\(\int \frac {4 x+e^{x+\frac {2 (10+2 x^2)}{x}} x^2+e^x (4-4 x^2+x^4)+e^{\frac {10+2 x^2}{x}} (-10 x-x^2+2 x^3+e^x (-4 x+2 x^3))}{4-4 x^2+e^{\frac {2 (10+2 x^2)}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} (-4 x+2 x^3)} \, dx\) [1199]

Optimal result

Integrand size = 133, antiderivative size = 28 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=1+e^x-\frac {x}{e^{\frac {10}{x}+2 x}-\frac {2}{x}+x} \] Output:

exp(x)-x/(exp(2*x+10/x)-2/x+x)+1
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.55 (sec) , antiderivative size = 901, normalized size of antiderivative = 32.18 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx =\text {Too large to display} \] Input:

Integrate[(4*x + E^(x + (2*(10 + 2*x^2))/x)*x^2 + E^x*(4 - 4*x^2 + x^4) + 
E^((10 + 2*x^2)/x)*(-10*x - x^2 + 2*x^3 + E^x*(-4*x + 2*x^3)))/(4 - 4*x^2 
+ E^((2*(10 + 2*x^2))/x)*x^2 + x^4 + E^((10 + 2*x^2)/x)*(-4*x + 2*x^3)),x]
 

Output:

((2095*E^x*(-1300 - 1030*x + 946*x^2 + 661*x^3))/(4*(-20 + 2*x + 14*x^2 + 
x^3 - 2*x^4)^2) - (16760*E^(10/x + 3*x)*x^3)/(20 - 2*x - 14*x^2 - x^3 + 2* 
x^4) + (E^x*(390015 + 82526*x - 110828*x^2 + 2612*x^3))/(80 - 8*x - 56*x^2 
 - 4*x^3 + 8*x^4) - 60060*RootSum[20 - 2*#1 - 14*#1^2 - #1^3 + 2*#1^4 & , 
(E^#1*ExpIntegralEi[x - #1])/(-2 - 28*#1 - 3*#1^2 + 8*#1^3) & ] - 52674*Ro 
otSum[20 - 2*#1 - 14*#1^2 - #1^3 + 2*#1^4 & , (E^#1*ExpIntegralEi[x - #1]* 
#1)/(-2 - 28*#1 - 3*#1^2 + 8*#1^3) & ] + 32550*RootSum[20 - 2*#1 - 14*#1^2 
 - #1^3 + 2*#1^4 & , (E^#1*ExpIntegralEi[x - #1]*#1^2)/(-2 - 28*#1 - 3*#1^ 
2 + 8*#1^3) & ] - 9033*RootSum[20 - 2*#1 - 14*#1^2 - #1^3 + 2*#1^4 & , (E^ 
#1*ExpIntegralEi[x - #1]*#1^3)/(-2 - 28*#1 - 3*#1^2 + 8*#1^3) & ])/16760 + 
 ((16760*E^(20/x + 5*x)*x^4)/((-2 + E^(10/x + 2*x)*x + x^2)*(20 - 2*x - 14 
*x^2 - x^3 + 2*x^4)) + (33520*E^(10/x + 3*x)*x^3*(-2 + x^2))/((-2 + E^(10/ 
x + 2*x)*x + x^2)*(20 - 2*x - 14*x^2 - x^3 + 2*x^4)) + (E^x*(5434800 - 101 
8960*x - 7183172*x^2 + 172792*x^3 + 3327352*x^4 + 275372*x^5 - 564053*x^6 
- 51586*x^7 + 33520*x^8))/(-20 + 2*x + 14*x^2 + x^3 - 2*x^4)^2 + (16760*x^ 
2*(-20 + 2*x + 14*x^2 + x^3 - 2*x^4 + E^x*(-2 + x^2)^2))/((-2 + E^(10/x + 
2*x)*x + x^2)*(20 - 2*x - 14*x^2 - x^3 + 2*x^4)) + 60060*RootSum[20 - 2*#1 
 - 14*#1^2 - #1^3 + 2*#1^4 & , (E^#1*ExpIntegralEi[x - #1])/(-2 - 28*#1 - 
3*#1^2 + 8*#1^3) & ] + 52674*RootSum[20 - 2*#1 - 14*#1^2 - #1^3 + 2*#1^4 & 
 , (E^#1*ExpIntegralEi[x - #1]*#1)/(-2 - 28*#1 - 3*#1^2 + 8*#1^3) & ] -...
 

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.

