\(\int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x (-30 x^3+e^{\frac {1+4 e^2}{e^2}} (250 x-250 x^2) \log (2))}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx\) [2036]

Optimal result

Integrand size = 139, antiderivative size = 30 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=x \left (-e^{-4-\frac {1}{e^2}} x+\frac {x \log (2)}{\left (e^x+\frac {x}{5}\right )^2}\right ) \] Output:

x*(x/(1/5*x+exp(x))^2*ln(2)-x/exp(1/exp(2)+4))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=-e^{-4-\frac {1}{e^2}} x^2 \left (1-\frac {25 e^{4+\frac {1}{e^2}} \log (2)}{\left (5 e^x+x\right )^2}\right ) \] Input:

Integrate[(-250*E^(3*x)*x - 150*E^(2*x)*x^2 - 2*x^4 + E^x*(-30*x^3 + E^((1 
 + 4*E^2)/E^2)*(250*x - 250*x^2)*Log[2]))/(125*E^((1 + 4*E^2)/E^2 + 3*x) + 
 75*E^((1 + 4*E^2)/E^2 + 2*x)*x + 15*E^((1 + 4*E^2)/E^2 + x)*x^2 + E^((1 + 
 4*E^2)/E^2)*x^3),x]
 

Output:

-(E^(-4 - E^(-2))*x^2*(1 - (25*E^(4 + E^(-2))*Log[2])/(5*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.

\(\displaystyle \int \frac {-2 x^4-150 e^{2 x} x^2+e^x \left (e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)-30 x^3\right )-250 e^{3 x} x}{e^{\frac {1+4 e^2}{e^2}} x^3+15 e^{x+\frac {1+4 e^2}{e^2}} x^2+75 e^{2 x+\frac {1+4 e^2}{e^2}} x+125 e^{3 x+\frac {1+4 e^2}{e^2}}} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {2 e^{-4-\frac {1}{e^2}} x \left (-x^3-15 e^x x^2-75 e^{2 x} x-125 e^{3 x}-125 e^{x+\frac {1}{e^2}+4} (x-1) \log (2)\right )}{\left (x+5 e^x\right )^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 e^{-4-\frac {1}{e^2}} \int -\frac {x \left (x^3+15 e^x x^2+75 e^{2 x} x+125 e^{3 x}-125 e^{x+\frac {1}{e^2}+4} (1-x) \log (2)\right )}{\left (x+5 e^x\right )^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -2 e^{-4-\frac {1}{e^2}} \int \frac {x \left (x^3+15 e^x x^2+75 e^{2 x} x+125 e^{3 x}-125 e^{x+\frac {1}{e^2}+4} (1-x) \log (2)\right )}{\left (x+5 e^x\right )^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -2 e^{-4-\frac {1}{e^2}} \int \left (-\frac {25 e^{4+\frac {1}{e^2}} (x-1) \log (2) x^2}{\left (x+5 e^x\right )^3}+\frac {25 e^{4+\frac {1}{e^2}} (x-1) \log (2) x}{\left (x+5 e^x\right )^2}+x\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 e^{-4-\frac {1}{e^2}} \left (-25 e^{4+\frac {1}{e^2}} \log (2) \int \frac {x^3}{\left (x+5 e^x\right )^3}dx+25 e^{4+\frac {1}{e^2}} \log (2) \int \frac {x^2}{\left (x+5 e^x\right )^3}dx+25 e^{4+\frac {1}{e^2}} \log (2) \int \frac {x^2}{\left (x+5 e^x\right )^2}dx-25 e^{4+\frac {1}{e^2}} \log (2) \int \frac {x}{\left (x+5 e^x\right )^2}dx+\frac {x^2}{2}\right )\)

Input:

