Integrand size = 89, antiderivative size = 29 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{-5+x+2 \left (1+\frac {1-\frac {x}{-5+x}}{x}\right )}}{x}} \] Output:
exp(exp(2*(1-x/(-5+x))/x+2)/x*exp(-5+x))
Time = 4.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{-3-\frac {2}{-5+x}+\frac {2}{x}+x}}{x}} \] Input:
Integrate[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(-50 - 5*x + 35*x^2 - 11*x^3 + x^4))/(2 5*x^3 - 10*x^4 + x^5),x]
Output:
E^(E^(-3 - 2/(-5 + x) + 2/x + 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.
| \(\displaystyle \int \frac {\left (x^4-11 x^3+35 x^2-5 x-50\right ) \exp \left (\frac {2 x^2-10 x-10}{x^2-5 x}+\frac {e^{\frac {2 x^2-10 x-10}{x^2-5 x}+x-5}}{x}+x-5\right )}{x^5-10 x^4+25 x^3} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (x^4-11 x^3+35 x^2-5 x-50\right ) \exp \left (\frac {2 x^2-10 x-10}{x^2-5 x}+\frac {e^{\frac {2 x^2-10 x-10}{x^2-5 x}+x-5}}{x}+x-5\right )}{x^3 \left (x^2-10 x+25\right )}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int -\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{4 (5-x)^2 x^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{(5-x)^2 x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\int \left (-\frac {27 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 x}+\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 (x-5)}-\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{5 (x-5)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{5} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{(x-5)^2}dx-\frac {2}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x-5}dx-\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}dx+\frac {27}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x}dx-2 \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}dx\) |
Input:
Int[(E^(-5 + E^(-5 + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))/x + x + (-10 - 10*x + 2*x^2)/(-5*x + x^2))*(-50 - 5*x + 35*x^2 - 11*x^3 + x^4))/(25*x^3 - 10*x^4 + x^5),x]
Output:
$Aborted
Time = 12.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {x^{3}-8 x^{2}+15 x -10}{\left (-5+x \right ) x}}}{x}}\) | \(29\) |
| parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {2 x^{2}-10 x -10}{\left (-5+x \right ) x}} {\mathrm e}^{-5+x}}{x}}\) | \(29\) |
Input:
int((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)*ex p(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x,method =_RETURNVERBOSE)
Output:
exp(1/x*exp((x^3-8*x^2+15*x-10)/(-5+x)/x))
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (23) = 46\).
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.66 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\left (\frac {x^{3} - 8 \, x^{2} + {\left (x - 5\right )} e^{\left (\frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} + 15 \, x - 10}{x^{2} - 5 \, x} - \frac {x^{3} - 8 \, x^{2} + 15 \, x - 10}{x^{2} - 5 \, x}\right )} \] Input:
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm="fricas")
Output:
e^((x^3 - 8*x^2 + (x - 5)*e^((x^3 - 8*x^2 + 15*x - 10)/(x^2 - 5*x)) + 15*x - 10)/(x^2 - 5*x) - (x^3 - 8*x^2 + 15*x - 10)/(x^2 - 5*x))
Time = 0.56 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=e^{\frac {e^{\frac {2 x^{2} - 10 x - 10}{x^{2} - 5 x}} e^{x - 5}}{x}} \] Input:
integrate((x**4-11*x**3+35*x**2-5*x-50)*exp((2*x**2-10*x-10)/(x**2-5*x))*e xp(-5+x)*exp(exp((2*x**2-10*x-10)/(x**2-5*x))*exp(-5+x)/x)/(x**5-10*x**4+2 5*x**3),x)
Output:
exp(exp((2*x**2 - 10*x - 10)/(x**2 - 5*x))*exp(x - 5)/x)
\[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=\int { \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}} \,d x } \] Input:
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm="maxima")
