Integrand size = 96, antiderivative size = 27 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 e^{e^{\frac {x^2}{5+x}}-x} \left (1+\frac {10}{e^4}+x\right ) \] Output:
2*(x+1+10/exp(4))/exp(-exp(x^2/(5+x))+x)
Time = 5.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=e^{-4+e^{\frac {x^2}{5+x}}-x} \left (20+2 e^4 (1+x)\right ) \] Input:
Integrate[(E^(-4 + E^(x^2/(5 + x)) - x)*(-500 - 200*x - 20*x^2 + E^4*(-50* x - 20*x^2 - 2*x^3) + E^(x^2/(5 + x))*(200*x + 20*x^2 + E^4*(20*x + 22*x^2 + 2*x^3))))/(25 + 10*x + x^2),x]
Output:
E^(-4 + E^(x^2/(5 + x)) - x)*(20 + 2*E^4*(1 + 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.
\(\displaystyle \int \frac {e^{e^{\frac {x^2}{x+5}}-x-4} \left (-20 x^2+e^4 \left (-2 x^3-20 x^2-50 x\right )+e^{\frac {x^2}{x+5}} \left (20 x^2+e^4 \left (2 x^3+22 x^2+20 x\right )+200 x\right )-200 x-500\right )}{x^2+10 x+25} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{e^{\frac {x^2}{x+5}}-x-4} \left (-20 x^2+e^4 \left (-2 x^3-20 x^2-50 x\right )+e^{\frac {x^2}{x+5}} \left (20 x^2+e^4 \left (2 x^3+22 x^2+20 x\right )+200 x\right )-200 x-500\right )}{(x+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{\frac {x^2}{x+5}+e^{\frac {x^2}{x+5}}-x-4} x (x+10) \left (e^4 x+e^4+10\right )}{(x+5)^2}-2 e^{e^{\frac {x^2}{x+5}}-x-4} \left (e^4 x+10\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \int e^{-x+e^{\frac {x^2}{x+5}}-4}dx+2 \left (10+e^4\right ) \int e^{\frac {x^2}{x+5}-x+e^{\frac {x^2}{x+5}}-4}dx-2 \int e^{e^{\frac {x^2}{x+5}}-x} xdx+2 \int e^{\frac {x^2}{x+5}-x+e^{\frac {x^2}{x+5}}} xdx-100 \left (5-2 e^4\right ) \int \frac {e^{\frac {x^2}{x+5}-x+e^{\frac {x^2}{x+5}}-4}}{(x+5)^2}dx-50 \int \frac {e^{\frac {x^2}{x+5}-x+e^{\frac {x^2}{x+5}}}}{x+5}dx\) |
Input:
Int[(E^(-4 + E^(x^2/(5 + x)) - x)*(-500 - 200*x - 20*x^2 + E^4*(-50*x - 20 *x^2 - 2*x^3) + E^(x^2/(5 + x))*(200*x + 20*x^2 + E^4*(20*x + 22*x^2 + 2*x ^3))))/(25 + 10*x + x^2),x]
Output:
$Aborted
Time = 0.80 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\left (2 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}+20\right ) {\mathrm e}^{-4+{\mathrm e}^{\frac {x^{2}}{5+x}}-x}\) | \(29\) |
parallelrisch | \({\mathrm e}^{-4} \left (2 x \,{\mathrm e}^{4}+2 \,{\mathrm e}^{4}+20\right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{5+x}}-x}\) | \(34\) |
norman | \(\frac {\left (2 x^{2}+10 \left (10+{\mathrm e}^{4}\right ) {\mathrm e}^{-4}+4 \left (5+3 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-4} x \right ) {\mathrm e}^{{\mathrm e}^{\frac {x^{2}}{5+x}}-x}}{5+x}\) | \(53\) |
Input:
int((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x ^2-50*x)*exp(4)-20*x^2-200*x-500)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x)) +x),x,method=_RETURNVERBOSE)
Output:
(2*x*exp(4)+2*exp(4)+20)*exp(-4+exp(x^2/(5+x))-x)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 \, {\left ({\left (x + 1\right )} e^{4} + 10\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )} \] Input:
integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^ 3-20*x^2-50*x)*exp(4)-20*x^2-200*x-500)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/ (5+x))+x),x, algorithm="fricas")
Output:
2*((x + 1)*e^4 + 10)*e^(-x + e^(x^2/(x + 5)) - 4)
Time = 11.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\frac {\left (2 x e^{4} + 20 + 2 e^{4}\right ) e^{- x + e^{\frac {x^{2}}{x + 5}}}}{e^{4}} \] Input:
integrate((((2*x**3+22*x**2+20*x)*exp(4)+20*x**2+200*x)*exp(x**2/(5+x))+(- 2*x**3-20*x**2-50*x)*exp(4)-20*x**2-200*x-500)/(x**2+10*x+25)/exp(4)/exp(- exp(x**2/(5+x))+x),x)
Output:
(2*x*exp(4) + 20 + 2*exp(4))*exp(-4)*exp(-x + exp(x**2/(x + 5)))
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=2 \, {\left (x e^{4} + e^{4} + 10\right )} e^{\left (-x + e^{\left (x + \frac {25}{x + 5} - 5\right )} - 4\right )} \] Input:
integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^ 3-20*x^2-50*x)*exp(4)-20*x^2-200*x-500)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/ (5+x))+x),x, algorithm="maxima")
Output:
2*(x*e^4 + e^4 + 10)*e^(-x + e^(x + 25/(x + 5) - 5) - 4)
\[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\int { -\frac {2 \, {\left (10 \, x^{2} + {\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{4} - {\left (10 \, x^{2} + {\left (x^{3} + 11 \, x^{2} + 10 \, x\right )} e^{4} + 100 \, x\right )} e^{\left (\frac {x^{2}}{x + 5}\right )} + 100 \, x + 250\right )} e^{\left (-x + e^{\left (\frac {x^{2}}{x + 5}\right )} - 4\right )}}{x^{2} + 10 \, x + 25} \,d x } \] Input:
integrate((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^ 3-20*x^2-50*x)*exp(4)-20*x^2-200*x-500)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/ (5+x))+x),x, algorithm="giac")
Output:
integrate(-2*(10*x^2 + (x^3 + 10*x^2 + 25*x)*e^4 - (10*x^2 + (x^3 + 11*x^2 + 10*x)*e^4 + 100*x)*e^(x^2/(x + 5)) + 100*x + 250)*e^(-x + e^(x^2/(x + 5 )) - 4)/(x^2 + 10*x + 25), x)
Time = 3.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {x^2}{x+5}}-x}\,\left (2\,x+{\mathrm {e}}^{-4}\,\left (2\,{\mathrm {e}}^4+20\right )\right ) \] Input:
int(-(exp(-4)*exp(exp(x^2/(x + 5)) - x)*(200*x - exp(x^2/(x + 5))*(200*x + exp(4)*(20*x + 22*x^2 + 2*x^3) + 20*x^2) + exp(4)*(50*x + 20*x^2 + 2*x^3) + 20*x^2 + 500))/(10*x + x^2 + 25),x)
Output:
exp(exp(x^2/(x + 5)) - x)*(2*x + exp(-4)*(2*exp(4) + 20))
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{-4+e^{\frac {x^2}{5+x}}-x} \left (-500-200 x-20 x^2+e^4 \left (-50 x-20 x^2-2 x^3\right )+e^{\frac {x^2}{5+x}} \left (200 x+20 x^2+e^4 \left (20 x+22 x^2+2 x^3\right )\right )\right )}{25+10 x+x^2} \, dx=\frac {2 e^{\frac {e^{\frac {x^{2}+5 x +25}{x +5}}}{e^{5}}} \left (e^{4} x +e^{4}+10\right )}{e^{x} e^{4}} \] Input:
int((((2*x^3+22*x^2+20*x)*exp(4)+20*x^2+200*x)*exp(x^2/(5+x))+(-2*x^3-20*x ^2-50*x)*exp(4)-20*x^2-200*x-500)/(x^2+10*x+25)/exp(4)/exp(-exp(x^2/(5+x)) +x),x)
Output:
(2*e**(e**((x**2 + 5*x + 25)/(x + 5))/e**5)*(e**4*x + e**4 + 10))/(e**x*e* *4)