Int[(-250*E^(3*x)*x - 150*E^(2*x)*x^2 - 2*x^4 + E^x*(-30*x^3 + E^((1 + 4*E 
^2)/E^2)*(250*x - 250*x^2)*Log[2]))/(125*E^((1 + 4*E^2)/E^2 + 3*x) + 75*E^ 
((1 + 4*E^2)/E^2 + 2*x)*x + 15*E^((1 + 4*E^2)/E^2 + x)*x^2 + E^((1 + 4*E^2 
)/E^2)*x^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
risch	\(-x^{2} {\mathrm e}^{-{\mathrm e}^{-2}-4}+\frac {25 \ln \left (2\right ) x^{2}}{\left (5 \,{\mathrm e}^{x}+x \right )^{2}}\)	\(29\)
parallelrisch	\(\frac {\left (625 x^{2} \ln \left (2\right ) {\mathrm e}^{\left (4 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2}}-25 x^{4}-250 \,{\mathrm e}^{x} x^{3}-625 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-\left (4 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2}}}{25 x^{2}+250 \,{\mathrm e}^{x} x +625 \,{\mathrm e}^{2 x}}\)	\(75\)
norman	\(\frac {-625 \ln \left (2\right ) {\mathrm e}^{2 x}-250 x \ln \left (2\right ) {\mathrm e}^{x}-{\mathrm e}^{-{\mathrm e}^{-2}} {\mathrm e}^{-4} x^{4}-25 \,{\mathrm e}^{-{\mathrm e}^{-2}} {\mathrm e}^{-4} x^{2} {\mathrm e}^{2 x}-10 \,{\mathrm e}^{-{\mathrm e}^{-2}} {\mathrm e}^{-4} x^{3} {\mathrm e}^{x}}{\left (5 \,{\mathrm e}^{x}+x \right )^{2}}\)	\(80\)

Input:

int((-250*x*exp(x)^3-150*exp(x)^2*x^2+((-250*x^2+250*x)*ln(2)*exp((4*exp(2 
)+1)/exp(2))-30*x^3)*exp(x)-2*x^4)/(125*exp((4*exp(2)+1)/exp(2))*exp(x)^3+ 
75*x*exp((4*exp(2)+1)/exp(2))*exp(x)^2+15*x^2*exp((4*exp(2)+1)/exp(2))*exp 
(x)+x^3*exp((4*exp(2)+1)/exp(2))),x,method=_RETURNVERBOSE)
 

Output:

-x^2*exp(-exp(-2)-4)+25*ln(2)*x^2/(5*exp(x)+x)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (26) = 52\).

Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 5.00 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=-\frac {x^{4} e^{\left (2 \, {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} + 10 \, x^{3} e^{\left ({\left ({\left (x + 4\right )} e^{2} + 1\right )} e^{\left (-2\right )} + {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} - 25 \, x^{2} e^{\left (3 \, {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} \log \left (2\right ) + 25 \, x^{2} e^{\left (2 \, {\left ({\left (x + 4\right )} e^{2} + 1\right )} e^{\left (-2\right )}\right )}}{x^{2} e^{\left (3 \, {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} + 10 \, x e^{\left ({\left ({\left (x + 4\right )} e^{2} + 1\right )} e^{\left (-2\right )} + 2 \, {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} + 25 \, e^{\left (2 \, {\left ({\left (x + 4\right )} e^{2} + 1\right )} e^{\left (-2\right )} + {\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )}} \] Input:

integrate((-250*x*exp(x)^3-150*exp(x)^2*x^2+((-250*x^2+250*x)*log(2)*exp(( 
4*exp(2)+1)/exp(2))-30*x^3)*exp(x)-2*x^4)/(125*exp((4*exp(2)+1)/exp(2))*ex 
p(x)^3+75*x*exp((4*exp(2)+1)/exp(2))*exp(x)^2+15*x^2*exp((4*exp(2)+1)/exp( 
2))*exp(x)+x^3*exp((4*exp(2)+1)/exp(2))),x, algorithm="fricas")
 

Output:

