Integrand size = 62, antiderivative size = 30 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=\frac {e^{-e^x+x}}{5 (4+2 x) \left (1+x-x^2\right )} \] Output:
exp(2*x-ln(5*(4+2*x)*exp(x))-exp(x))/(-x^2+x+1)
Time = 2.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=-\frac {e^{-e^x+x}}{10 \left (-2-3 x+x^2+x^3\right )} \] Input:
Integrate[(E^(-E^x + x)*(-1 + 5*x + 2*x^2 - x^3 + E^x*(-2 - 3*x + x^2 + x^ 3)))/((20 + 10*x)*(2 + 5*x - 5*x^3 + x^5)),x]
Output:
-1/10*E^(-E^x + x)/(-2 - 3*x + x^2 + x^3)
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.
\(\displaystyle \int \frac {e^{x-e^x} \left (-x^3+2 x^2+e^x \left (x^3+x^2-3 x-2\right )+5 x-1\right )}{(10 x+20) \left (x^5-5 x^3+5 x+2\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^{x-e^x} (3-x) \left (-x^3+2 x^2+e^x \left (x^3+x^2-3 x-2\right )+5 x-1\right )}{25 (10 x+20) \left (x^2-x-1\right )}+\frac {2 e^{x-e^x} \left (-x^3+2 x^2+e^x \left (x^3+x^2-3 x-2\right )+5 x-1\right )}{5 (10 x+20)^2}+\frac {e^{x-e^x} (3-x) \left (-x^3+2 x^2+e^x \left (x^3+x^2-3 x-2\right )+5 x-1\right )}{5 (10 x+20) \left (x^2-x-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{x-e^x} \left (-x^3+2 x^2+e^x \left (x^3+x^2-3 x-2\right )+5 x-1\right )}{10 \left (-x^3-x^2+3 x+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{10} \int -\frac {e^{x-e^x} \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{\left (-x^3-x^2+3 x+2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{10} \int \frac {e^{x-e^x} \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{\left (-x^3-x^2+3 x+2\right )^2}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle -\frac {1}{10} \int \left (\frac {2 e^{x-e^x} \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{25 (x+2)}+\frac {e^{x-e^x} (5-2 x) \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{25 \left (x^2-x-1\right )}+\frac {e^{x-e^x} \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{25 (x+2)^2}+\frac {e^{x-e^x} (2-x) \left (x^3-2 x^2-5 x+e^x \left (-x^3-x^2+3 x+2\right )+1\right )}{5 \left (x^2-x-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{10} \left (\frac {2}{5} \left (1+\sqrt {5}\right ) \int \frac {e^{x-e^x}}{\left (-2 x+\sqrt {5}+1\right )^2}dx-\frac {4}{5} \int \frac {e^{x-e^x}}{\left (-2 x+\sqrt {5}+1\right )^2}dx+\frac {6 \int \frac {e^{x-e^x}}{-2 x+\sqrt {5}+1}dx}{5 \sqrt {5}}+\frac {1}{5} \int \frac {e^{x-e^x}}{(x+2)^2}dx-\frac {1}{5} \int \frac {e^{x-e^x}}{x+2}dx+\frac {1}{5} \int \frac {e^{2 x-e^x}}{x+2}dx+\frac {1}{25} \left (15-\sqrt {5}\right ) \int \frac {e^{x-e^x}}{2 x-\sqrt {5}-1}dx-\frac {2}{25} \left (5-\sqrt {5}\right ) \int \frac {e^{x-e^x}}{2 x-\sqrt {5}-1}dx-\frac {1}{5} \left (1-\sqrt {5}\right ) \int \frac {e^{2 x-e^x}}{2 x-\sqrt {5}-1}dx+\frac {2}{5} \left (1-\sqrt {5}\right ) \int \frac {e^{x-e^x}}{\left (2 x+\sqrt {5}-1\right )^2}dx-\frac {4}{5} \int \frac {e^{x-e^x}}{\left (2 x+\sqrt {5}-1\right )^2}dx+\frac {1}{25} \left (15+\sqrt {5}\right ) \int \frac {e^{x-e^x}}{2 x+\sqrt {5}-1}dx-\frac {2}{25} \left (5+\sqrt {5}\right ) \int \frac {e^{x-e^x}}{2 x+\sqrt {5}-1}dx+\frac {6 \int \frac {e^{x-e^x}}{2 x+\sqrt {5}-1}dx}{5 \sqrt {5}}-\frac {1}{5} \left (1+\sqrt {5}\right ) \int \frac {e^{2 x-e^x}}{2 x+\sqrt {5}-1}dx\right )\) |
Input:
Int[(E^(-E^x + x)*(-1 + 5*x + 2*x^2 - x^3 + E^x*(-2 - 3*x + x^2 + x^3)))/( (20 + 10*x)*(2 + 5*x - 5*x^3 + x^5)),x]
