\(\int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} (-1-20 e^5 x^2+20 x^4)}{2 x^2} \, dx\) [2664]

Optimal result

Integrand size = 83, antiderivative size = 29 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 \left (2+e^3+\left (-e^5+x^2\right )^2\right )}}{2 x}} \] Output:

exp(1/2*exp(10+5*exp(3)+5*(x^2-exp(5))^2)/x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 \left (2+e^3+e^{10}-2 e^5 x^2+x^4\right )}}{2 x}} \] Input:

Integrate[(E^(10 + 5*E^3 + 5*E^10 + E^(10 + 5*E^3 + 5*E^10 - 10*E^5*x^2 + 
5*x^4)/(2*x) - 10*E^5*x^2 + 5*x^4)*(-1 - 20*E^5*x^2 + 20*x^4))/(2*x^2),x]
 

Output:

E^(E^(5*(2 + E^3 + E^10 - 2*E^5*x^2 + x^4))/(2*x))
 

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.

\(\displaystyle \int \frac {\left (20 x^4-20 e^5 x^2-1\right ) \exp \left (5 x^4-10 e^5 x^2+\frac {e^{5 x^4-10 e^5 x^2+5 e^{10}+5 e^3+10}}{2 x}+5 e^{10}+5 e^3+10\right )}{2 x^2} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) \left (-20 x^4+20 e^5 x^2+1\right )}{x^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) \left (-20 x^4+20 e^5 x^2+1\right )}{x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (-20 \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) x^2+20 \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+5+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )+\frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-20 \int \exp \left (5 x^4-10 e^5 x^2+5 \left (3+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )dx-\int \frac {\exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right )}{x^2}dx+20 \int \exp \left (5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )+\frac {e^{5 x^4-10 e^5 x^2+5 \left (2+e^3+e^{10}\right )}}{2 x}\right ) x^2dx\right )\)

Input:

Int[(E^(10 + 5*E^3 + 5*E^10 + E^(10 + 5*E^3 + 5*E^10 - 10*E^5*x^2 + 5*x^4) 
/(2*x) - 10*E^5*x^2 + 5*x^4)*(-1 - 20*E^5*x^2 + 20*x^4))/(2*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\)	\(30\)
norman	\({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\)	\(32\)
parallelrisch	\({\mathrm e}^{\frac {{\mathrm e}^{5 \,{\mathrm e}^{10}-10 x^{2} {\mathrm e}^{5}+5 \,{\mathrm e}^{3}+5 x^{4}+10}}{2 x}}\)	\(32\)

Input:

int(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5* 
x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x^2,x,m 
ethod=_RETURNVERBOSE)
 

Output:

exp(1/2*exp(5*exp(10)-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (25) = 50\).

Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.66 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\left (-5 \, x^{4} + 10 \, x^{2} e^{5} + \frac {10 \, x^{5} - 20 \, x^{3} e^{5} + 10 \, x e^{10} + 10 \, x e^{3} + 20 \, x + e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x} - 5 \, e^{10} - 5 \, e^{3} - 10\right )} \] Input:

integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp 
(3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x 
^2,x, algorithm="fricas")
 

Output:

e^(-5*x^4 + 10*x^2*e^5 + 1/2*(10*x^5 - 20*x^3*e^5 + 10*x*e^10 + 10*x*e^3 + 
 20*x + e^(5*x^4 - 10*x^2*e^5 + 5*e^10 + 5*e^3 + 10))/x - 5*e^10 - 5*e^3 - 
 10)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\frac {e^{5 x^{4} - 10 x^{2} e^{5} + 10 + 5 e^{3} + 5 e^{10}}}{2 x}} \] Input:

integrate(1/2*(-20*x**2*exp(5)+20*x**4-1)*exp(5*exp(5)**2-10*x**2*exp(5)+5 
*exp(3)+5*x**4+10)*exp(1/2*exp(5*exp(5)**2-10*x**2*exp(5)+5*exp(3)+5*x**4+ 
10)/x)/x**2,x)
 

Output:

exp(exp(5*x**4 - 10*x**2*exp(5) + 10 + 5*exp(3) + 5*exp(10))/(2*x))
 

Maxima [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=e^{\left (\frac {e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x}\right )} \] Input:

integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp 
(3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x 
^2,x, algorithm="maxima")
 

