3.27.10 \(\int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x)))+e^{e^{4 x-x \log (x)} x} (8 x+2 x^2+e^{4 x-x \log (x)} (4 x^2+12 x^3-4 x^3 \log (x)))}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx\) [2610]

3.27.10.1 Optimal result

Integrand size = 155, antiderivative size = 29 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x+\log \left (\left (x-\left (e^{e^{x (4-\log (x))} x}+x^2\right )^2\right )^2\right ) \]

output

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

3.27.10.2 Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x+2 \log \left (e^{2 e^{4 x} x^{1-x}}-x+2 e^{e^{4 x} x^{1-x}} x^2+x^4\right ) \]

input

Integrate[(-2 - x + 8*x^3 + x^4 + E^(2*E^(4*x - x*Log[x])*x)*(1 + E^(4*x - 
 x*Log[x])*(4 + 12*x - 4*x*Log[x])) + E^(E^(4*x - x*Log[x])*x)*(8*x + 2*x^ 
2 + E^(4*x - x*Log[x])*(4*x^2 + 12*x^3 - 4*x^3*Log[x])))/(E^(2*E^(4*x - x* 
Log[x])*x) - x + 2*E^(E^(4*x - x*Log[x])*x)*x^2 + x^4),x]
 

output

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

3.27.10.3 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.

\(\displaystyle \int \frac {x^4+8 x^3+e^{x e^{4 x-x \log (x)}} \left (2 x^2+e^{4 x-x \log (x)} \left (12 x^3-4 x^3 \log (x)+4 x^2\right )+8 x\right )-x+e^{2 x e^{4 x-x \log (x)}} \left (e^{4 x-x \log (x)} (12 x-4 x \log (x)+4)+1\right )-2}{x^4+2 x^2 e^{x e^{4 x-x \log (x)}}-x+e^{2 x e^{4 x-x \log (x)}}} \, dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -2 \int \frac {1}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+\int \frac {e^{2 e^{4 x} x^{1-x}}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx-\int \frac {x}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+8 \int \frac {e^{e^{4 x} x^{1-x}} x}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+2 \int \frac {e^{e^{4 x} x^{1-x}} x^2}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+8 \int \frac {x^3}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+\int \frac {x^4}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx-4 \log (x) \int \frac {e^{2 x \left (e^{4 x} x^{-x}+2\right )} x^{1-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+12 \int \frac {e^{2 x \left (e^{4 x} x^{-x}+2\right )} x^{1-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+4 \int \frac {e^{x^{1-x} \left (4 x^x+e^{4 x}\right )} x^{2-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx-4 \log (x) \int \frac {e^{x^{1-x} \left (4 x^x+e^{4 x}\right )} x^{3-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+12 \int \frac {e^{x^{1-x} \left (4 x^x+e^{4 x}\right )} x^{3-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+4 \int \frac {e^{2 x \left (e^{4 x} x^{-x}+2\right )} x^{-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx+4 \int \frac {\int \frac {e^{2 x \left (e^{4 x} x^{-x}+2\right )} x^{1-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx}{x}dx+4 \int \frac {\int \frac {e^{x^{1-x} \left (4 x^x+e^{4 x}\right )} x^{3-x}}{x^4+2 e^{e^{4 x} x^{1-x}} x^2-x+e^{2 e^{4 x} x^{1-x}}}dx}{x}dx\)

input

Int[(-2 - x + 8*x^3 + x^4 + E^(2*E^(4*x - x*Log[x])*x)*(1 + E^(4*x - x*Log 
[x])*(4 + 12*x - 4*x*Log[x])) + E^(E^(4*x - x*Log[x])*x)*(8*x + 2*x^2 + E^ 
(4*x - x*Log[x])*(4*x^2 + 12*x^3 - 4*x^3*Log[x])))/(E^(2*E^(4*x - x*Log[x] 
)*x) - x + 2*E^(E^(4*x - x*Log[x])*x)*x^2 + x^4),x]
 

output

$Aborted
 

3.27.10.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

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

3.27.10.4 Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48

method	result	size
risch	\(x +2 \ln \left ({\mathrm e}^{2 x^{-x} {\mathrm e}^{4 x} x}+2 \,{\mathrm e}^{x^{-x} {\mathrm e}^{4 x} x} x^{2}+x^{4}-x \right )\)	\(43\)
parallelrisch	\(x +2 \ln \left ({\mathrm e}^{2 x \,{\mathrm e}^{-x \ln \left (x \right )+4 x}}+2 x^{2} {\mathrm e}^{x \,{\mathrm e}^{-x \ln \left (x \right )+4 x}}+x^{4}-x \right )\)	\(46\)

input

int((((-4*x*ln(x)+12*x+4)*exp(-x*ln(x)+4*x)+1)*exp(x*exp(-x*ln(x)+4*x))^2+ 
((-4*x^3*ln(x)+12*x^3+4*x^2)*exp(-x*ln(x)+4*x)+2*x^2+8*x)*exp(x*exp(-x*ln( 
x)+4*x))+x^4+8*x^3-x-2)/(exp(x*exp(-x*ln(x)+4*x))^2+2*x^2*exp(x*exp(-x*ln( 
x)+4*x))+x^4-x),x,method=_RETURNVERBOSE)
 

