\(\int \frac {e^{e^{\frac {1}{4} (1+4 x)}} (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} (4 x-6 x^2+x^3)+(2-2 x+e^{\frac {1}{4} (1+4 x)} (6 x-2 x^2)) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x))}{2 x} \, dx\) [1109]

Optimal result

Integrand size = 100, antiderivative size = 24 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (-5+(3-x+\log (x))^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} e^{e^{\frac {1}{4}+x}} \left (4-6 x+x^2+(6-2 x) \log (x)+\log ^2(x)\right ) \] Input:

Integrate[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x 
^2 + x^3) + (2 - 2*x + E^((1 + 4*x)/4)*(6*x - 2*x^2))*Log[x] + E^((1 + 4*x 
)/4)*x*Log[x]^2))/(2*x),x]
 

Output:

(E^E^(1/4 + x)*(4 - 6*x + x^2 + (6 - 2*x)*Log[x] + Log[x]^2))/2
 

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.

\(\displaystyle \int \frac {e^{e^{\frac {1}{4} (4 x+1)}} \left (2 x^2+\left (e^{\frac {1}{4} (4 x+1)} \left (6 x-2 x^2\right )-2 x+2\right ) \log (x)+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )-8 x+e^{\frac {1}{4} (4 x+1)} x \log ^2(x)+6\right )}{2 x} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {e^{e^{\frac {1}{4} (4 x+1)}} \left (2 x^2+e^{\frac {1}{4} (4 x+1)} \log ^2(x) x-8 x+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )+2 \left (-x+e^{\frac {1}{4} (4 x+1)} \left (3 x-x^2\right )+1\right ) \log (x)+6\right )}{x}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {1}{2} \int \frac {e^{e^{x+\frac {1}{4}}} \left (2 x^2+e^{\frac {1}{4} (4 x+1)} \log ^2(x) x-8 x+e^{\frac {1}{4} (4 x+1)} \left (x^3-6 x^2+4 x\right )+2 \left (-x+e^{\frac {1}{4} (4 x+1)} \left (3 x-x^2\right )+1\right ) \log (x)+6\right )}{x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} \int \left (\frac {2 e^{e^{x+\frac {1}{4}}} (x-1) (x-\log (x)-3)}{x}+e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} \left (x^2-2 \log (x) x-6 x+\log ^2(x)+6 \log (x)+4\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (2 \int \frac {\operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right )}{x}dx+\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} x^2dx+2 \int e^{e^{x+\frac {1}{4}}} xdx-6 \int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx-2 \int \frac {\int \frac {e^{e^{x+\frac {1}{4}}}}{x}dx}{x}dx+2 \int \frac {\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx}{x}dx+\int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} \log ^2(x)dx+2 \log (x) \int \frac {e^{e^{x+\frac {1}{4}}}}{x}dx-2 \log (x) \int e^{x+e^{x+\frac {1}{4}}+\frac {1}{4}} xdx-8 \operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right )-2 \operatorname {ExpIntegralEi}\left (e^{x+\frac {1}{4}}\right ) \log (x)+4 e^{e^{x+\frac {1}{4}}}+6 e^{e^{x+\frac {1}{4}}} \log (x)\right )\)

Input:

Int[(E^E^((1 + 4*x)/4)*(6 - 8*x + 2*x^2 + E^((1 + 4*x)/4)*(4*x - 6*x^2 + x 
^3) + (2 - 2*x + E^((1 + 4*x)/4)*(6*x - 2*x^2))*Log[x] + E^((1 + 4*x)/4)*x 
*Log[x]^2))/(2*x),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
risch	\(\frac {\left (x^{2}-2 x \ln \left (x \right )+\ln \left (x \right )^{2}-6 x +6 \ln \left (x \right )+4\right ) {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}}{2}\)	\(29\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}}{2}-{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} \ln \left (x \right ) x +\frac {{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} \ln \left (x \right )^{2}}{2}-3 \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} x +3 \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}} \ln \left (x \right )+2 \,{\mathrm e}^{{\mathrm e}^{x +\frac {1}{4}}}\)	\(57\)

Input:

int(1/2*(x*exp(x+1/4)*ln(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*ln(x)+(x^3-6 
*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x,method=_RETURNVERBOS 
E)
 

Output:

1/2*(x^2-2*x*ln(x)+ln(x)^2-6*x+6*ln(x)+4)*exp(exp(x+1/4))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, {\left (x^{2} - 2 \, {\left (x - 3\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 6 \, x + 4\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \] Input:

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x 
)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=" 
fricas")
 

Output:

1/2*(x^2 - 2*(x - 3)*log(x) + log(x)^2 - 6*x + 4)*e^(e^(x + 1/4))
 

Sympy [A] (verification not implemented)

Time = 28.67 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {\left (x^{2} - 2 x \log {\left (x \right )} - 6 x + \log {\left (x \right )}^{2} + 6 \log {\left (x \right )} + 4\right ) e^{e^{x + \frac {1}{4}}}}{2} \] Input:

integrate(1/2*(x*exp(x+1/4)*ln(x)**2+((-2*x**2+6*x)*exp(x+1/4)-2*x+2)*ln(x 
)+(x**3-6*x**2+4*x)*exp(x+1/4)+2*x**2-8*x+6)*exp(exp(x+1/4))/x,x)
 

Output:

(x**2 - 2*x*log(x) - 6*x + log(x)**2 + 6*log(x) + 4)*exp(exp(x + 1/4))/2
 

Maxima [F]

\[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\int { \frac {{\left (x e^{\left (x + \frac {1}{4}\right )} \log \left (x\right )^{2} + 2 \, x^{2} + {\left (x^{3} - 6 \, x^{2} + 4 \, x\right )} e^{\left (x + \frac {1}{4}\right )} - 2 \, {\left ({\left (x^{2} - 3 \, x\right )} e^{\left (x + \frac {1}{4}\right )} + x - 1\right )} \log \left (x\right ) - 8 \, x + 6\right )} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )}}{2 \, x} \,d x } \] Input:

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x 
)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=" 
maxima")
 

Output:

1/2*(x^2 - 2*(x - 3)*log(x) + log(x)^2 - 6*x + 4)*e^(e^(x + 1/4)) - 4*Ei(e 
^(x + 1/4)) + 4*integrate(e^(e^(x + 1/4)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (18) = 36\).

Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {1}{2} \, x^{2} e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} - x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + \frac {1}{2} \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right )^{2} - 3 \, x e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} + 3 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \log \left (x\right ) + 2 \, e^{\left (e^{\left (x + \frac {1}{4}\right )}\right )} \] Input:

integrate(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x 
)+(x^3-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x, algorithm=" 
giac")
 

Output:

1/2*x^2*e^(e^(x + 1/4)) - x*e^(e^(x + 1/4))*log(x) + 1/2*e^(e^(x + 1/4))*l 
og(x)^2 - 3*x*e^(e^(x + 1/4)) + 3*e^(e^(x + 1/4))*log(x) + 2*e^(e^(x + 1/4 
))
 

Mupad [B] (verification not implemented)

Time = 7.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{1/4}\,{\mathrm {e}}^x}\,\left (\frac {x^2}{2}-x\,\ln \left (x\right )-3\,x+\frac {{\ln \left (x\right )}^2}{2}+3\,\ln \left (x\right )+2\right ) \] Input:

int((exp(exp(x + 1/4))*(log(x)*(exp(x + 1/4)*(6*x - 2*x^2) - 2*x + 2) - 8* 
x + 2*x^2 + exp(x + 1/4)*(4*x - 6*x^2 + x^3) + x*exp(x + 1/4)*log(x)^2 + 6 
))/(2*x),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{e^{\frac {1}{4} (1+4 x)}} \left (6-8 x+2 x^2+e^{\frac {1}{4} (1+4 x)} \left (4 x-6 x^2+x^3\right )+\left (2-2 x+e^{\frac {1}{4} (1+4 x)} \left (6 x-2 x^2\right )\right ) \log (x)+e^{\frac {1}{4} (1+4 x)} x \log ^2(x)\right )}{2 x} \, dx=\frac {e^{e^{x +\frac {1}{4}}} \left (\mathrm {log}\left (x \right )^{2}-2 \,\mathrm {log}\left (x \right ) x +6 \,\mathrm {log}\left (x \right )+x^{2}-6 x +4\right )}{2} \] Input:

int(1/2*(x*exp(x+1/4)*log(x)^2+((-2*x^2+6*x)*exp(x+1/4)-2*x+2)*log(x)+(x^3 
-6*x^2+4*x)*exp(x+1/4)+2*x^2-8*x+6)*exp(exp(x+1/4))/x,x)
 

Output:

(e**(e**((4*x + 1)/4))*(log(x)**2 - 2*log(x)*x + 6*log(x) + x**2 - 6*x + 4 
))/2