\(\int \frac {e^{-e^x} (e^{e^x} (-4 x+10 x^3)+(2-5 x^2) \log ^2(2-5 x^2)+\log (x) (-20 x^2 \log (2-5 x^2)+e^x (-2 x+5 x^3) \log ^2(2-5 x^2)))}{-2 x+5 x^3} \, dx\) [650]

Optimal result

Integrand size = 93, antiderivative size = 26 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-4+2 x-e^{-e^x} \log (x) \log ^2\left (2-5 x^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=2 x-e^{-e^x} \log (x) \log ^2\left (2-5 x^2\right ) \] Input:

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

Output:

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

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

\(\Big \downarrow \) 2026

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 25.70 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(2 x -\ln \left (x \right ) \ln \left (-5 x^{2}+2\right )^{2} {\mathrm e}^{-{\mathrm e}^{x}}\)	\(24\)
parallelrisch	\(-\frac {\left (100 \ln \left (-5 x^{2}+2\right )^{2} \ln \left (x \right )-200 x \,{\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-{\mathrm e}^{x}}}{100}\)	\(29\)

Input:

int(((10*x^3-4*x)*exp(exp(x))+((5*x^3-2*x)*exp(x)*ln(-5*x^2+2)^2-20*x^2*ln 
(-5*x^2+2))*ln(x)+(-5*x^2+2)*ln(-5*x^2+2)^2)/(5*x^3-2*x)/exp(exp(x)),x,met 
hod=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-{\left (\log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) - 2 \, x e^{\left (e^{x}\right )}\right )} e^{\left (-e^{x}\right )} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-e^{\left (-e^{x}\right )} \log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) + 2 \, x \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=-{\left (e^{\left (x - e^{x}\right )} \log \left (-5 \, x^{2} + 2\right )^{2} \log \left (x\right ) - 2 \, x e^{x}\right )} e^{\left (-x\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 2.84 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-e^x} \left (e^{e^x} \left (-4 x+10 x^3\right )+\left (2-5 x^2\right ) \log ^2\left (2-5 x^2\right )+\log (x) \left (-20 x^2 \log \left (2-5 x^2\right )+e^x \left (-2 x+5 x^3\right ) \log ^2\left (2-5 x^2\right )\right )\right )}{-2 x+5 x^3} \, dx=\frac {2 e^{e^{x}} x -\mathrm {log}\left (-5 x^{2}+2\right )^{2} \mathrm {log}\left (x \right )}{e^{e^{x}}} \] Input:

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

Output:

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