\(\int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} (-10 x^3+e^5 (1+2 x^2))-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+(-8 x^2+2 x^3) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} (8 x^2-2 x^3-2 x^2 (i \pi +\log (25)))} \, dx\) [2423]

Optimal result

Integrand size = 143, antiderivative size = 30 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {e^5-5 x}{x \left (-4-e^{x^2}+i \pi +x+\log (25)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {e^5-5 x}{x \left (-4-e^{x^2}+i \pi +x+\log (25)\right )} \] Input:

Integrate[(E^5*(4 - 2*x) + 5*x^2 + E^x^2*(-10*x^3 + E^5*(1 + 2*x^2)) - E^5 
*(I*Pi + Log[25]))/(16*x^2 + E^(2*x^2)*x^2 - 8*x^3 + x^4 + (-8*x^2 + 2*x^3 
)*(I*Pi + Log[25]) + x^2*(I*Pi + Log[25])^2 + E^x^2*(8*x^2 - 2*x^3 - 2*x^2 
*(I*Pi + Log[25]))),x]
 

Output:

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

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.

\(\displaystyle \int \frac {5 x^2+e^{x^2} \left (e^5 \left (2 x^2+1\right )-10 x^3\right )+e^5 (4-2 x)-e^5 (\log (25)+i \pi )}{x^4-8 x^3+e^{2 x^2} x^2+16 x^2+x^2 (\log (25)+i \pi )^2+e^{x^2} \left (-2 x^3+8 x^2-2 x^2 (\log (25)+i \pi )\right )+\left (2 x^3-8 x^2\right ) (\log (25)+i \pi )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {5 x^2+e^{x^2} \left (e^5 \left (2 x^2+1\right )-10 x^3\right )+e^5 (4-2 x)-e^5 (\log (25)+i \pi )}{x^4-8 x^3+e^{2 x^2} x^2+x^2 \left (16+(\log (25)+i \pi )^2\right )+e^{x^2} \left (-2 x^3+8 x^2-2 x^2 (\log (25)+i \pi )\right )+\left (2 x^3-8 x^2\right ) (\log (25)+i \pi )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {5 x^2+e^{x^2+5} \left (2 x^2+1\right )-10 e^{x^2} x^3+e^5 (-2 x-i \pi +4-\log (25))}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {\left (e^5-5 x\right ) \left (2 x^2-x (8-2 i \pi -\log (625))-1\right )}{x \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}+\frac {-10 x^3+2 e^5 x^2+e^5}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 10 \int \frac {x}{x-e^{x^2}-4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )}dx-\left (5-e^5 (8-2 i \pi -\log (625))\right ) \int \frac {1}{\left (-i x+i e^{x^2}+4 i \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}dx-e^5 \int \frac {1}{x \left (-x+e^{x^2}+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}dx+\left (40+2 e^5-10 i \pi -5 \log (625)\right ) \int \frac {x}{\left (-x+e^{x^2}+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}dx-10 \int \frac {x^2}{\left (-x+e^{x^2}+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}dx+2 e^5 \int \frac {1}{-x+e^{x^2}+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )}dx+e^5 \int \frac {1}{x^2 \left (-x+e^{x^2}+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )}dx\)

Input:

Int[(E^5*(4 - 2*x) + 5*x^2 + E^x^2*(-10*x^3 + E^5*(1 + 2*x^2)) - E^5*(I*Pi 
 + Log[25]))/(16*x^2 + E^(2*x^2)*x^2 - 8*x^3 + x^4 + (-8*x^2 + 2*x^3)*(I*P 
i + Log[25]) + x^2*(I*Pi + Log[25])^2 + E^x^2*(8*x^2 - 2*x^3 - 2*x^2*(I*Pi 
 + Log[25]))),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{5}-5 x}{x \left (2 \ln \left (5\right )+i \pi -4-{\mathrm e}^{x^{2}}+x \right )}\)	\(30\)
parallelrisch	\(\frac {{\mathrm e}^{5}-5 x}{x \left (2 \ln \left (5\right )+i \pi -4-{\mathrm e}^{x^{2}}+x \right )}\)	\(30\)

