\(\int \frac {e^{-\frac {x^2}{e^x+x^2}} (20 x^5-45 x^6+e^{2 x} (20 x-45 x^2)+e^x (20 x^3-50 x^4-15 x^5))}{e^{2 x}+2 e^x x^2+x^4} \, dx\) [99]

Optimal result

Integrand size = 82, antiderivative size = 29 \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=e^{-\frac {x^2}{e^x+x^2}} (1+3 (3-5 x)) x^2 \] Output:

x^2*(10-15*x)/exp(x^2/(x^2+exp(x)))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=-5 e^{-\frac {x^2}{e^x+x^2}} x^2 (-2+3 x) \] Input:

Integrate[(20*x^5 - 45*x^6 + E^(2*x)*(20*x - 45*x^2) + E^x*(20*x^3 - 50*x^ 
4 - 15*x^5))/(E^(x^2/(E^x + x^2))*(E^(2*x) + 2*E^x*x^2 + x^4)),x]
 

Output:

(-5*x^2*(-2 + 3*x))/E^(x^2/(E^x + x^2))
 

Rubi [F]

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.

\(\displaystyle \int \frac {e^{-\frac {x^2}{x^2+e^x}} \left (-45 x^6+20 x^5+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (-15 x^5-50 x^4+20 x^3\right )\right )}{x^4+2 e^x x^2+e^{2 x}} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-\frac {x^2}{x^2+e^x}} \left (-45 x^6+20 x^5+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (-15 x^5-50 x^4+20 x^3\right )\right )}{\left (x^2+e^x\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-5 e^{-\frac {x^2}{x^2+e^x}} (9 x-4) x+\frac {5 e^{-\frac {x^2}{x^2+e^x}} \left (3 x^2-8 x+4\right ) x^5}{\left (x^2+e^x\right )^2}-\frac {5 e^{-\frac {x^2}{x^2+e^x}} \left (3 x^2-8 x+4\right ) x^3}{x^2+e^x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 20 \int e^{-\frac {x^2}{x^2+e^x}} xdx-45 \int e^{-\frac {x^2}{x^2+e^x}} x^2dx+15 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^7}{\left (x^2+e^x\right )^2}dx-40 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^6}{\left (x^2+e^x\right )^2}dx+20 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^5}{\left (x^2+e^x\right )^2}dx-15 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^5}{x^2+e^x}dx+40 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^4}{x^2+e^x}dx-20 \int \frac {e^{-\frac {x^2}{x^2+e^x}} x^3}{x^2+e^x}dx\)

Input:

Int[(20*x^5 - 45*x^6 + E^(2*x)*(20*x - 45*x^2) + E^x*(20*x^3 - 50*x^4 - 15 
*x^5))/(E^(x^2/(E^x + x^2))*(E^(2*x) + 2*E^x*x^2 + x^4)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
risch	\(\left (-15 x^{3}+10 x^{2}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}+{\mathrm e}^{x}}}\)	\(27\)
parallelrisch	\(-\frac {\left (90 x^{5}-60 x^{4}+90 \,{\mathrm e}^{x} x^{3}-60 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-\frac {x^{2}}{x^{2}+{\mathrm e}^{x}}}}{6 \left (x^{2}+{\mathrm e}^{x}\right )}\)	\(51\)

Input:

int(((-45*x^2+20*x)*exp(x)^2+(-15*x^5-50*x^4+20*x^3)*exp(x)-45*x^6+20*x^5) 
/(exp(x)^2+2*exp(x)*x^2+x^4)/exp(x^2/(x^2+exp(x))),x,method=_RETURNVERBOSE 
)
 

Output:

(-15*x^3+10*x^2)*exp(-x^2/(x^2+exp(x)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=-5 \, {\left (3 \, x^{3} - 2 \, x^{2}\right )} e^{\left (-\frac {x^{2}}{x^{2} + e^{x}}\right )} \] Input:

integrate(((-45*x^2+20*x)*exp(x)^2+(-15*x^5-50*x^4+20*x^3)*exp(x)-45*x^6+2 
0*x^5)/(exp(x)^2+2*exp(x)*x^2+x^4)/exp(x^2/(x^2+exp(x))),x, algorithm="fri 
cas")
 

Output:

-5*(3*x^3 - 2*x^2)*e^(-x^2/(x^2 + e^x))
 

Sympy [A] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=\left (- 15 x^{3} + 10 x^{2}\right ) e^{- \frac {x^{2}}{x^{2} + e^{x}}} \] Input:

integrate(((-45*x**2+20*x)*exp(x)**2+(-15*x**5-50*x**4+20*x**3)*exp(x)-45* 
x**6+20*x**5)/(exp(x)**2+2*exp(x)*x**2+x**4)/exp(x**2/(x**2+exp(x))),x)
 

Output:

(-15*x**3 + 10*x**2)*exp(-x**2/(x**2 + exp(x)))
 

Maxima [F]

\[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=\int { -\frac {5 \, {\left (9 \, x^{6} - 4 \, x^{5} + {\left (9 \, x^{2} - 4 \, x\right )} e^{\left (2 \, x\right )} + {\left (3 \, x^{5} + 10 \, x^{4} - 4 \, x^{3}\right )} e^{x}\right )} e^{\left (-\frac {x^{2}}{x^{2} + e^{x}}\right )}}{x^{4} + 2 \, x^{2} e^{x} + e^{\left (2 \, x\right )}} \,d x } \] Input:

integrate(((-45*x^2+20*x)*exp(x)^2+(-15*x^5-50*x^4+20*x^3)*exp(x)-45*x^6+2 
0*x^5)/(exp(x)^2+2*exp(x)*x^2+x^4)/exp(x^2/(x^2+exp(x))),x, algorithm="max 
ima")
 

