3.17.66 \(\int \frac {e^{-x-\frac {2 e^{-x} (e^{\frac {1}{x}} x^2-x^3)}{\log (x)}} (-5 e^{\frac {1}{x}} x+5 x^2+(-15 x^2+5 x^3+e^{\frac {1}{x}} (-5+10 x-5 x^2)) \log (x)+e^{x+\frac {2 e^{-x} (e^{\frac {1}{x}} x^2-x^3)}{\log (x)}} \log ^2(x))}{\log ^2(x)} \, dx\) [1666]

3.17.66.1 Optimal result

Integrand size = 116, antiderivative size = 31 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=-\frac {5}{2} e^{-\frac {2 e^{-x} \left (e^{\frac {1}{x}}-x\right ) x^2}{\log (x)}}+x \]

output

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

3.17.66.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=-\frac {5}{2} e^{-\frac {2 e^{-x} \left (e^{\frac {1}{x}}-x\right ) x^2}{\log (x)}}+x \]

input

Integrate[(E^(-x - (2*(E^x^(-1)*x^2 - x^3))/(E^x*Log[x]))*(-5*E^x^(-1)*x + 
 5*x^2 + (-15*x^2 + 5*x^3 + E^x^(-1)*(-5 + 10*x - 5*x^2))*Log[x] + E^(x + 
(2*(E^x^(-1)*x^2 - x^3))/(E^x*Log[x]))*Log[x]^2))/Log[x]^2,x]
 

output

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

3.17.66.3 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.

\(\displaystyle \int \frac {e^{-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}-x} \left (5 x^2+e^{\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}+x} \log ^2(x)+\left (5 x^3-15 x^2+e^{\frac {1}{x}} \left (-5 x^2+10 x-5\right )\right ) \log (x)-5 e^{\frac {1}{x}} x\right )}{\log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (5 x^2+e^{\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}+x} \log ^2(x)+\left (5 x^3-15 x^2+e^{\frac {1}{x}} \left (-5 x^2+10 x-5\right )\right ) \log (x)-5 e^{\frac {1}{x}} x\right ) \exp \left (-\frac {e^{-x} x \left (-2 x^2+2 e^{\frac {1}{x}} x+e^x \log (x)\right )}{\log (x)}\right )}{\log ^2(x)}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 7299

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

input

Int[(E^(-x - (2*(E^x^(-1)*x^2 - x^3))/(E^x*Log[x]))*(-5*E^x^(-1)*x + 5*x^2 
 + (-15*x^2 + 5*x^3 + E^x^(-1)*(-5 + 10*x - 5*x^2))*Log[x] + E^(x + (2*(E^ 
x^(-1)*x^2 - x^3))/(E^x*Log[x]))*Log[x]^2))/Log[x]^2,x]
 

output

$Aborted
 

3.17.66.3.1 Defintions of rubi rules used

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

3.17.66.4 Maple [F(-1)]

Timed out.

\[\int \frac {\left ({\mathrm e}^{x} \ln \left (x \right )^{2} {\mathrm e}^{\frac {2 \left (x^{2} {\mathrm e}^{\frac {1}{x}}-x^{3}\right ) {\mathrm e}^{-x}}{\ln \left (x \right )}}+\left (\left (-5 x^{2}+10 x -5\right ) {\mathrm e}^{\frac {1}{x}}+5 x^{3}-15 x^{2}\right ) \ln \left (x \right )-5 x \,{\mathrm e}^{\frac {1}{x}}+5 x^{2}\right ) {\mathrm e}^{-x} {\mathrm e}^{-\frac {2 \left (x^{2} {\mathrm e}^{\frac {1}{x}}-x^{3}\right ) {\mathrm e}^{-x}}{\ln \left (x \right )}}}{\ln \left (x \right )^{2}}d x\]

input

int((exp(x)*ln(x)^2*exp((x^2*exp(1/x)-x^3)/exp(x)/ln(x))^2+((-5*x^2+10*x-5 
)*exp(1/x)+5*x^3-15*x^2)*ln(x)-5*x*exp(1/x)+5*x^2)/exp(x)/ln(x)^2/exp((x^2 
*exp(1/x)-x^3)/exp(x)/ln(x))^2,x)
 

output

int((exp(x)*ln(x)^2*exp((x^2*exp(1/x)-x^3)/exp(x)/ln(x))^2+((-5*x^2+10*x-5 
)*exp(1/x)+5*x^3-15*x^2)*ln(x)-5*x*exp(1/x)+5*x^2)/exp(x)/ln(x)^2/exp((x^2 
*exp(1/x)-x^3)/exp(x)/ln(x))^2,x)
 