\(\displaystyle \int \frac {e^{\frac {2 \left (2 x^2+10\right )}{x}+x} x^2+e^x \left (x^4-4 x^2+4\right )+e^{\frac {2 x^2+10}{x}} \left (2 x^3+e^x \left (2 x^3-4 x\right )-x^2-10 x\right )+4 x}{x^4+e^{\frac {2 \left (2 x^2+10\right )}{x}} x^2-4 x^2+e^{\frac {2 x^2+10}{x}} \left (2 x^3-4 x\right )+4} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {2 \left (2 x^2+10\right )}{x}+x} x^2+e^x \left (x^4-4 x^2+4\right )+e^{\frac {2 x^2+10}{x}} \left (2 x^3+e^x \left (2 x^3-4 x\right )-x^2-10 x\right )+4 x}{\left (-x^2-e^{2 x+\frac {10}{x}} x+2\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x^2-x-10}{x^2+e^{2 x+\frac {10}{x}} x-2}-\frac {2 x^4-x^3-14 x^2-2 x+20}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}+e^x\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -20 \int \frac {1}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}dx+2 \int \frac {x}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}dx+14 \int \frac {x^2}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}dx-10 \int \frac {1}{x^2+e^{2 x+\frac {10}{x}} x-2}dx-\int \frac {x}{x^2+e^{2 x+\frac {10}{x}} x-2}dx+2 \int \frac {x^2}{x^2+e^{2 x+\frac {10}{x}} x-2}dx-2 \int \frac {x^4}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}dx+\int \frac {x^3}{\left (x^2+e^{2 x+\frac {10}{x}} x-2\right )^2}dx+e^x\)

Input:

Int[(4*x + E^(x + (2*(10 + 2*x^2))/x)*x^2 + E^x*(4 - 4*x^2 + x^4) + E^((10 
 + 2*x^2)/x)*(-10*x - x^2 + 2*x^3 + E^x*(-4*x + 2*x^3)))/(4 - 4*x^2 + E^(( 
2*(10 + 2*x^2))/x)*x^2 + x^4 + E^((10 + 2*x^2)/x)*(-4*x + 2*x^3)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\({\mathrm e}^{x}-\frac {x^{2}}{x^{2}+x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2}\)	\(29\)
parts	\(\frac {x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2}{x^{2}+x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2}+{\mathrm e}^{x}\)	\(42\)
parallelrisch	\(\frac {{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}+{\mathrm e}^{x} x^{2}-x^{2}-2 \,{\mathrm e}^{x}}{x^{2}+x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2}\)	\(53\)
norman	\(\frac {{\mathrm e}^{x} x^{2}+x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}+{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2 \,{\mathrm e}^{x}-2}{x^{2}+x \,{\mathrm e}^{\frac {2 x^{2}+10}{x}}-2}\)	\(65\)

Input:

int((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10*x)*ex 
p((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+(2*x^3- 
4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x,method=_RETURNVERBOSE)
 

Output:

exp(x)-x^2/(exp(2*(x^2+5)/x)*x+x^2-2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (26) = 52\).

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.93 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=-\frac {x^{2} e^{\left (\frac {4 \, {\left (x^{2} + 5\right )}}{x}\right )} - {\left (x^{2} + x e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - 2\right )} e^{\left (\frac {5 \, {\left (x^{2} + 4\right )}}{x}\right )}}{x e^{\left (\frac {6 \, {\left (x^{2} + 5\right )}}{x}\right )} + {\left (x^{2} - 2\right )} e^{\left (\frac {4 \, {\left (x^{2} + 5\right )}}{x}\right )}} \] Input:

integrate((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10 
*x)*exp((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+( 
2*x^3-4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x, algorithm="fricas")
 

Output:

-(x^2*e^(4*(x^2 + 5)/x) - (x^2 + x*e^(2*(x^2 + 5)/x) - 2)*e^(5*(x^2 + 4)/x 
))/(x*e^(6*(x^2 + 5)/x) + (x^2 - 2)*e^(4*(x^2 + 5)/x))
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=- \frac {x^{2}}{x^{2} + x e^{\frac {2 x^{2} + 10}{x}} - 2} + e^{x} \] Input:

integrate((x**2*exp(x)*exp((2*x**2+10)/x)**2+((2*x**3-4*x)*exp(x)+2*x**3-x 
**2-10*x)*exp((2*x**2+10)/x)+(x**4-4*x**2+4)*exp(x)+4*x)/(x**2*exp((2*x**2 
+10)/x)**2+(2*x**3-4*x)*exp((2*x**2+10)/x)+x**4-4*x**2+4),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=-\frac {x^{2} - x e^{\left (3 \, x + \frac {10}{x}\right )} - {\left (x^{2} - 2\right )} e^{x}}{x^{2} + x e^{\left (2 \, x + \frac {10}{x}\right )} - 2} \] Input:

integrate((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10 
*x)*exp((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+( 
2*x^3-4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (26) = 52\).