Output:
integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)
\[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=\int { \frac {{\left (x^{4} - 11 \, x^{3} + 35 \, x^{2} - 5 \, x - 50\right )} e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} + \frac {e^{\left (x + \frac {2 \, {\left (x^{2} - 5 \, x - 5\right )}}{x^{2} - 5 \, x} - 5\right )}}{x} - 5\right )}}{x^{5} - 10 \, x^{4} + 25 \, x^{3}} \,d x } \] Input:
integrate((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5 +x)*exp(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x, algorithm="giac")
Output:
integrate((x^4 - 11*x^3 + 35*x^2 - 5*x - 50)*e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) + e^(x + 2*(x^2 - 5*x - 5)/(x^2 - 5*x) - 5)/x - 5)/(x^5 - 10*x^4 + 25*x^3), x)
Time = 3.62 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{\frac {2\,x}{x-5}}\,{\mathrm {e}}^{\frac {10}{5\,x-x^2}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {10}{x-5}}}{x}} \] Input:
int(-(exp(x - 5)*exp((exp(x - 5)*exp((10*x - 2*x^2 + 10)/(5*x - x^2)))/x)* exp((10*x - 2*x^2 + 10)/(5*x - x^2))*(5*x - 35*x^2 + 11*x^3 - x^4 + 50))/( 25*x^3 - 10*x^4 + x^5),x)
Output:
exp((exp(-5)*exp((2*x)/(x - 5))*exp(10/(5*x - x^2))*exp(x)*exp(-10/(x - 5) ))/x)
\[ \int \frac {e^{-5+\frac {e^{-5+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}}}{x}+x+\frac {-10-10 x+2 x^2}{-5 x+x^2}} \left (-50-5 x+35 x^2-11 x^3+x^4\right )}{25 x^3-10 x^4+x^5} \, dx=\frac {-50 \left (\int \frac {e^{\frac {e^{\frac {10}{x^{2}-5 x}} e^{3} x^{2}+e^{x}}{e^{\frac {10}{x^{2}-5 x}} e^{3} x}}}{e^{\frac {10}{x^{2}-5 x}} x^{5}-10 e^{\frac {10}{x^{2}-5 x}} x^{4}+25 e^{\frac {10}{x^{2}-5 x}} x^{3}}d x \right )-5 \left (\int \frac {e^{\frac {e^{\frac {10}{x^{2}-5 x}} e^{3} x^{2}+e^{x}}{e^{\frac {10}{x^{2}-5 x}} e^{3} x}}}{e^{\frac {10}{x^{2}-5 x}} x^{4}-10 e^{\frac {10}{x^{2}-5 x}} x^{3}+25 e^{\frac {10}{x^{2}-5 x}} x^{2}}d x \right )+35 \left (\int \frac {e^{\frac {e^{\frac {10}{x^{2}-5 x}} e^{3} x^{2}+e^{x}}{e^{\frac {10}{x^{2}-5 x}} e^{3} x}}}{e^{\frac {10}{x^{2}-5 x}} x^{3}-10 e^{\frac {10}{x^{2}-5 x}} x^{2}+25 e^{\frac {10}{x^{2}-5 x}} x}d x \right )-11 \left (\int \frac {e^{\frac {e^{\frac {10}{x^{2}-5 x}} e^{3} x^{2}+e^{x}}{e^{\frac {10}{x^{2}-5 x}} e^{3} x}}}{e^{\frac {10}{x^{2}-5 x}} x^{2}-10 e^{\frac {10}{x^{2}-5 x}} x +25 e^{\frac {10}{x^{2}-5 x}}}d x \right )+\int \frac {e^{\frac {e^{\frac {10}{x^{2}-5 x}} e^{3} x^{2}+e^{x}}{e^{\frac {10}{x^{2}-5 x}} e^{3} x}} x}{e^{\frac {10}{x^{2}-5 x}} x^{2}-10 e^{\frac {10}{x^{2}-5 x}} x +25 e^{\frac {10}{x^{2}-5 x}}}d x}{e^{3}} \] Input:
int((x^4-11*x^3+35*x^2-5*x-50)*exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)*ex p(exp((2*x^2-10*x-10)/(x^2-5*x))*exp(-5+x)/x)/(x^5-10*x^4+25*x^3),x)
Output:
( - 50*int(e**((e**(10/(x**2 - 5*x))*e**3*x**2 + e**x)/(e**(10/(x**2 - 5*x ))*e**3*x))/(e**(10/(x**2 - 5*x))*x**5 - 10*e**(10/(x**2 - 5*x))*x**4 + 25 *e**(10/(x**2 - 5*x))*x**3),x) - 5*int(e**((e**(10/(x**2 - 5*x))*e**3*x**2 + e**x)/(e**(10/(x**2 - 5*x))*e**3*x))/(e**(10/(x**2 - 5*x))*x**4 - 10*e* *(10/(x**2 - 5*x))*x**3 + 25*e**(10/(x**2 - 5*x))*x**2),x) + 35*int(e**((e **(10/(x**2 - 5*x))*e**3*x**2 + e**x)/(e**(10/(x**2 - 5*x))*e**3*x))/(e**( 10/(x**2 - 5*x))*x**3 - 10*e**(10/(x**2 - 5*x))*x**2 + 25*e**(10/(x**2 - 5 *x))*x),x) - 11*int(e**((e**(10/(x**2 - 5*x))*e**3*x**2 + e**x)/(e**(10/(x **2 - 5*x))*e**3*x))/(e**(10/(x**2 - 5*x))*x**2 - 10*e**(10/(x**2 - 5*x))* x + 25*e**(10/(x**2 - 5*x))),x) + int((e**((e**(10/(x**2 - 5*x))*e**3*x**2 + e**x)/(e**(10/(x**2 - 5*x))*e**3*x))*x)/(e**(10/(x**2 - 5*x))*x**2 - 10 *e**(10/(x**2 - 5*x))*x + 25*e**(10/(x**2 - 5*x))),x))/e**3