Integrand size = 89, antiderivative size = 23 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 e^{-\frac {8}{\left (4+e^{\frac {x^2}{5}}\right ) x}} x \] Output:
4*exp(-4/(4+exp(1/5*x^2))/x)^2*x
Time = 0.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} x \] Input:
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]
Output:
(4*x)/E^(8/(4*x + E^(x^2/5)*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 |
\(\displaystyle \frac {4}{5} \left (5 \int e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}}dx+40 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}}}{\left (4+e^{\frac {x^2}{5}}\right ) x}dx-64 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} x}{\left (4+e^{\frac {x^2}{5}}\right )^2}dx+16 \int \frac {e^{-\frac {8}{e^{\frac {x^2}{5}} x+4 x}} x}{4+e^{\frac {x^2}{5}}}dx\right )\) |
Input:
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]
Output:
$Aborted
Time = 0.42 (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 | \(\frac {16 x \,{\mathrm e}^{-\frac {8}{x \,{\mathrm e}^{\frac {x^{2}}{5}}+4 x}}+4 x \,{\mathrm e}^{\frac {x^{2}}{5}} {\mathrm e}^{-\frac {8}{x \,{\mathrm e}^{\frac {x^{2}}{5}}+4 x}}}{4+{\mathrm e}^{\frac {x^{2}}{5}}}\) | \(63\) |
Input:
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)
Output:
4*x*exp(-8/(4+exp(1/5*x^2))/x)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 \, x e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )} \] Input:
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="fricas")
Output:
4*x*e^(-8/(x*e^(1/5*x^2) + 4*x))
Time = 11.84 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 x e^{- \frac {8}{x e^{\frac {x^{2}}{5}} + 4 x}} \] Input:
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)
Output:
4*x*exp(-8/(x*exp(x**2/5) + 4*x))
\[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=\int { \frac {4 \, {\left (5 \, x e^{\left (\frac {2}{5} \, x^{2}\right )} + 8 \, {\left (2 \, x^{2} + 5 \, x + 5\right )} e^{\left (\frac {1}{5} \, x^{2}\right )} + 80 \, x + 160\right )} e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )}}{5 \, {\left (x e^{\left (\frac {2}{5} \, x^{2}\right )} + 8 \, x e^{\left (\frac {1}{5} \, x^{2}\right )} + 16 \, x\right )}} \,d x } \] Input:
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="maxima")
Output:
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 = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4 \, x e^{\left (-\frac {8}{x e^{\left (\frac {1}{5} \, x^{2}\right )} + 4 \, x}\right )} \] Input:
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="giac")
Output:
4*x*e^(-8/(x*e^(1/5*x^2) + 4*x))
Time = 4.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=4\,x\,{\mathrm {e}}^{-\frac {8}{4\,x+x\,{\mathrm {e}}^{\frac {x^2}{5}}}} \] Input:
int((exp(-8/(4*x + x*exp(x^2/5)))*(320*x + exp(x^2/5)*(160*x + 64*x^2 + 16 0) + 20*x*exp((2*x^2)/5) + 640))/(80*x + 40*x*exp(x^2/5) + 5*x*exp((2*x^2) /5)),x)
Output:
4*x*exp(-8/(4*x + x*exp(x^2/5)))
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-\frac {8}{4 x+e^{\frac {x^2}{5}} x}} \left (640+320 x+20 e^{\frac {2 x^2}{5}} x+e^{\frac {x^2}{5}} \left (160+160 x+64 x^2\right )\right )}{80 x+40 e^{\frac {x^2}{5}} x+5 e^{\frac {2 x^2}{5}} x} \, dx=\frac {4 x}{e^{\frac {8}{e^{\frac {x^{2}}{5}} x +4 x}}} \] Input:
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)
Output:
(4*x)/e**(8/(e**(x**2/5)*x + 4*x))