Input:

int((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*ln(5)+I*Pi)+(4-2*x)*exp( 
5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2*ln(5)+I*Pi)-2*x^3+8*x^2)*exp(x^2)+x^2 
*(2*ln(5)+I*Pi)^2+(2*x^3-8*x^2)*(2*ln(5)+I*Pi)+x^4-8*x^3+16*x^2),x,method= 
_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {5 \, x - e^{5}}{{\left (-i \, \pi + 4\right )} x - x^{2} + x e^{\left (x^{2}\right )} - 2 \, x \log \left (5\right )} \] Input:

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2* 
x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp( 
x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^2) 
,x, algorithm="fricas")
 

Output:

(5*x - e^5)/((-I*pi + 4)*x - x^2 + x*e^(x^2) - 2*x*log(5))
 

Sympy [A] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {5 x - e^{5}}{- x^{2} + x e^{x^{2}} - 2 x \log {\left (5 \right )} + 4 x - i \pi x} \] Input:

integrate((((2*x**2+1)*exp(5)-10*x**3)*exp(x**2)-exp(5)*(2*ln(5)+I*pi)+(4- 
2*x)*exp(5)+5*x**2)/(x**2*exp(x**2)**2+(-2*x**2*(2*ln(5)+I*pi)-2*x**3+8*x* 
*2)*exp(x**2)+x**2*(2*ln(5)+I*pi)**2+(2*x**3-8*x**2)*(2*ln(5)+I*pi)+x**4-8 
*x**3+16*x**2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {5 \, x - e^{5}}{{\left (-i \, \pi - 2 \, \log \left (5\right ) + 4\right )} x - x^{2} + x e^{\left (x^{2}\right )}} \] Input:

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2* 
x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp( 
x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^2) 
,x, algorithm="maxima")
 

Output:

(5*x - e^5)/((-I*pi - 2*log(5) + 4)*x - x^2 + x*e^(x^2))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\frac {5 i \, x - i \, e^{5}}{\pi x - i \, x^{2} + i \, x e^{\left (x^{2}\right )} - 2 i \, x \log \left (5\right ) + 4 i \, x} \] Input:

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2* 
x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp( 
x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^2) 
,x, algorithm="giac")
 

Output:

(5*I*x - I*e^5)/(pi*x - I*x^2 + I*x*e^(x^2) - 2*I*x*log(5) + 4*I*x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx=\int \frac {{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^5\,\left (2\,x^2+1\right )-10\,x^3\right )-{\mathrm {e}}^5\,\left (2\,\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )+5\,x^2-{\mathrm {e}}^5\,\left (2\,x-4\right )}{x^2\,{\mathrm {e}}^{2\,x^2}+x^2\,{\left (2\,\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )}^2-{\mathrm {e}}^{x^2}\,\left (2\,x^2\,\left (2\,\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )-8\,x^2+2\,x^3\right )-\left (8\,x^2-2\,x^3\right )\,\left (2\,\ln \left (5\right )+\Pi \,1{}\mathrm {i}\right )+16\,x^2-8\,x^3+x^4} \,d x \] Input:

int((exp(x^2)*(exp(5)*(2*x^2 + 1) - 10*x^3) - exp(5)*(Pi*1i + 2*log(5)) + 
5*x^2 - exp(5)*(2*x - 4))/(x^2*(Pi*1i + 2*log(5))^2 - (8*x^2 - 2*x^3)*(Pi* 
1i + 2*log(5)) - exp(x^2)*(2*x^2*(Pi*1i + 2*log(5)) - 8*x^2 + 2*x^3) + x^2 
*exp(2*x^2) + 16*x^2 - 8*x^3 + x^4),x)
 