Output:

e^(1/2*e^(5*x^4 - 10*x^2*e^5 + 5*e^10 + 5*e^3 + 10)/x)
 

Giac [F]

\[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=\int { \frac {{\left (20 \, x^{4} - 20 \, x^{2} e^{5} - 1\right )} e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + \frac {e^{\left (5 \, x^{4} - 10 \, x^{2} e^{5} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x} + 5 \, e^{10} + 5 \, e^{3} + 10\right )}}{2 \, x^{2}} \,d x } \] Input:

integrate(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp 
(3)+5*x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x 
^2,x, algorithm="giac")
 

Output:

integrate(1/2*(20*x^4 - 20*x^2*e^5 - 1)*e^(5*x^4 - 10*x^2*e^5 + 1/2*e^(5*x 
^4 - 10*x^2*e^5 + 5*e^10 + 5*e^3 + 10)/x + 5*e^10 + 5*e^3 + 10)/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 2.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-10\,x^2\,{\mathrm {e}}^5}\,{\mathrm {e}}^{5\,{\mathrm {e}}^3}\,{\mathrm {e}}^{5\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{10}\,{\mathrm {e}}^{5\,x^4}}{2\,x}} \] Input:

int(-(exp(5*exp(3) + 5*exp(10) - 10*x^2*exp(5) + 5*x^4 + 10)*exp(exp(5*exp 
(3) + 5*exp(10) - 10*x^2*exp(5) + 5*x^4 + 10)/(2*x))*(20*x^2*exp(5) - 20*x 
^4 + 1))/(2*x^2),x)
 

Output:

exp((exp(-10*x^2*exp(5))*exp(5*exp(3))*exp(5*exp(10))*exp(10)*exp(5*x^4))/ 
(2*x))
 

Reduce [F]

\[ \int \frac {e^{10+5 e^3+5 e^{10}+\frac {e^{10+5 e^3+5 e^{10}-10 e^5 x^2+5 x^4}}{2 x}-10 e^5 x^2+5 x^4} \left (-1-20 e^5 x^2+20 x^4\right )}{2 x^2} \, dx=\frac {e^{5 e^{10}+5 e^{3}} e^{10} \left (-20 \left (\int \frac {e^{\frac {e^{5 e^{10}+5 x^{4}+5 e^{3}} e^{10}+10 e^{10 e^{5} x^{2}} x^{5}}{2 e^{10 e^{5} x^{2}} x}}}{e^{10 e^{5} x^{2}}}d x \right ) e^{5}-\left (\int \frac {e^{\frac {e^{5 e^{10}+5 x^{4}+5 e^{3}} e^{10}+10 e^{10 e^{5} x^{2}} x^{5}}{2 e^{10 e^{5} x^{2}} x}}}{e^{10 e^{5} x^{2}} x^{2}}d x \right )+20 \left (\int \frac {e^{\frac {e^{5 e^{10}+5 x^{4}+5 e^{3}} e^{10}+10 e^{10 e^{5} x^{2}} x^{5}}{2 e^{10 e^{5} x^{2}} x}} x^{2}}{e^{10 e^{5} x^{2}}}d x \right )\right )}{2} \] Input:

int(1/2*(-20*x^2*exp(5)+20*x^4-1)*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5* 
x^4+10)*exp(1/2*exp(5*exp(5)^2-10*x^2*exp(5)+5*exp(3)+5*x^4+10)/x)/x^2,x)
 

Output:

(e**(5*e**10 + 5*e**3)*e**10*( - 20*int(e**((e**(5*e**10 + 5*e**3 + 5*x**4 
)*e**10 + 10*e**(10*e**5*x**2)*x**5)/(2*e**(10*e**5*x**2)*x))/e**(10*e**5* 
x**2),x)*e**5 - int(e**((e**(5*e**10 + 5*e**3 + 5*x**4)*e**10 + 10*e**(10* 
e**5*x**2)*x**5)/(2*e**(10*e**5*x**2)*x))/(e**(10*e**5*x**2)*x**2),x) + 20 
*int((e**((e**(5*e**10 + 5*e**3 + 5*x**4)*e**10 + 10*e**(10*e**5*x**2)*x** 
5)/(2*e**(10*e**5*x**2)*x))*x**2)/e**(10*e**5*x**2),x)))/2