\(\int \frac {2 x^3+(e^{-2-2 x} x^4)^{\frac {1}{x^2}} (4-2 x+x^2-2 \log (e^{-2-2 x} x^4))}{2 x^2} \, dx\) [359]

Optimal result

Integrand size = 51, antiderivative size = 22 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} x \left (x+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} x \left (x+\left (e^{-2 (1+x)} x^4\right )^{\frac {1}{x^2}}\right ) \] Input:

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

Output:

(x*(x + (x^4/E^(2*(1 + x)))^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.

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2010

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\)	\(26\)
parallelrisch	\(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\)	\(26\)
parts	\(\frac {{\mathrm e}^{\frac {\ln \left (x^{4} {\mathrm e}^{-2-2 x}\right )}{x^{2}}} x}{2}+\frac {x^{2}}{2}\)	\(26\)
risch	\(\frac {x^{2}}{2}+\frac {x \,x^{\frac {4}{x^{2}}} \left ({\mathrm e}^{1+x}\right )^{-\frac {2}{x^{2}}} {\mathrm e}^{-\frac {i \pi \left (-\operatorname {csgn}\left (i {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right ) \operatorname {csgn}\left (i x^{4}\right )-\operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right )^{3}+2 \operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{1+x}\right )-\operatorname {csgn}\left (i {\mathrm e}^{2+2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{1+x}\right )^{2}+\operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{3}-\operatorname {csgn}\left (i x^{4} {\mathrm e}^{-2-2 x}\right )^{2} \operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{4}\right )^{3}-\operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )-\operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{4}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )^{3}-2 \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )-\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}-\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\operatorname {csgn}\left (i x^{3}\right )^{3}\right )}{2 x^{2}}}}{2}\)	\(357\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} \, \left (x^{4} e^{\left (-2 \, x - 2\right )}\right )^{\left (\frac {1}{x^{2}}\right )} x + \frac {1}{2} \, x^{2} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

(Integral(2*x, x) + Integral(exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2), x 
) + Integral(8*exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2)/x**2, x) + Integ 
ral(-2*exp(-2/x**2)*exp(log(x**4*exp(-2*x))/x**2)/x, x) + Integral(-2*exp( 
-2/x**2)*exp(log(x**4*exp(-2*x))/x**2)*log(x**4*exp(-2*x))/x**2, x))/2
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{\left (-\frac {2}{x} + \frac {4 \, \log \left (x\right )}{x^{2}} - \frac {2}{x^{2}}\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {x^2}{2}+\frac {x\,{\mathrm {e}}^{-\frac {2}{x}}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\left (x^4\right )}^{\frac {1}{x^2}}}{2} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {2 x^3+\left (e^{-2-2 x} x^4\right )^{\frac {1}{x^2}} \left (4-2 x+x^2-2 \log \left (e^{-2-2 x} x^4\right )\right )}{2 x^2} \, dx=\frac {x \left (e^{\frac {\mathrm {log}\left (\frac {x^{4}}{e^{2 x} e^{2}}\right )}{x^{2}}}+x \right )}{2} \] Input:

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

Output:

(x*(e**(log(x**4/(e**(2*x)*e**2))/x**2) + x))/2