3.4.10 \(\int \frac {(x+e^{1-x} x-2 x^2) \log (x)+(-2-2 e^{1-x}) \log ^2(x)+8 \log ^3(x)+(-x-e^{1-x} x+2 x^2+(-2 x^2-e^{1-x} x^2) \log (x)+(1+e^{1-x}-2 x) \log ^2(x)+(2 x+e^{1-x} x) \log ^3(x)) \log (-x+\log ^2(x))}{((x^2+e^{1-x} x^2-2 x^3) \log (x)-4 x^2 \log ^2(x)+(-x-e^{1-x} x+2 x^2) \log ^3(x)+4 x \log ^4(x)) \log (-x+\log ^2(x))} \, dx\) [310]

3.4.10.1 Optimal result

Integrand size = 205, antiderivative size = 32 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=\log \left (\left (4+\frac {-1-e^{1-x}+2 x}{\log (x)}\right ) \log \left (-x+\log ^2(x)\right )\right ) \]

output

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

3.4.10.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=-x-\log (\log (x))+\log \left (-e-e^x+2 e^x x+4 e^x \log (x)\right )+\log \left (\log \left (-x+\log ^2(x)\right )\right ) \]

input

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

output

-x - Log[Log[x]] + Log[-E - E^x + 2*E^x*x + 4*E^x*Log[x]] + Log[Log[-x + L 
og[x]^2]]
 

3.4.10.3 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.4.10.3.1 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] 
]
 

3.4.10.4 Maple [A] (verified)

Time = 13.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03

method	result	size
risch	\(\ln \left (\frac {x}{2}-\frac {{\mathrm e}^{1-x}}{4}+\ln \left (x \right )-\frac {1}{4}\right )-\ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (\ln \left (x \right )^{2}-x \right )\right )\)	\(33\)
parallelrisch	\(-\ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (\ln \left (x \right )^{2}-x \right )\right )+\ln \left (x +2 \ln \left (x \right )-\frac {{\mathrm e}^{1-x}}{2}-\frac {1}{2}\right )\)	\(33\)

input

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

output

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

3.4.10.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=\log \left (2 \, x - e^{\left (-x + 1\right )} + 4 \, \log \left (x\right ) - 1\right ) + \log \left (\log \left (\log \left (x\right )^{2} - x\right )\right ) - \log \left (\log \left (x\right )\right ) \]

input

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

output

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

3.4.10.6 Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=\log {\left (- 2 x + e^{1 - x} - 4 \log {\left (x \right )} + 1 \right )} - \log {\left (\log {\left (x \right )} \right )} + \log {\left (\log {\left (- x + \log {\left (x \right )}^{2} \right )} \right )} \]

input

integrate((((x*exp(1-x)+2*x)*ln(x)**3+(exp(1-x)+1-2*x)*ln(x)**2+(-x**2*exp 
(1-x)-2*x**2)*ln(x)-x*exp(1-x)+2*x**2-x)*ln(ln(x)**2-x)+8*ln(x)**3+(-2*exp 
(1-x)-2)*ln(x)**2+(x*exp(1-x)-2*x**2+x)*ln(x))/(4*x*ln(x)**4+(-x*exp(1-x)+ 
2*x**2-x)*ln(x)**3-4*x**2*ln(x)**2+(x**2*exp(1-x)-2*x**3+x**2)*ln(x))/ln(l 
n(x)**2-x),x)
 

output

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

3.4.10.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.78 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=-x + \log \left (\frac {1}{2} \, x + \log \left (x\right ) - \frac {1}{4}\right ) + \log \left (\frac {{\left (2 \, x + 4 \, \log \left (x\right ) - 1\right )} e^{x} - e}{2 \, x + 4 \, \log \left (x\right ) - 1}\right ) + \log \left (\log \left (\log \left (x\right )^{2} - x\right )\right ) - \log \left (\log \left (x\right )\right ) \]

input

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

output

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

3.4.10.8 Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=\log \left (2 \, x - e^{\left (-x + 1\right )} + 4 \, \log \left (x\right ) - 1\right ) + \log \left (\log \left (\log \left (x\right )^{2} - x\right )\right ) - \log \left (\log \left (x\right )\right ) \]

input

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

output

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

3.4.10.9 Mupad [B] (verification not implemented)

Time = 8.48 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.50 \[ \int \frac {\left (x+e^{1-x} x-2 x^2\right ) \log (x)+\left (-2-2 e^{1-x}\right ) \log ^2(x)+8 \log ^3(x)+\left (-x-e^{1-x} x+2 x^2+\left (-2 x^2-e^{1-x} x^2\right ) \log (x)+\left (1+e^{1-x}-2 x\right ) \log ^2(x)+\left (2 x+e^{1-x} x\right ) \log ^3(x)\right ) \log \left (-x+\log ^2(x)\right )}{\left (\left (x^2+e^{1-x} x^2-2 x^3\right ) \log (x)-4 x^2 \log ^2(x)+\left (-x-e^{1-x} x+2 x^2\right ) \log ^3(x)+4 x \log ^4(x)\right ) \log \left (-x+\log ^2(x)\right )} \, dx=\ln \left (\frac {2\,x-{\mathrm {e}}^{1-x}+4\,\ln \left (x\right )-1}{x}\right )+\ln \left (\frac {2\,x+x\,{\mathrm {e}}^{1-x}+4}{x}\right )+\ln \left (\ln \left ({\ln \left (x\right )}^2-x\right )\right )-\ln \left (\frac {4\,\ln \left (x\right )+2\,x\,\ln \left (x\right )+x\,{\mathrm {e}}^{1-x}\,\ln \left (x\right )}{x}\right )+\ln \left (x\right ) \]

input

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

output

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