Output:

-5*integrate((9*x^6 - 4*x^5 + (9*x^2 - 4*x)*e^(2*x) + (3*x^5 + 10*x^4 - 4* 
x^3)*e^x)*e^(-x^2/(x^2 + e^x))/(x^4 + 2*x^2*e^x + e^(2*x)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=-5 \, {\left (3 \, x^{3} e^{\left (\frac {x^{3} - x^{2} + x e^{x}}{x^{2} + e^{x}}\right )} - 2 \, x^{2} e^{\left (\frac {x^{3} - x^{2} + x e^{x}}{x^{2} + e^{x}}\right )}\right )} e^{\left (-x\right )} \] Input:

integrate(((-45*x^2+20*x)*exp(x)^2+(-15*x^5-50*x^4+20*x^3)*exp(x)-45*x^6+2 
0*x^5)/(exp(x)^2+2*exp(x)*x^2+x^4)/exp(x^2/(x^2+exp(x))),x, algorithm="gia 
c")
 

Output:

-5*(3*x^3*e^((x^3 - x^2 + x*e^x)/(x^2 + e^x)) - 2*x^2*e^((x^3 - x^2 + x*e^ 
x)/(x^2 + e^x)))*e^(-x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=\int \frac {{\mathrm {e}}^{-\frac {x^2}{{\mathrm {e}}^x+x^2}}\,\left ({\mathrm {e}}^{2\,x}\,\left (20\,x-45\,x^2\right )-{\mathrm {e}}^x\,\left (15\,x^5+50\,x^4-20\,x^3\right )+20\,x^5-45\,x^6\right )}{{\mathrm {e}}^{2\,x}+2\,x^2\,{\mathrm {e}}^x+x^4} \,d x \] Input:

int((exp(-x^2/(exp(x) + x^2))*(exp(2*x)*(20*x - 45*x^2) - exp(x)*(50*x^4 - 
 20*x^3 + 15*x^5) + 20*x^5 - 45*x^6))/(exp(2*x) + 2*x^2*exp(x) + x^4),x)
 

Output:

int((exp(-x^2/(exp(x) + x^2))*(exp(2*x)*(20*x - 45*x^2) - exp(x)*(50*x^4 - 
 20*x^3 + 15*x^5) + 20*x^5 - 45*x^6))/(exp(2*x) + 2*x^2*exp(x) + x^4), x)
 

Reduce [F]

\[ \int \frac {e^{-\frac {x^2}{e^x+x^2}} \left (20 x^5-45 x^6+e^{2 x} \left (20 x-45 x^2\right )+e^x \left (20 x^3-50 x^4-15 x^5\right )\right )}{e^{2 x}+2 e^x x^2+x^4} \, dx=-45 \left (\int \frac {x^{6}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )+20 \left (\int \frac {x^{5}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )-45 \left (\int \frac {e^{2 x} x^{2}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )+20 \left (\int \frac {e^{2 x} x}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )-15 \left (\int \frac {e^{x} x^{5}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )-50 \left (\int \frac {e^{x} x^{4}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right )+20 \left (\int \frac {e^{x} x^{3}}{e^{\frac {2 e^{x} x +2 x^{3}+x^{2}}{e^{x}+x^{2}}}+2 e^{\frac {e^{x} x +x^{3}+x^{2}}{e^{x}+x^{2}}} x^{2}+e^{\frac {x^{2}}{e^{x}+x^{2}}} x^{4}}d x \right ) \] Input:

int(((-45*x^2+20*x)*exp(x)^2+(-15*x^5-50*x^4+20*x^3)*exp(x)-45*x^6+20*x^5) 
/(exp(x)^2+2*exp(x)*x^2+x^4)/exp(x^2/(x^2+exp(x))),x)
 

Output:

5*( - 9*int(x**6/(e**((2*e**x*x + 2*x**3 + x**2)/(e**x + x**2)) + 2*e**((e 
**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x**2/(e**x + x**2))*x**4),x 
) + 4*int(x**5/(e**((2*e**x*x + 2*x**3 + x**2)/(e**x + x**2)) + 2*e**((e** 
x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x**2/(e**x + x**2))*x**4),x) 
- 9*int((e**(2*x)*x**2)/(e**((2*e**x*x + 2*x**3 + x**2)/(e**x + x**2)) + 2 
*e**((e**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x**2/(e**x + x**2))* 
x**4),x) + 4*int((e**(2*x)*x)/(e**((2*e**x*x + 2*x**3 + x**2)/(e**x + x**2 
)) + 2*e**((e**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x**2/(e**x + x 
**2))*x**4),x) - 3*int((e**x*x**5)/(e**((2*e**x*x + 2*x**3 + x**2)/(e**x + 
 x**2)) + 2*e**((e**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x**2/(e** 
x + x**2))*x**4),x) - 10*int((e**x*x**4)/(e**((2*e**x*x + 2*x**3 + x**2)/( 
e**x + x**2)) + 2*e**((e**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e**(x** 
2/(e**x + x**2))*x**4),x) + 4*int((e**x*x**3)/(e**((2*e**x*x + 2*x**3 + x* 
*2)/(e**x + x**2)) + 2*e**((e**x*x + x**3 + x**2)/(e**x + x**2))*x**2 + e* 
*(x**2/(e**x + x**2))*x**4),x))