3.17.66.5 Fricas [B] (verification not implemented)

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

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.65 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {1}{2} \, {\left (2 \, x e^{\left (-x - \frac {2 \, x^{3} e^{\left (-x\right )} - 2 \, x^{2} e^{\left (-x + \frac {1}{x}\right )} - x \log \left (x\right )}{\log \left (x\right )}\right )} - 5\right )} e^{\left (x + \frac {2 \, x^{3} e^{\left (-x\right )} - 2 \, x^{2} e^{\left (-x + \frac {1}{x}\right )} - x \log \left (x\right )}{\log \left (x\right )}\right )} \]

input

integrate((exp(x)*log(x)^2*exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2+((-5*x^ 
2+10*x-5)*exp(1/x)+5*x^3-15*x^2)*log(x)-5*x*exp(1/x)+5*x^2)/exp(x)/log(x)^ 
2/exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2,x, algorithm=\
 

output

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

3.17.66.6 Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=x - \frac {5 e^{- \frac {2 \left (- x^{3} + x^{2} e^{\frac {1}{x}}\right ) e^{- x}}{\log {\left (x \right )}}}}{2} \]

input

integrate((exp(x)*ln(x)**2*exp((x**2*exp(1/x)-x**3)/exp(x)/ln(x))**2+((-5* 
x**2+10*x-5)*exp(1/x)+5*x**3-15*x**2)*ln(x)-5*x*exp(1/x)+5*x**2)/exp(x)/ln 
(x)**2/exp((x**2*exp(1/x)-x**3)/exp(x)/ln(x))**2,x)
 

output

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

3.17.66.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=x - \frac {5}{2} \, e^{\left (\frac {2 \, x^{3} e^{\left (-x\right )}}{\log \left (x\right )} - \frac {2 \, x^{2} e^{\left (-x + \frac {1}{x}\right )}}{\log \left (x\right )}\right )} \]

input

integrate((exp(x)*log(x)^2*exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2+((-5*x^ 
2+10*x-5)*exp(1/x)+5*x^3-15*x^2)*log(x)-5*x*exp(1/x)+5*x^2)/exp(x)/log(x)^ 
2/exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2,x, algorithm=\
 

output

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

3.17.66.8 Giac [F]

\[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=\int { \frac {{\left (e^{\left (x - \frac {2 \, {\left (x^{3} - x^{2} e^{\frac {1}{x}}\right )} e^{\left (-x\right )}}{\log \left (x\right )}\right )} \log \left (x\right )^{2} + 5 \, x^{2} - 5 \, x e^{\frac {1}{x}} + 5 \, {\left (x^{3} - 3 \, x^{2} - {\left (x^{2} - 2 \, x + 1\right )} e^{\frac {1}{x}}\right )} \log \left (x\right )\right )} e^{\left (-x + \frac {2 \, {\left (x^{3} - x^{2} e^{\frac {1}{x}}\right )} e^{\left (-x\right )}}{\log \left (x\right )}\right )}}{\log \left (x\right )^{2}} \,d x } \]

input

integrate((exp(x)*log(x)^2*exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2+((-5*x^ 
2+10*x-5)*exp(1/x)+5*x^3-15*x^2)*log(x)-5*x*exp(1/x)+5*x^2)/exp(x)/log(x)^ 
2/exp((x^2*exp(1/x)-x^3)/exp(x)/log(x))^2,x, algorithm=\
 

output

integrate((e^(x - 2*(x^3 - x^2*e^(1/x))*e^(-x)/log(x))*log(x)^2 + 5*x^2 - 
5*x*e^(1/x) + 5*(x^3 - 3*x^2 - (x^2 - 2*x + 1)*e^(1/x))*log(x))*e^(-x + 2* 
(x^3 - x^2*e^(1/x))*e^(-x)/log(x))/log(x)^2, x)
 

3.17.66.9 Mupad [B] (verification not implemented)

Time = 10.68 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x-\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \left (-5 e^{\frac {1}{x}} x+5 x^2+\left (-15 x^2+5 x^3+e^{\frac {1}{x}} \left (-5+10 x-5 x^2\right )\right ) \log (x)+e^{x+\frac {2 e^{-x} \left (e^{\frac {1}{x}} x^2-x^3\right )}{\log (x)}} \log ^2(x)\right )}{\log ^2(x)} \, dx=x-\frac {5\,{\mathrm {e}}^{\frac {2\,x^3\,{\mathrm {e}}^{-x}}{\ln \left (x\right )}}\,{\mathrm {e}}^{-\frac {2\,x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{1/x}}{\ln \left (x\right )}}}{2} \]

input

int(-(exp(-x)*exp(-(2*exp(-x)*(x^2*exp(1/x) - x^3))/log(x))*(5*x*exp(1/x) 
+ log(x)*(exp(1/x)*(5*x^2 - 10*x + 5) + 15*x^2 - 5*x^3) - 5*x^2 - exp((2*e 
xp(-x)*(x^2*exp(1/x) - x^3))/log(x))*exp(x)*log(x)^2))/log(x)^2,x)
 

output

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