Time = 0.16 (sec) , antiderivative size = 304, normalized size of antiderivative = 10.86 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=\frac {2 \, x^{6} e^{x} - 2 \, x^{6} - x^{5} e^{x} + 2 \, x^{5} e^{\left (\frac {3 \, x^{2} + 10}{x}\right )} + x^{5} - x^{4} e^{\left (x + \frac {4 \, {\left (x^{2} + 5\right )}}{x}\right )} - x^{4} e^{\left (x + \frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - 18 \, x^{4} e^{x} + x^{4} e^{\left (\frac {5 \, {\left (x^{2} + 4\right )}}{x}\right )} + 14 \, x^{4} - 10 \, x^{3} e^{\left (x + \frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - 4 \, x^{3} e^{\left (\frac {3 \, x^{2} + 10}{x}\right )} + 2 \, x^{3} - 2 \, x^{2} e^{\left (x + \frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} + 48 \, x^{2} e^{x} - 20 \, x^{2} + 20 \, x e^{\left (x + \frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} + 4 \, x e^{x} - 40 \, e^{x}}{2 \, x^{6} + 2 \, x^{5} e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - x^{5} - x^{4} e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - 18 \, x^{4} - 14 \, x^{3} e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} - 2 \, x^{2} e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} + 48 \, x^{2} + 20 \, x e^{\left (\frac {2 \, {\left (x^{2} + 5\right )}}{x}\right )} + 4 \, x - 40} \] Input:

integrate((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10 
*x)*exp((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+( 
2*x^3-4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x, algorithm="giac")
 

Output:

(2*x^6*e^x - 2*x^6 - x^5*e^x + 2*x^5*e^((3*x^2 + 10)/x) + x^5 - x^4*e^(x + 
 4*(x^2 + 5)/x) - x^4*e^(x + 2*(x^2 + 5)/x) - 18*x^4*e^x + x^4*e^(5*(x^2 + 
 4)/x) + 14*x^4 - 10*x^3*e^(x + 2*(x^2 + 5)/x) - 4*x^3*e^((3*x^2 + 10)/x) 
+ 2*x^3 - 2*x^2*e^(x + 2*(x^2 + 5)/x) + 48*x^2*e^x - 20*x^2 + 20*x*e^(x + 
2*(x^2 + 5)/x) + 4*x*e^x - 40*e^x)/(2*x^6 + 2*x^5*e^(2*(x^2 + 5)/x) - x^5 
- x^4*e^(2*(x^2 + 5)/x) - 18*x^4 - 14*x^3*e^(2*(x^2 + 5)/x) - 2*x^2*e^(2*( 
x^2 + 5)/x) + 48*x^2 + 20*x*e^(2*(x^2 + 5)/x) + 4*x - 40)
 

Mupad [B] (verification not implemented)

Time = 3.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx={\mathrm {e}}^x-\frac {x^2}{x^2+x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{10/x}-2} \] Input:

int((4*x - exp((2*x^2 + 10)/x)*(10*x + exp(x)*(4*x - 2*x^3) + x^2 - 2*x^3) 
 + exp(x)*(x^4 - 4*x^2 + 4) + x^2*exp((2*(2*x^2 + 10))/x)*exp(x))/(x^2*exp 
((2*(2*x^2 + 10))/x) - 4*x^2 + x^4 - exp((2*x^2 + 10)/x)*(4*x - 2*x^3) + 4 
),x)
 

Output:

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

Reduce [F]

\[ \int \frac {4 x+e^{x+\frac {2 \left (10+2 x^2\right )}{x}} x^2+e^x \left (4-4 x^2+x^4\right )+e^{\frac {10+2 x^2}{x}} \left (-10 x-x^2+2 x^3+e^x \left (-4 x+2 x^3\right )\right )}{4-4 x^2+e^{\frac {2 \left (10+2 x^2\right )}{x}} x^2+x^4+e^{\frac {10+2 x^2}{x}} \left (-4 x+2 x^3\right )} \, dx=\int \frac {x^{2} {\mathrm e}^{x} \left ({\mathrm e}^{\frac {2 x^{2}+10}{x}}\right )^{2}+\left (\left (2 x^{3}-4 x \right ) {\mathrm e}^{x}+2 x^{3}-x^{2}-10 x \right ) {\mathrm e}^{\frac {2 x^{2}+10}{x}}+\left (x^{4}-4 x^{2}+4\right ) {\mathrm e}^{x}+4 x}{x^{2} \left ({\mathrm e}^{\frac {2 x^{2}+10}{x}}\right )^{2}+\left (2 x^{3}-4 x \right ) {\mathrm e}^{\frac {2 x^{2}+10}{x}}+x^{4}-4 x^{2}+4}d x \] Input:

int((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10*x)*ex 
p((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+(2*x^3- 
4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x)
 

Output:

int((x^2*exp(x)*exp((2*x^2+10)/x)^2+((2*x^3-4*x)*exp(x)+2*x^3-x^2-10*x)*ex 
p((2*x^2+10)/x)+(x^4-4*x^2+4)*exp(x)+4*x)/(x^2*exp((2*x^2+10)/x)^2+(2*x^3- 
4*x)*exp((2*x^2+10)/x)+x^4-4*x^2+4),x)