Output:

int((exp(x^2)*(exp(5)*(2*x^2 + 1) - 10*x^3) - exp(5)*(Pi*1i + 2*log(5)) + 
5*x^2 - exp(5)*(2*x - 4))/(x^2*(Pi*1i + 2*log(5))^2 - (8*x^2 - 2*x^3)*(Pi* 
1i + 2*log(5)) - exp(x^2)*(2*x^2*(Pi*1i + 2*log(5)) - 8*x^2 + 2*x^3) + x^2 
*exp(2*x^2) + 16*x^2 - 8*x^3 + x^4), x)
 

Reduce [F]

\[ \int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx =\text {Too large to display} \] Input:

int((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*Pi)+(4-2*x)*exp 
(5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2*log(5)+I*Pi)-2*x^3+8*x^2)*exp(x^2)+x 
^2*(2*log(5)+I*Pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*Pi)+x^4-8*x^3+16*x^2),x)
 

Output:

2*int(e**(x**2)/(e**(2*x**2) - 4*e**(x**2)*log(5) - 2*e**(x**2)*i*pi - 2*e 
**(x**2)*x + 8*e**(x**2) + 4*log(5)**2 + 4*log(5)*i*pi + 4*log(5)*x - 16*l 
og(5) + 2*i*pi*x - 8*i*pi - pi**2 + x**2 - 8*x + 16),x)*e**5 + int(e**(x** 
2)/(e**(2*x**2)*x**2 - 4*e**(x**2)*log(5)*x**2 - 2*e**(x**2)*i*pi*x**2 - 2 
*e**(x**2)*x**3 + 8*e**(x**2)*x**2 + 4*log(5)**2*x**2 + 4*log(5)*i*pi*x**2 
 + 4*log(5)*x**3 - 16*log(5)*x**2 + 2*i*pi*x**3 - 8*i*pi*x**2 - pi**2*x**2 
 + x**4 - 8*x**3 + 16*x**2),x)*e**5 - 10*int((e**(x**2)*x)/(e**(2*x**2) - 
4*e**(x**2)*log(5) - 2*e**(x**2)*i*pi - 2*e**(x**2)*x + 8*e**(x**2) + 4*lo 
g(5)**2 + 4*log(5)*i*pi + 4*log(5)*x - 16*log(5) + 2*i*pi*x - 8*i*pi - pi* 
*2 + x**2 - 8*x + 16),x) + 5*int(1/(e**(2*x**2) - 4*e**(x**2)*log(5) - 2*e 
**(x**2)*i*pi - 2*e**(x**2)*x + 8*e**(x**2) + 4*log(5)**2 + 4*log(5)*i*pi 
+ 4*log(5)*x - 16*log(5) + 2*i*pi*x - 8*i*pi - pi**2 + x**2 - 8*x + 16),x) 
 - 2*int(1/(e**(2*x**2)*x**2 - 4*e**(x**2)*log(5)*x**2 - 2*e**(x**2)*i*pi* 
x**2 - 2*e**(x**2)*x**3 + 8*e**(x**2)*x**2 + 4*log(5)**2*x**2 + 4*log(5)*i 
*pi*x**2 + 4*log(5)*x**3 - 16*log(5)*x**2 + 2*i*pi*x**3 - 8*i*pi*x**2 - pi 
**2*x**2 + x**4 - 8*x**3 + 16*x**2),x)*log(5)*e**5 - int(1/(e**(2*x**2)*x* 
*2 - 4*e**(x**2)*log(5)*x**2 - 2*e**(x**2)*i*pi*x**2 - 2*e**(x**2)*x**3 + 
8*e**(x**2)*x**2 + 4*log(5)**2*x**2 + 4*log(5)*i*pi*x**2 + 4*log(5)*x**3 - 
 16*log(5)*x**2 + 2*i*pi*x**3 - 8*i*pi*x**2 - pi**2*x**2 + x**4 - 8*x**3 + 
 16*x**2),x)*e**5*i*pi + 4*int(1/(e**(2*x**2)*x**2 - 4*e**(x**2)*log(5)...