Output:
$Aborted
Time = 0.81 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{-\ln \left (\left (10 x +20\right ) {\mathrm e}^{x}\right )+2 x -{\mathrm e}^{x}}}{x^{2}-x -1}\) | \(33\) |
risch | \(-\frac {{\mathrm e}^{x +\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{3}}{2}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right )^{2} \operatorname {csgn}\left (i \left (2+x \right )\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{x} \left (2+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \left (2+x \right )\right )}{2}-{\mathrm e}^{x}}}{10 \left (x^{2}-x -1\right ) \left (2+x \right )}\) | \(109\) |
Input:
int(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-ln((10*x+20)*exp(x))+2*x -exp(x))/(x^5-5*x^3+5*x+2),x,method=_RETURNVERBOSE)
Output:
-1/(x^2-x-1)*exp(-ln((10*x+20)*exp(x))+2*x-exp(x))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=-\frac {e^{\left (2 \, x - e^{x} - \log \left (10 \, {\left (x + 2\right )} e^{x}\right )\right )}}{x^{2} - x - 1} \] Input:
integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp( x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x, algorithm="fricas")
Output:
-e^(2*x - e^x - log(10*(x + 2)*e^x))/(x^2 - x - 1)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=- \frac {e^{- x} e^{2 x - e^{x}}}{10 x^{3} + 10 x^{2} - 30 x - 20} \] Input:
integrate(((x**3+x**2-3*x-2)*exp(x)-x**3+2*x**2+5*x-1)*exp(-ln((10*x+20)*e xp(x))+2*x-exp(x))/(x**5-5*x**3+5*x+2),x)
Output:
-exp(-x)*exp(2*x - exp(x))/(10*x**3 + 10*x**2 - 30*x - 20)
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=-\frac {e^{\left (x - e^{x}\right )}}{10 \, {\left (x^{3} + x^{2} - 3 \, x - 2\right )}} \] Input:
integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp( x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x, algorithm="maxima")
Output:
-1/10*e^(x - e^x)/(x^3 + x^2 - 3*x - 2)
\[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=\int { -\frac {{\left (x^{3} - 2 \, x^{2} - {\left (x^{3} + x^{2} - 3 \, x - 2\right )} e^{x} - 5 \, x + 1\right )} e^{\left (2 \, x - e^{x} - \log \left (10 \, {\left (x + 2\right )} e^{x}\right )\right )}}{x^{5} - 5 \, x^{3} + 5 \, x + 2} \,d x } \] Input:
integrate(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp( x))+2*x-exp(x))/(x^5-5*x^3+5*x+2),x, algorithm="giac")
Output:
integrate(-(x^3 - 2*x^2 - (x^3 + x^2 - 3*x - 2)*e^x - 5*x + 1)*e^(2*x - e^ x - log(10*(x + 2)*e^x))/(x^5 - 5*x^3 + 5*x + 2), x)
Time = 2.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=\frac {{\mathrm {e}}^{2\,x-{\mathrm {e}}^x}}{20\,{\mathrm {e}}^x-10\,x^2\,{\mathrm {e}}^x-10\,x^3\,{\mathrm {e}}^x+30\,x\,{\mathrm {e}}^x} \] Input:
int(-(exp(2*x - exp(x) - log(exp(x)*(10*x + 20)))*(x^3 - 2*x^2 - 5*x + exp (x)*(3*x - x^2 - x^3 + 2) + 1))/(5*x - 5*x^3 + x^5 + 2),x)
Output:
exp(2*x - exp(x))/(20*exp(x) - 10*x^2*exp(x) - 10*x^3*exp(x) + 30*x*exp(x) )
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-e^x+x} \left (-1+5 x+2 x^2-x^3+e^x \left (-2-3 x+x^2+x^3\right )\right )}{(20+10 x) \left (2+5 x-5 x^3+x^5\right )} \, dx=-\frac {e^{x}}{10 e^{e^{x}} \left (x^{3}+x^{2}-3 x -2\right )} \] Input:
int(((x^3+x^2-3*x-2)*exp(x)-x^3+2*x^2+5*x-1)*exp(-log((10*x+20)*exp(x))+2* x-exp(x))/(x^5-5*x^3+5*x+2),x)
Output:
( - e**x)/(10*e**(e**x)*(x**3 + x**2 - 3*x - 2))