output

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

3.27.10.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x + 2 \, \log \left (x^{4} + 2 \, x^{2} e^{\left (x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )} - x + e^{\left (2 \, x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )}\right ) \]

input

integrate((((-4*x*log(x)+12*x+4)*exp(-x*log(x)+4*x)+1)*exp(x*exp(-x*log(x) 
+4*x))^2+((-4*x^3*log(x)+12*x^3+4*x^2)*exp(-x*log(x)+4*x)+2*x^2+8*x)*exp(x 
*exp(-x*log(x)+4*x))+x^4+8*x^3-x-2)/(exp(x*exp(-x*log(x)+4*x))^2+2*x^2*exp 
(x*exp(-x*log(x)+4*x))+x^4-x),x, algorithm=\
 

output

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

3.27.10.6 Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x + 2 \log {\left (x^{4} + 2 x^{2} e^{x e^{- x \log {\left (x \right )} + 4 x}} - x + e^{2 x e^{- x \log {\left (x \right )} + 4 x}} \right )} \]

input

integrate((((-4*x*ln(x)+12*x+4)*exp(-x*ln(x)+4*x)+1)*exp(x*exp(-x*ln(x)+4* 
x))**2+((-4*x**3*ln(x)+12*x**3+4*x**2)*exp(-x*ln(x)+4*x)+2*x**2+8*x)*exp(x 
*exp(-x*ln(x)+4*x))+x**4+8*x**3-x-2)/(exp(x*exp(-x*ln(x)+4*x))**2+2*x**2*e 
xp(x*exp(-x*ln(x)+4*x))+x**4-x),x)
 

output

x + 2*log(x**4 + 2*x**2*exp(x*exp(-x*log(x) + 4*x)) - x + exp(2*x*exp(-x*l 
og(x) + 4*x)))
 

3.27.10.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x + 2 \, \log \left (x^{4} + 2 \, x^{2} e^{\left (x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )} - x + e^{\left (2 \, x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )}\right ) \]

input

integrate((((-4*x*log(x)+12*x+4)*exp(-x*log(x)+4*x)+1)*exp(x*exp(-x*log(x) 
+4*x))^2+((-4*x^3*log(x)+12*x^3+4*x^2)*exp(-x*log(x)+4*x)+2*x^2+8*x)*exp(x 
*exp(-x*log(x)+4*x))+x^4+8*x^3-x-2)/(exp(x*exp(-x*log(x)+4*x))^2+2*x^2*exp 
(x*exp(-x*log(x)+4*x))+x^4-x),x, algorithm=\
 

output

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

3.27.10.8 Giac [A] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x + 2 \, \log \left (x^{4} + 2 \, x^{2} e^{\left (x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )} - x + e^{\left (2 \, x e^{\left (-x \log \left (x\right ) + 4 \, x\right )}\right )}\right ) \]

input

integrate((((-4*x*log(x)+12*x+4)*exp(-x*log(x)+4*x)+1)*exp(x*exp(-x*log(x) 
+4*x))^2+((-4*x^3*log(x)+12*x^3+4*x^2)*exp(-x*log(x)+4*x)+2*x^2+8*x)*exp(x 
*exp(-x*log(x)+4*x))+x^4+8*x^3-x-2)/(exp(x*exp(-x*log(x)+4*x))^2+2*x^2*exp 
(x*exp(-x*log(x)+4*x))+x^4-x),x, algorithm=\
 

output

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

3.27.10.9 Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {-2-x+8 x^3+x^4+e^{2 e^{4 x-x \log (x)} x} \left (1+e^{4 x-x \log (x)} (4+12 x-4 x \log (x))\right )+e^{e^{4 x-x \log (x)} x} \left (8 x+2 x^2+e^{4 x-x \log (x)} \left (4 x^2+12 x^3-4 x^3 \log (x)\right )\right )}{e^{2 e^{4 x-x \log (x)} x}-x+2 e^{e^{4 x-x \log (x)} x} x^2+x^4} \, dx=x+2\,\ln \left ({\mathrm {e}}^{2\,x^{1-x}\,{\mathrm {e}}^{4\,x}}-x+2\,x^2\,{\mathrm {e}}^{x^{1-x}\,{\mathrm {e}}^{4\,x}}+x^4\right ) \]

input

int((exp(x*exp(4*x - x*log(x)))*(8*x + exp(4*x - x*log(x))*(4*x^2 - 4*x^3* 
log(x) + 12*x^3) + 2*x^2) - x + 8*x^3 + x^4 + exp(2*x*exp(4*x - x*log(x))) 
*(exp(4*x - x*log(x))*(12*x - 4*x*log(x) + 4) + 1) - 2)/(exp(2*x*exp(4*x - 
 x*log(x))) - x + 2*x^2*exp(x*exp(4*x - x*log(x))) + x^4),x)
 

output

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