\(\int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} (1+e^{\frac {24 e^5}{5 x}} (9 e^x-9 x))-\frac {24 e^5}{5 x}} (8 e^5+e^{\frac {24 e^5}{5 x}} (-15 x^2+15 e^x x^2))}{15 x^2} \, dx\) [2793]

Optimal result

Integrand size = 93, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x} \] Output:

exp(1/9/exp(3/5*exp(5)/x)^8+exp(x)-x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x-x} \] Input:

Integrate[(E^((1 + E^((24*E^5)/(5*x))*(9*E^x - 9*x))/(9*E^((24*E^5)/(5*x)) 
) - (24*E^5)/(5*x))*(8*E^5 + E^((24*E^5)/(5*x))*(-15*x^2 + 15*E^x*x^2)))/( 
15*x^2),x]
 

Output:

E^(1/(9*E^((24*E^5)/(5*x))) + E^x - 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.

\(\displaystyle \int \frac {\left (e^{\frac {24 e^5}{5 x}} \left (15 e^x x^2-15 x^2\right )+8 e^5\right ) \exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )+1\right )-\frac {24 e^5}{5 x}\right )}{15 x^2} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{15} \int \frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (9 e^{\frac {24 e^5}{5 x}} \left (e^x-x\right )+1\right )-\frac {24 e^5}{5 x}\right ) \left (8 e^5-15 e^{\frac {24 e^5}{5 x}} \left (x^2-e^x x^2\right )\right )}{x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{15} \int \left (15 \exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (9 e^{\frac {24 e^5}{5 x}} \left (e^x-x\right )+1\right )+x\right )-\frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (9 e^{\frac {24 e^5}{5 x}} \left (e^x-x\right )+1\right )-\frac {24 e^5}{5 x}\right ) \left (15 e^{\frac {24 e^5}{5 x}} x^2-8 e^5\right )}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{15} \left (8 \int \frac {\exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (9 e^{\frac {24 e^5}{5 x}} \left (e^x-x\right )+1\right )-\frac {24 e^5}{5 x}+5\right )}{x^2}dx-15 \int \exp \left (\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (9 e^{\frac {24 e^5}{5 x}} \left (e^x-x\right )+1\right )\right )dx+15 \int e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}}+e^x}dx\right )\)

Input:

Int[(E^((1 + E^((24*E^5)/(5*x))*(9*E^x - 9*x))/(9*E^((24*E^5)/(5*x))) - (2 
4*E^5)/(5*x))*(8*E^5 + E^((24*E^5)/(5*x))*(-15*x^2 + 15*E^x*x^2)))/(15*x^2 
),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(42\) vs. \(2(19)=38\).

Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72

\[{\mathrm e}^{-\frac {\left (9 \,{\mathrm e}^{\frac {24 \,{\mathrm e}^{5}}{5 x}} x -9 \,{\mathrm e}^{\frac {5 x^{2}+24 \,{\mathrm e}^{5}}{5 x}}-1\right ) {\mathrm e}^{-\frac {24 \,{\mathrm e}^{5}}{5 x}}}{9}}\]

Input:

int(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9 
*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5 
)/x)^8,x)
 

Output:

exp(-1/9*(9*exp(24/5*exp(5)/x)*x-9*exp(1/5*(5*x^2+24*exp(5))/x)-1)*exp(-24 
/5*exp(5)/x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=e^{\left (-\frac {{\left (9 \, {\left (5 \, x^{2} - 5 \, x e^{x} + 24 \, e^{5}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 5 \, x\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )}}{45 \, x} + \frac {24 \, e^{5}}{5 \, x}\right )} \] Input:

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1 
/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5 
*exp(5)/x)^8,x, algorithm="fricas")
 

Output:

e^(-1/45*(9*(5*x^2 - 5*x*e^x + 24*e^5)*e^(24/5*e^5/x) - 5*x)*e^(-24/5*e^5/ 
x)/x + 24/5*e^5/x)
 

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=e^{\left (\frac {\left (- 9 x + 9 e^{x}\right ) e^{\frac {24 e^{5}}{5 x}}}{9} + \frac {1}{9}\right ) e^{- \frac {24 e^{5}}{5 x}}} \] Input:

integrate(1/15*((15*exp(x)*x**2-15*x**2)*exp(3/5*exp(5)/x)**8+8*exp(5))*ex 
p(1/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)**8+1)/exp(3/5*exp(5)/x)**8)/x**2/e 
xp(3/5*exp(5)/x)**8,x)
 

