^2}{5}}\right )^2}dx+16 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} x}{4+e^{\frac {x^2}{5}}}dx\right )\) |
Int[(640 + 320*x + 20*E^((2*x^2)/5)*x + E^(x^2/5)*(160 + 160*x + 64*x^2))/ (E^(8/(4*x + E^(x^2/5)*x))*(80*x + 40*E^(x^2/5)*x + 5*E^((2*x^2)/5)*x)),x]
3.4.95.3.1 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 = 1.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(4 x \,{\mathrm e}^{-\frac {8}{\left (4+{\mathrm e}^{\frac {x^{2}}{5}}\right ) x}}\) | \(20\) |
parallelrisch | \(4 x \,{\mathrm e}^{-\frac {8}{\left (4+{\mathrm e}^{\frac {x^{2}}{5}}\right ) x}}\) | \(22\) |
norman | \(\frac {16 x \,{\mathrm e}^{-\frac {8}{x \,{\mathrm e}^{\frac {x^{2}}{5}}+4 x}}+4 x \,{\mathrm e}^{\frac {x^{2}}{5}} {\mathrm e}^{-\frac {8}{x \,{\mathrm e}^{\frac {x^{2}}{5}}+4 x}}}{4+{\mathrm e}^{\frac {x^{2}}{5}}}\) | \(63\) |
int((20*x*exp(1/5*x^2)^2+(64*x^2+160*x+160)*exp(1/5*x^2)+320*x+640)*exp(-4 /(x*exp(1/5*x^2)+4*x))^2/(5*x*exp(1/5*x^2)^2+40*x*exp(1/5*x^2)+80*x),x,met hod=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 \, x e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )} \end {dmath*}
integrate((20*x*exp(1/5*x^2)^2+(64*x^2+160*x+160)*exp(1/5*x^2)+320*x+640)* exp(-4/(x*exp(1/5*x^2)+4*x))^2/(5*x*exp(1/5*x^2)^2+40*x*exp(1/5*x^2)+80*x) ,x, algorithm=\
Time = 11.77 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 x e^{- \frac {8}{x e^{\frac {x^{2}}{5}} + 4 x}} \end {dmath*}
integrate((20*x*exp(1/5*x**2)**2+(64*x**2+160*x+160)*exp(1/5*x**2)+320*x+6 40)*exp(-4/(x*exp(1/5*x**2)+4*x))**2/(5*x*exp(1/5*x**2)**2+40*x*exp(1/5*x* *2)+80*x),x)
\begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=\int { \frac {4 \, {\left (5 \, x e^{\left (\frac {2}{5} \, x^{2}\right )} + 8 \, {\left (2 \, x^{2} + 5 \, x + 5\right )} e^{\left (\frac {1}{5} \, x^{2}\right )} + 80 \, x + 160\right )} e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )}}{5 \, {\left (x e^{\left (\frac {2}{5} \, x^{2}\right )} + 8 \, x e^{\left (\frac {1}{5} \, x^{2}\right )} + 16 \, x\right )}} \,d x } \end {dmath*}
integrate((20*x*exp(1/5*x^2)^2+(64*x^2+160*x+160)*exp(1/5*x^2)+320*x+640)* exp(-4/(x*exp(1/5*x^2)+4*x))^2/(5*x*exp(1/5*x^2)^2+40*x*exp(1/5*x^2)+80*x) ,x, algorithm=\
4/5*integrate((5*x*e^(2/5*x^2) + 8*(2*x^2 + 5*x + 5)*e^(1/5*x^2) + 80*x + 160)*e^(-8/(x*e^(1/5*x^2) + 4*x))/(x*e^(2/5*x^2) + 8*x*e^(1/5*x^2) + 16*x) , x)
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 \, x e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )} \end {dmath*}
integrate((20*x*exp(1/5*x^2)^2+(64*x^2+160*x+160)*exp(1/5*x^2)+320*x+640)* exp(-4/(x*exp(1/5*x^2)+4*x))^2/(5*x*exp(1/5*x^2)^2+40*x*exp(1/5*x^2)+80*x) ,x, algorithm=\
Time = 14.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \begin {dmath*} \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4\,x\,{\mathrm {e}}^{-\frac {8}{4\,x+x\,{\mathrm {e}}^{\frac {x^2}{5}}}} \end {dmath*}