-(x^4*e^(2*(4*e^2 + 1)*e^(-2)) + 10*x^3*e^(((x + 4)*e^2 + 1)*e^(-2) + (4*e 
^2 + 1)*e^(-2)) - 25*x^2*e^(3*(4*e^2 + 1)*e^(-2))*log(2) + 25*x^2*e^(2*((x 
 + 4)*e^2 + 1)*e^(-2)))/(x^2*e^(3*(4*e^2 + 1)*e^(-2)) + 10*x*e^(((x + 4)*e 
^2 + 1)*e^(-2) + 2*(4*e^2 + 1)*e^(-2)) + 25*e^(2*((x + 4)*e^2 + 1)*e^(-2) 
+ (4*e^2 + 1)*e^(-2)))
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=- \frac {x^{2}}{e^{4} e^{e^{-2}}} + \frac {x^{2} \log {\left (2 \right )}}{\frac {x^{2}}{25} + \frac {2 x e^{x}}{5} + e^{2 x}} \] Input:

integrate((-250*x*exp(x)**3-150*exp(x)**2*x**2+((-250*x**2+250*x)*ln(2)*ex 
p((4*exp(2)+1)/exp(2))-30*x**3)*exp(x)-2*x**4)/(125*exp((4*exp(2)+1)/exp(2 
))*exp(x)**3+75*x*exp((4*exp(2)+1)/exp(2))*exp(x)**2+15*x**2*exp((4*exp(2) 
+1)/exp(2))*exp(x)+x**3*exp((4*exp(2)+1)/exp(2))),x)
 

Output:

-x**2*exp(-4)*exp(-exp(-2)) + x**2*log(2)/(x**2/25 + 2*x*exp(x)/5 + exp(2* 
x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=-\frac {x^{4} + 10 \, x^{3} e^{x} - 25 \, x^{2} e^{\left (e^{\left (-2\right )} + 4\right )} \log \left (2\right ) + 25 \, x^{2} e^{\left (2 \, x\right )}}{x^{2} e^{\left (e^{\left (-2\right )} + 4\right )} + 10 \, x e^{\left (x + e^{\left (-2\right )} + 4\right )} + 25 \, e^{\left (2 \, x + e^{\left (-2\right )} + 4\right )}} \] Input:

integrate((-250*x*exp(x)^3-150*exp(x)^2*x^2+((-250*x^2+250*x)*log(2)*exp(( 
4*exp(2)+1)/exp(2))-30*x^3)*exp(x)-2*x^4)/(125*exp((4*exp(2)+1)/exp(2))*ex 
p(x)^3+75*x*exp((4*exp(2)+1)/exp(2))*exp(x)^2+15*x^2*exp((4*exp(2)+1)/exp( 
2))*exp(x)+x^3*exp((4*exp(2)+1)/exp(2))),x, algorithm="maxima")
 

Output:

-(x^4 + 10*x^3*e^x - 25*x^2*e^(e^(-2) + 4)*log(2) + 25*x^2*e^(2*x))/(x^2*e 
^(e^(-2) + 4) + 10*x*e^(x + e^(-2) + 4) + 25*e^(2*x + e^(-2) + 4))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (26) = 52\).

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.90 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=-\frac {x^{4} + 10 \, x^{3} e^{x} - 25 \, x^{2} e^{\left ({\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} \log \left (2\right ) + 25 \, x^{2} e^{\left (2 \, x\right )}}{x^{2} e^{\left ({\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )}\right )} + 10 \, x e^{\left ({\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )} + x\right )} + 25 \, e^{\left ({\left (4 \, e^{2} + 1\right )} e^{\left (-2\right )} + 2 \, x\right )}} \] Input:

integrate((-250*x*exp(x)^3-150*exp(x)^2*x^2+((-250*x^2+250*x)*log(2)*exp(( 
4*exp(2)+1)/exp(2))-30*x^3)*exp(x)-2*x^4)/(125*exp((4*exp(2)+1)/exp(2))*ex 
p(x)^3+75*x*exp((4*exp(2)+1)/exp(2))*exp(x)^2+15*x^2*exp((4*exp(2)+1)/exp( 
2))*exp(x)+x^3*exp((4*exp(2)+1)/exp(2))),x, algorithm="giac")
 

