3.20.66 \(\int \frac {e^{\frac {2+8 x+2 x^2+(x+2 x^2+x^3) \log (\frac {1}{x})}{2+2 x+(x+x^2) \log (\frac {1}{x})}} (12+12 x+8 x^2+(4 x+4 x^2+4 x^3) \log (\frac {1}{x})+(x^2+2 x^3+x^4) \log ^2(\frac {1}{x}))}{4+8 x+4 x^2+(4 x+8 x^2+4 x^3) \log (\frac {1}{x})+(x^2+2 x^3+x^4) \log ^2(\frac {1}{x})} \, dx\) [1966]

3.20.66.1 Optimal result

Integrand size = 144, antiderivative size = 24 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{1+x+\frac {4}{(1+x) \left (\frac {2}{x}+\log \left (\frac {1}{x}\right )\right )}} \]

output

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

3.20.66.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\frac {2 \left (1+4 x+x^2\right )+x (1+x)^2 \log \left (\frac {1}{x}\right )}{(1+x) \left (2+x \log \left (\frac {1}{x}\right )\right )}} \]

input

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

output

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

3.20.66.3 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.20.66.3.1 Defintions of rubi rules used

rule 2009

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

rule 7292

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

rule 7293

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

3.20.66.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(48\) vs. \(2(23)=46\).

Time = 8.81 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.04

method	result	size
parallelrisch	\({\mathrm e}^{\frac {\left (x^{3}+2 x^{2}+x \right ) \ln \left (\frac {1}{x}\right )+2 x^{2}+8 x +2}{x^{2} \ln \left (\frac {1}{x}\right )+x \ln \left (\frac {1}{x}\right )+2 x +2}}\)	\(49\)
risch	\({\mathrm e}^{\frac {x^{3} \ln \left (\frac {1}{x}\right )+2 x^{2} \ln \left (\frac {1}{x}\right )+x \ln \left (\frac {1}{x}\right )+2 x^{2}+8 x +2}{\left (1+x \right ) \left (x \ln \left (\frac {1}{x}\right )+2\right )}}\)	\(51\)

input

int(((x^4+2*x^3+x^2)*ln(1/x)^2+(4*x^3+4*x^2+4*x)*ln(1/x)+8*x^2+12*x+12)*ex 
p(((x^3+2*x^2+x)*ln(1/x)+2*x^2+8*x+2)/((x^2+x)*ln(1/x)+2*x+2))/((x^4+2*x^3 
+x^2)*ln(1/x)^2+(4*x^3+8*x^2+4*x)*ln(1/x)+4*x^2+8*x+4),x,method=_RETURNVER 
BOSE)
 

output

exp(((x^3+2*x^2+x)*ln(1/x)+2*x^2+8*x+2)/(x^2*ln(1/x)+x*ln(1/x)+2*x+2))
 

3.20.66.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (\frac {2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + x\right )} \log \left (\frac {1}{x}\right ) + 8 \, x + 2}{{\left (x^{2} + x\right )} \log \left (\frac {1}{x}\right ) + 2 \, x + 2}\right )} \]

input

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

output

e^((2*x^2 + (x^3 + 2*x^2 + x)*log(1/x) + 8*x + 2)/((x^2 + x)*log(1/x) + 2* 
x + 2))
 

3.20.66.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\frac {2 x^{2} + 8 x + \left (x^{3} + 2 x^{2} + x\right ) \log {\left (\frac {1}{x} \right )} + 2}{2 x + \left (x^{2} + x\right ) \log {\left (\frac {1}{x} \right )} + 2}} \]

input

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

output

exp((2*x**2 + 8*x + (x**3 + 2*x**2 + x)*log(1/x) + 2)/(2*x + (x**2 + x)*lo 
g(1/x) + 2))
 

3.20.66.7 Maxima [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (x - \frac {8}{x \log \left (x\right )^{2} + 2 \, {\left (x - 1\right )} \log \left (x\right ) - 4} - \frac {4}{{\left (x + 1\right )} \log \left (x\right ) + 2 \, x + 2} + 1\right )} \]

input

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

output

e^(x - 8/(x*log(x)^2 + 2*(x - 1)*log(x) - 4) - 4/((x + 1)*log(x) + 2*x + 2 
) + 1)
 

3.20.66.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (23) = 46\).

Time = 0.46 (sec) , antiderivative size = 131, normalized size of antiderivative = 5.46 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx=e^{\left (\frac {x^{3} \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} + \frac {2 \, x^{2} \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {2 \, x^{2}}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} + \frac {x \log \left (x\right )}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {8 \, x}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2} - \frac {2}{x^{2} \log \left (x\right ) + x \log \left (x\right ) - 2 \, x - 2}\right )} \]

input

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

output

e^(x^3*log(x)/(x^2*log(x) + x*log(x) - 2*x - 2) + 2*x^2*log(x)/(x^2*log(x) 
 + x*log(x) - 2*x - 2) - 2*x^2/(x^2*log(x) + x*log(x) - 2*x - 2) + x*log(x 
)/(x^2*log(x) + x*log(x) - 2*x - 2) - 8*x/(x^2*log(x) + x*log(x) - 2*x - 2 
) - 2/(x^2*log(x) + x*log(x) - 2*x - 2))
 

3.20.66.9 Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.04 \[ \int \frac {e^{\frac {2+8 x+2 x^2+\left (x+2 x^2+x^3\right ) \log \left (\frac {1}{x}\right )}{2+2 x+\left (x+x^2\right ) \log \left (\frac {1}{x}\right )}} \left (12+12 x+8 x^2+\left (4 x+4 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )\right )}{4+8 x+4 x^2+\left (4 x+8 x^2+4 x^3\right ) \log \left (\frac {1}{x}\right )+\left (x^2+2 x^3+x^4\right ) \log ^2\left (\frac {1}{x}\right )} \, dx={\mathrm {e}}^{\frac {8\,x}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\mathrm {e}}^{\frac {2\,x^2}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\mathrm {e}}^{\frac {2}{2\,x+x\,\ln \left (\frac {1}{x}\right )+x^2\,\ln \left (\frac {1}{x}\right )+2}}\,{\left (\frac {1}{x}\right )}^{\frac {x^2+x}{x\,\ln \left (\frac {1}{x}\right )+2}} \]

input

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

output

exp((8*x)/(2*x + x*log(1/x) + x^2*log(1/x) + 2))*exp((2*x^2)/(2*x + x*log( 
1/x) + x^2*log(1/x) + 2))*exp(2/(2*x + x*log(1/x) + x^2*log(1/x) + 2))*(1/ 
x)^((x + x^2)/(x*log(1/x) + 2))