Output:

exp(((-9*x + 9*exp(x))*exp(24*exp(5)/(5*x))/9 + 1/9)*exp(-24*exp(5)/(5*x)) 
)
 

Maxima [F]

\[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=\int { \frac {{\left (15 \, {\left (x^{2} e^{x} - x^{2}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} + 8 \, e^{5}\right )} e^{\left (-\frac {1}{9} \, {\left (9 \, {\left (x - e^{x}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 1\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )} - \frac {24 \, e^{5}}{5 \, x}\right )}}{15 \, x^{2}} \,d x } \] Input:

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1 
/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5 
*exp(5)/x)^8,x, algorithm="maxima")
 

Output:

1/15*integrate((15*(x^2*e^x - x^2)*e^(24/5*e^5/x) + 8*e^5)*e^(-1/9*(9*(x - 
 e^x)*e^(24/5*e^5/x) - 1)*e^(-24/5*e^5/x) - 24/5*e^5/x)/x^2, x)
 

Giac [F]

\[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=\int { \frac {{\left (15 \, {\left (x^{2} e^{x} - x^{2}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} + 8 \, e^{5}\right )} e^{\left (-\frac {1}{9} \, {\left (9 \, {\left (x - e^{x}\right )} e^{\left (\frac {24 \, e^{5}}{5 \, x}\right )} - 1\right )} e^{\left (-\frac {24 \, e^{5}}{5 \, x}\right )} - \frac {24 \, e^{5}}{5 \, x}\right )}}{15 \, x^{2}} \,d x } \] Input:

integrate(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1 
/9*((9*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5 
*exp(5)/x)^8,x, algorithm="giac")
 

Output:

integrate(1/15*(15*(x^2*e^x - x^2)*e^(24/5*e^5/x) + 8*e^5)*e^(-1/9*(9*(x - 
 e^x)*e^(24/5*e^5/x) - 1)*e^(-24/5*e^5/x) - 24/5*e^5/x)/x^2, x)
 

Mupad [B] (verification not implemented)

Time = 2.75 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {24\,{\mathrm {e}}^5}{5\,x}}}{9}} \] Input:

int((exp(-(24*exp(5))/(5*x))*exp(-exp(-(24*exp(5))/(5*x))*((exp((24*exp(5) 
)/(5*x))*(9*x - 9*exp(x)))/9 - 1/9))*(8*exp(5) + exp((24*exp(5))/(5*x))*(1 
5*x^2*exp(x) - 15*x^2)))/(15*x^2),x)
 

Output:

exp(-x)*exp(exp(x))*exp(exp(-(24*exp(5))/(5*x))/9)
 

Reduce [F]

\[ \int \frac {e^{\frac {1}{9} e^{-\frac {24 e^5}{5 x}} \left (1+e^{\frac {24 e^5}{5 x}} \left (9 e^x-9 x\right )\right )-\frac {24 e^5}{5 x}} \left (8 e^5+e^{\frac {24 e^5}{5 x}} \left (-15 x^2+15 e^x x^2\right )\right )}{15 x^2} \, dx=\int \frac {\left (\left (15 \,{\mathrm e}^{x} x^{2}-15 x^{2}\right ) \left ({\mathrm e}^{\frac {3 \,{\mathrm e}^{5}}{5 x}}\right )^{8}+8 \,{\mathrm e}^{5}\right ) {\mathrm e}^{\frac {\left (9 \,{\mathrm e}^{x}-9 x \right ) \left ({\mathrm e}^{\frac {3 \,{\mathrm e}^{5}}{5 x}}\right )^{8}+1}{9 \left ({\mathrm e}^{\frac {3 \,{\mathrm e}^{5}}{5 x}}\right )^{8}}}}{15 x^{2} \left ({\mathrm e}^{\frac {3 \,{\mathrm e}^{5}}{5 x}}\right )^{8}}d x \] Input:

int(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9 
*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5 
)/x)^8,x)
 

Output:

int(1/15*((15*exp(x)*x^2-15*x^2)*exp(3/5*exp(5)/x)^8+8*exp(5))*exp(1/9*((9 
*exp(x)-9*x)*exp(3/5*exp(5)/x)^8+1)/exp(3/5*exp(5)/x)^8)/x^2/exp(3/5*exp(5 
)/x)^8,x)