\(\int \frac {x^2+e^{\frac {2}{x^2}} (4+x^2)+(x^2+e^{\frac {2}{x^2}} x^2) \log (\frac {1+e^{\frac {2}{x^2}}}{5 x})+\log ^2(\frac {1+e^{\frac {2}{x^2}}}{5 x}) (x^2+e^{\frac {2}{x^2}} x^2+e^x (x^2+x^3+e^{\frac {2}{x^2}} (x^2+x^3))+(x^2+e^{\frac {2}{x^2}} x^2) \log (x))}{(x^2+e^{\frac {2}{x^2}} x^2) \log ^2(\frac {1+e^{\frac {2}{x^2}}}{5 x})} \, dx\) [1956]

Optimal result

Integrand size = 166, antiderivative size = 27 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=x \left (e^x+\frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\log (x)\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=x \left (e^x+\frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}+\log (x)\right ) \] Input:

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

Output:

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

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

\(\Big \downarrow \) 7239

\(\displaystyle \int \left (\frac {x^2+e^{\frac {2}{x^2}} \left (x^2+4\right )}{\left (e^{\frac {2}{x^2}}+1\right ) x^2 \log ^2\left (\frac {e^{\frac {2}{x^2}}+1}{5 x}\right )}+\frac {1}{\log \left (\frac {e^{\frac {2}{x^2}}+1}{5 x}\right )}+e^x x+e^x+\log (x)+1\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \text {Subst}\left (\int \frac {1}{\log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )}dx,x,\frac {1}{x}\right )+4 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x^2}\right ) \log ^2\left (\frac {1}{5} \left (1+e^{2 x^2}\right ) x\right )}dx,x,\frac {1}{x}\right )+\int \frac {1}{\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}dx+\int \frac {1}{\log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )}dx+e^x x+x \log (x)\)

Input:

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

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(23)=46\).

Time = 36.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.26

method	result	size
parallelrisch	\(-\frac {-2 \ln \left (\frac {{\mathrm e}^{\frac {2}{x^{2}}}+1}{5 x}\right ) x \ln \left (x \right )-2 \ln \left (\frac {{\mathrm e}^{\frac {2}{x^{2}}}+1}{5 x}\right ) {\mathrm e}^{x} x -2 x}{2 \ln \left (\frac {{\mathrm e}^{\frac {2}{x^{2}}}+1}{5 x}\right )}\)	\(61\)
risch	\({\mathrm e}^{x} x +x \ln \left (x \right )+\frac {2 i x}{-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )}^{2}+\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )}^{3}-\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )-2 i \ln \left (5\right )-2 i \ln \left (x \right )+2 i \ln \left ({\mathrm e}^{\frac {2}{x^{2}}}+1\right )}\)	\(153\)

Input:

int((((x^2*exp(2/x^2)+x^2)*ln(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^2 
*exp(2/x^2)+x^2)*ln(1/5*(exp(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*ln(1/5*(e 
xp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/ln(1/5*(exp(2 
/x^2)+1)/x)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=\frac {{\left (x e^{x} + x \log \left (x\right )\right )} \log \left (\frac {e^{\left (\frac {2}{x^{2}}\right )} + 1}{5 \, x}\right ) + x}{\log \left (\frac {e^{\left (\frac {2}{x^{2}}\right )} + 1}{5 \, x}\right )} \] Input:

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp 
(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*l 
og(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/log( 
1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=x e^{x} + x \log {\left (x \right )} + \frac {x}{\log {\left (\frac {\frac {e^{\frac {2}{x^{2}}}}{5} + \frac {1}{5}}{x} \right )}} \] Input:

integrate((((x**2*exp(2/x**2)+x**2)*ln(x)+((x**3+x**2)*exp(2/x**2)+x**3+x* 
*2)*exp(x)+x**2*exp(2/x**2)+x**2)*ln(1/5*(exp(2/x**2)+1)/x)**2+(x**2*exp(2 
/x**2)+x**2)*ln(1/5*(exp(2/x**2)+1)/x)+(x**2+4)*exp(2/x**2)+x**2)/(x**2*ex 
p(2/x**2)+x**2)/ln(1/5*(exp(2/x**2)+1)/x)**2,x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (23) = 46\).

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.48 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=\frac {x \log \left (5\right ) \log \left (x\right ) + x \log \left (x\right )^{2} + {\left (x \log \left (5\right ) + x \log \left (x\right )\right )} e^{x} - {\left (x e^{x} + x \log \left (x\right )\right )} \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x}{\log \left (5\right ) + \log \left (x\right ) - \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right )} \] Input:

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp 
(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*l 
og(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/log( 
1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (23) = 46\).

Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.78 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=\frac {x e^{x} \log \left (5\right ) + x e^{x} \log \left (x\right ) + x \log \left (5\right ) \log \left (x\right ) + x \log \left (x\right )^{2} - x e^{x} \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x \log \left (x\right ) \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right ) - x}{\log \left (5\right ) + \log \left (x\right ) - \log \left (e^{\left (\frac {2}{x^{2}}\right )} + 1\right )} \] Input:

integrate((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp 
(x)+x^2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*l 
og(1/5*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/log( 
1/5*(exp(2/x^2)+1)/x)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 5.33 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=\frac {x+\frac {x^3\,\ln \left (\frac {\frac {{\mathrm {e}}^{\frac {2}{x^2}}}{5}+\frac {1}{5}}{x}\right )\,\left ({\mathrm {e}}^{\frac {2}{x^2}}+1\right )}{4\,{\mathrm {e}}^{\frac {2}{x^2}}+x^2\,{\mathrm {e}}^{\frac {2}{x^2}}+x^2}}{\ln \left (\frac {\frac {{\mathrm {e}}^{\frac {2}{x^2}}}{5}+\frac {1}{5}}{x}\right )}-x+\frac {4\,x}{x^2+4}+x\,{\mathrm {e}}^x+x\,\ln \left (x\right )-\frac {4\,\left (3\,x^{10}+4\,x^8\right )}{\left (x^2+4\right )\,\left ({\mathrm {e}}^{\frac {2}{x^2}}\,\left (x^2+4\right )+x^2\right )\,\left (3\,x^7+4\,x^5\right )} \] Input:

int((exp(2/x^2)*(x^2 + 4) + log((exp(2/x^2)/5 + 1/5)/x)*(x^2*exp(2/x^2) + 
x^2) + log((exp(2/x^2)/5 + 1/5)/x)^2*(log(x)*(x^2*exp(2/x^2) + x^2) + exp( 
x)*(exp(2/x^2)*(x^2 + x^3) + x^2 + x^3) + x^2*exp(2/x^2) + x^2) + x^2)/(lo 
g((exp(2/x^2)/5 + 1/5)/x)^2*(x^2*exp(2/x^2) + x^2)),x)
 

Output:

(x + (x^3*log((exp(2/x^2)/5 + 1/5)/x)*(exp(2/x^2) + 1))/(4*exp(2/x^2) + x^ 
2*exp(2/x^2) + x^2))/log((exp(2/x^2)/5 + 1/5)/x) - x + (4*x)/(x^2 + 4) + x 
*exp(x) + x*log(x) - (4*(4*x^8 + 3*x^10))/((x^2 + 4)*(exp(2/x^2)*(x^2 + 4) 
 + x^2)*(4*x^5 + 3*x^7))
 

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {x^2+e^{\frac {2}{x^2}} \left (4+x^2\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log \left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )+\log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right ) \left (x^2+e^{\frac {2}{x^2}} x^2+e^x \left (x^2+x^3+e^{\frac {2}{x^2}} \left (x^2+x^3\right )\right )+\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log (x)\right )}{\left (x^2+e^{\frac {2}{x^2}} x^2\right ) \log ^2\left (\frac {1+e^{\frac {2}{x^2}}}{5 x}\right )} \, dx=\frac {x \left (e^{x} \mathrm {log}\left (\frac {e^{\frac {2}{x^{2}}}+1}{5 x}\right )+\mathrm {log}\left (\frac {e^{\frac {2}{x^{2}}}+1}{5 x}\right ) \mathrm {log}\left (x \right )+1\right )}{\mathrm {log}\left (\frac {e^{\frac {2}{x^{2}}}+1}{5 x}\right )} \] Input:

int((((x^2*exp(2/x^2)+x^2)*log(x)+((x^3+x^2)*exp(2/x^2)+x^3+x^2)*exp(x)+x^ 
2*exp(2/x^2)+x^2)*log(1/5*(exp(2/x^2)+1)/x)^2+(x^2*exp(2/x^2)+x^2)*log(1/5 
*(exp(2/x^2)+1)/x)+(x^2+4)*exp(2/x^2)+x^2)/(x^2*exp(2/x^2)+x^2)/log(1/5*(e 
xp(2/x^2)+1)/x)^2,x)
 

Output:

(x*(e**x*log((e**(2/x**2) + 1)/(5*x)) + log((e**(2/x**2) + 1)/(5*x))*log(x 
) + 1))/log((e**(2/x**2) + 1)/(5*x))