Integrand size = 95, antiderivative size = 24 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=e^{x-\left (-\frac {10+e^{225 x/16}}{x}+x\right )^2} \] Output:
exp(x-(x-(10+exp(225/16*x))/x)^2)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=e^{20-\frac {100}{x^2}-\frac {e^{225 x/8}}{x^2}+x-x^2+\frac {2 e^{225 x/16} \left (-10+x^2\right )}{x^2}} \] Input:
Integrate[(E^((-100 - E^((225*x)/8) + 20*x^2 + x^3 - x^4 + E^((225*x)/16)* (-20 + 2*x^2))/x^2)*(1600 + E^((225*x)/8)*(16 - 225*x) + 8*x^3 - 16*x^4 + E^((225*x)/16)*(320 - 2250*x + 225*x^3)))/(8*x^3),x]
Output:
E^(20 - 100/x^2 - E^((225*x)/8)/x^2 + x - x^2 + (2*E^((225*x)/16)*(-10 + x ^2))/x^2)
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.
| \(\displaystyle \int \frac {\left (-16 x^4+8 x^3+e^{225 x/16} \left (225 x^3-2250 x+320\right )+e^{225 x/8} (16-225 x)+1600\right ) \exp \left (\frac {-x^4+x^3+20 x^2+e^{225 x/16} \left (2 x^2-20\right )-e^{225 x/8}-100}{x^2}\right )}{8 x^3} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {\exp \left (-\frac {x^4-x^3-20 x^2+e^{225 x/8}+2 e^{225 x/16} \left (10-x^2\right )+100}{x^2}\right ) \left (-16 x^4+8 x^3+e^{225 x/8} (16-225 x)+5 e^{225 x/16} \left (45 x^3-450 x+64\right )+1600\right )}{x^3}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {1}{8} \int \frac {\exp \left (-x^2+x+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right ) \left (-16 x^4+8 x^3+e^{225 x/8} (16-225 x)+5 e^{225 x/16} \left (45 x^3-450 x+64\right )+1600\right )}{x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{8} \int \left (-\frac {\exp \left (-x^2+\frac {233 x}{8}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right ) (225 x-16)}{x^3}+\frac {5 \exp \left (-x^2+\frac {241 x}{16}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right ) \left (45 x^3-450 x+64\right )}{x^3}-\frac {8 \exp \left (-x^2+x+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right ) \left (2 x^4-x^3-200\right )}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{8} \left (8 \int \exp \left (-x^2+x+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )dx+225 \int \exp \left (-x^2+\frac {241 x}{16}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )dx-2250 \int \frac {\exp \left (-x^2+\frac {241 x}{16}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )}{x^2}dx-225 \int \frac {\exp \left (-x^2+\frac {233 x}{8}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )}{x^2}dx-16 \int \exp \left (-x^2+x+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right ) xdx+1600 \int \frac {\exp \left (-x^2+x+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )}{x^3}dx+320 \int \frac {\exp \left (-x^2+\frac {241 x}{16}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )}{x^3}dx+16 \int \frac {\exp \left (-x^2+\frac {233 x}{8}+20-\frac {e^{225 x/8}}{x^2}-\frac {2 e^{225 x/16} \left (10-x^2\right )}{x^2}-\frac {100}{x^2}\right )}{x^3}dx\right )\) |
Input:
Int[(E^((-100 - E^((225*x)/8) + 20*x^2 + x^3 - x^4 + E^((225*x)/16)*(-20 + 2*x^2))/x^2)*(1600 + E^((225*x)/8)*(16 - 225*x) + 8*x^3 - 16*x^4 + E^((22 5*x)/16)*(320 - 2250*x + 225*x^3)))/(8*x^3),x]
Output:
$Aborted
Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71
| method | result | size |
| norman | \({\mathrm e}^{\frac {-{\mathrm e}^{\frac {225 x}{8}}+\left (2 x^{2}-20\right ) {\mathrm e}^{\frac {225 x}{16}}-x^{4}+x^{3}+20 x^{2}-100}{x^{2}}}\) | \(41\) |
| risch | \({\mathrm e}^{-\frac {x^{4}-2 \,{\mathrm e}^{\frac {225 x}{16}} x^{2}-x^{3}-20 x^{2}+20 \,{\mathrm e}^{\frac {225 x}{16}}+{\mathrm e}^{\frac {225 x}{8}}+100}{x^{2}}}\) | \(41\) |
| parallelrisch | \({\mathrm e}^{\frac {-{\mathrm e}^{\frac {225 x}{8}}+\left (2 x^{2}-20\right ) {\mathrm e}^{\frac {225 x}{16}}-x^{4}+x^{3}+20 x^{2}-100}{x^{2}}}\) | \(41\) |
Input:
int(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16*x)-16 *x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+x^3+20 *x^2-100)/x^2)/x^3,x,method=_RETURNVERBOSE)
Output:
exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+x^3+20*x^2-100)/x^2)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=e^{\left (-\frac {x^{4} - x^{3} - 20 \, x^{2} - 2 \, {\left (x^{2} - 10\right )} e^{\left (\frac {225}{16} \, x\right )} + e^{\left (\frac {225}{8} \, x\right )} + 100}{x^{2}}\right )} \] Input:
integrate(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16 *x)-16*x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+ x^3+20*x^2-100)/x^2)/x^3,x, algorithm="fricas")
Output:
e^(-(x^4 - x^3 - 20*x^2 - 2*(x^2 - 10)*e^(225/16*x) + e^(225/8*x) + 100)/x ^2)
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).
Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=e^{\frac {- x^{4} + x^{3} + 20 x^{2} + \left (2 x^{2} - 20\right ) e^{\frac {225 x}{16}} - e^{\frac {225 x}{8}} - 100}{x^{2}}} \] Input:
integrate(1/8*((-225*x+16)*exp(225/16*x)**2+(225*x**3-2250*x+320)*exp(225/ 16*x)-16*x**4+8*x**3+1600)*exp((-exp(225/16*x)**2+(2*x**2-20)*exp(225/16*x )-x**4+x**3+20*x**2-100)/x**2)/x**3,x)
Output:
exp((-x**4 + x**3 + 20*x**2 + (2*x**2 - 20)*exp(225*x/16) - exp(225*x/8) - 100)/x**2)
\[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=\int { -\frac {{\left (16 \, x^{4} - 8 \, x^{3} + {\left (225 \, x - 16\right )} e^{\left (\frac {225}{8} \, x\right )} - 5 \, {\left (45 \, x^{3} - 450 \, x + 64\right )} e^{\left (\frac {225}{16} \, x\right )} - 1600\right )} e^{\left (-\frac {x^{4} - x^{3} - 20 \, x^{2} - 2 \, {\left (x^{2} - 10\right )} e^{\left (\frac {225}{16} \, x\right )} + e^{\left (\frac {225}{8} \, x\right )} + 100}{x^{2}}\right )}}{8 \, x^{3}} \,d x } \] Input:
integrate(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16 *x)-16*x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+ x^3+20*x^2-100)/x^2)/x^3,x, algorithm="maxima")
Output:
-1/8*integrate((16*x^4 - 8*x^3 + (225*x - 16)*e^(225/8*x) - 5*(45*x^3 - 45 0*x + 64)*e^(225/16*x) - 1600)*e^(-(x^4 - x^3 - 20*x^2 - 2*(x^2 - 10)*e^(2 25/16*x) + e^(225/8*x) + 100)/x^2)/x^3, x)
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=e^{\left (-x^{2} + x - \frac {e^{\left (\frac {225}{8} \, x\right )}}{x^{2}} - \frac {20 \, e^{\left (\frac {225}{16} \, x\right )}}{x^{2}} - \frac {100}{x^{2}} + 2 \, e^{\left (\frac {225}{16} \, x\right )} + 20\right )} \] Input:
integrate(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16 *x)-16*x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+ x^3+20*x^2-100)/x^2)/x^3,x, algorithm="giac")
Output:
e^(-x^2 + x - e^(225/8*x)/x^2 - 20*e^(225/16*x)/x^2 - 100/x^2 + 2*e^(225/1 6*x) + 20)
Time = 3.98 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx={\mathrm {e}}^{2\,{\mathrm {e}}^{\frac {225\,x}{16}}}\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-\frac {100}{x^2}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{\frac {225\,x}{8}}}{x^2}}\,{\mathrm {e}}^{-\frac {20\,{\mathrm {e}}^{\frac {225\,x}{16}}}{x^2}}\,{\mathrm {e}}^x \] Input:
int((exp(-(exp((225*x)/8) - exp((225*x)/16)*(2*x^2 - 20) - 20*x^2 - x^3 + x^4 + 100)/x^2)*(exp((225*x)/16)*(225*x^3 - 2250*x + 320) - exp((225*x)/8) *(225*x - 16) + 8*x^3 - 16*x^4 + 1600))/(8*x^3),x)
Output:
exp(2*exp((225*x)/16))*exp(20)*exp(-x^2)*exp(-100/x^2)*exp(-exp((225*x)/8) /x^2)*exp(-(20*exp((225*x)/16))/x^2)*exp(x)
\[ \int \frac {e^{\frac {-100-e^{225 x/8}+20 x^2+x^3-x^4+e^{225 x/16} \left (-20+2 x^2\right )}{x^2}} \left (1600+e^{225 x/8} (16-225 x)+8 x^3-16 x^4+e^{225 x/16} \left (320-2250 x+225 x^3\right )\right )}{8 x^3} \, dx=\int \frac {\left (\left (-225 x +16\right ) \left ({\mathrm e}^{\frac {225 x}{16}}\right )^{2}+\left (225 x^{3}-2250 x +320\right ) {\mathrm e}^{\frac {225 x}{16}}-16 x^{4}+8 x^{3}+1600\right ) {\mathrm e}^{\frac {-\left ({\mathrm e}^{\frac {225 x}{16}}\right )^{2}+\left (2 x^{2}-20\right ) {\mathrm e}^{\frac {225 x}{16}}-x^{4}+x^{3}+20 x^{2}-100}{x^{2}}}}{8 x^{3}}d x \] Input:
int(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16*x)-16 *x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+x^3+20 *x^2-100)/x^2)/x^3,x)
Output:
int(1/8*((-225*x+16)*exp(225/16*x)^2+(225*x^3-2250*x+320)*exp(225/16*x)-16 *x^4+8*x^3+1600)*exp((-exp(225/16*x)^2+(2*x^2-20)*exp(225/16*x)-x^4+x^3+20 *x^2-100)/x^2)/x^3,x)