Output:

-(x^4 + 10*x^3*e^x - 25*x^2*e^((4*e^2 + 1)*e^(-2))*log(2) + 25*x^2*e^(2*x) 
)/(x^2*e^((4*e^2 + 1)*e^(-2)) + 10*x*e^((4*e^2 + 1)*e^(-2) + x) + 25*e^((4 
*e^2 + 1)*e^(-2) + 2*x))
 

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.17 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=-\frac {x^2\,\left (25\,{\mathrm {e}}^{2\,x}-25\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,\left (4\,{\mathrm {e}}^2+1\right )}\,\ln \left (2\right )+10\,x\,{\mathrm {e}}^x+x^2\right )}{25\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{-2}+4}+x^2\,{\mathrm {e}}^{{\mathrm {e}}^{-2}+4}+10\,x\,{\mathrm {e}}^{x+{\mathrm {e}}^{-2}+4}} \] Input:

int(-(250*x*exp(3*x) + exp(x)*(30*x^3 - exp(exp(-2)*(4*exp(2) + 1))*log(2) 
*(250*x - 250*x^2)) + 150*x^2*exp(2*x) + 2*x^4)/(x^3*exp(exp(-2)*(4*exp(2) 
 + 1)) + 125*exp(3*x)*exp(exp(-2)*(4*exp(2) + 1)) + 75*x*exp(2*x)*exp(exp( 
-2)*(4*exp(2) + 1)) + 15*x^2*exp(exp(-2)*(4*exp(2) + 1))*exp(x)),x)
 

Output:

-(x^2*(25*exp(2*x) - 25*exp(exp(-2)*(4*exp(2) + 1))*log(2) + 10*x*exp(x) + 
 x^2))/(25*exp(2*x + exp(-2) + 4) + x^2*exp(exp(-2) + 4) + 10*x*exp(x + ex 
p(-2) + 4))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.13 \[ \int \frac {-250 e^{3 x} x-150 e^{2 x} x^2-2 x^4+e^x \left (-30 x^3+e^{\frac {1+4 e^2}{e^2}} \left (250 x-250 x^2\right ) \log (2)\right )}{125 e^{\frac {1+4 e^2}{e^2}+3 x}+75 e^{\frac {1+4 e^2}{e^2}+2 x} x+15 e^{\frac {1+4 e^2}{e^2}+x} x^2+e^{\frac {1+4 e^2}{e^2}} x^3} \, dx=\frac {x^{2} \left (25 e^{\frac {1}{e^{2}}} \mathrm {log}\left (2\right ) e^{4}-25 e^{2 x}-10 e^{x} x -x^{2}\right )}{e^{\frac {1}{e^{2}}} e^{4} \left (25 e^{2 x}+10 e^{x} x +x^{2}\right )} \] Input:

int((-250*x*exp(x)^3-150*exp(x)^2*x^2+((-250*x^2+250*x)*log(2)*exp((4*exp( 
2)+1)/exp(2))-30*x^3)*exp(x)-2*x^4)/(125*exp((4*exp(2)+1)/exp(2))*exp(x)^3 
+75*x*exp((4*exp(2)+1)/exp(2))*exp(x)^2+15*x^2*exp((4*exp(2)+1)/exp(2))*ex 
p(x)+x^3*exp((4*exp(2)+1)/exp(2))),x)
 

Output:

(x**2*(25*e**(1/e**2)*log(2)*e**4 - 25*e**(2*x) - 10*e**x*x - x**2))/(e**( 
1/e**2)*e**4*(25*e**(2*x) + 10*e**x*x + x**2))