3.15.72 \(\int \frac {(4 x^3+(32 x^3-4 x^3 \log (x)) \log (16-2 \log (x))+(-16+18 \log (x)-2 \log ^2(x)) \log ^3(16-2 \log (x))) \log (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))})}{(8 x^4-x^4 \log (x)) \log (16-2 \log (x))+(-8 x^2+(-8 x+x^2) \log (x)+x \log ^2(x)) \log ^3(16-2 \log (x))} \, dx\) [1472]

3.15.72.1 Optimal result
3.15.72.2 Mathematica [A] (verified)
3.15.72.3 Rubi [F]
3.15.72.4 Maple [F]
3.15.72.5 Fricas [A] (verification not implemented)
3.15.72.6 Sympy [A] (verification not implemented)
3.15.72.7 Maxima [F]
3.15.72.8 Giac [F(-1)]
3.15.72.9 Mupad [B] (verification not implemented)

3.15.72.1 Optimal result

Integrand size = 142, antiderivative size = 28 \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\log ^2\left (-\frac {x+\log (x)}{x}+\frac {x^2}{\log ^2(2 (8-\log (x)))}\right ) \]

output
ln(x^2/ln(-2*ln(x)+16)^2-(x+ln(x))/x)^2
 
3.15.72.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\log ^2\left (-1-\frac {\log (x)}{x}+\frac {x^2}{\log ^2(-2 (-8+\log (x)))}\right ) \]

input
Integrate[((4*x^3 + (32*x^3 - 4*x^3*Log[x])*Log[16 - 2*Log[x]] + (-16 + 18 
*Log[x] - 2*Log[x]^2)*Log[16 - 2*Log[x]]^3)*Log[(x^3 + (-x - Log[x])*Log[1 
6 - 2*Log[x]]^2)/(x*Log[16 - 2*Log[x]]^2)])/((8*x^4 - x^4*Log[x])*Log[16 - 
 2*Log[x]] + (-8*x^2 + (-8*x + x^2)*Log[x] + x*Log[x]^2)*Log[16 - 2*Log[x] 
]^3),x]
 
output
Log[-1 - Log[x]/x + x^2/Log[-2*(-8 + Log[x])]^2]^2
 
3.15.72.3 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.

\(\displaystyle \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-2 \log ^2(x)+18 \log (x)-16\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (x^2-8 x\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-2 \log ^2(x)+18 \log (x)-16\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(-2 (\log (x)-8))}\right )}{x (8-\log (x)) \log (-2 (\log (x)-8)) \left (x^3-x \log ^2(-2 (\log (x)-8))-\log (x) \log ^2(-2 (\log (x)-8))\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {4 x^2 \log (x) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{(\log (x)-8) \left (-x^3+x \log ^2(-2 (\log (x)-8))+\log (x) \log ^2(-2 (\log (x)-8))\right )}+\frac {32 x^2 \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{(\log (x)-8) \left (-x^3+x \log ^2(-2 (\log (x)-8))+\log (x) \log ^2(-2 (\log (x)-8))\right )}+\frac {4 x^2 \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{(\log (x)-8) \log (-2 (\log (x)-8)) \left (-x^3+x \log ^2(-2 (\log (x)-8))+\log (x) \log ^2(-2 (\log (x)-8))\right )}+\frac {2 \log ^2(x) \log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{x (\log (x)-8) \left (x^3-x \log ^2(-2 (\log (x)-8))-\log (x) \log ^2(-2 (\log (x)-8))\right )}-\frac {18 \log (x) \log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{x (\log (x)-8) \left (x^3-x \log ^2(-2 (\log (x)-8))-\log (x) \log ^2(-2 (\log (x)-8))\right )}+\frac {16 \log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-\frac {\log (x)}{x}-1\right )}{x (\log (x)-8) \left (x^3-x \log ^2(-2 (\log (x)-8))-\log (x) \log ^2(-2 (\log (x)-8))\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 16 \int \frac {\log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{x (\log (x)-8) \left (x^3-\log ^2(-2 (\log (x)-8)) x-\log (x) \log ^2(-2 (\log (x)-8))\right )}dx-18 \int \frac {\log (x) \log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{x (\log (x)-8) \left (x^3-\log ^2(-2 (\log (x)-8)) x-\log (x) \log ^2(-2 (\log (x)-8))\right )}dx+2 \int \frac {\log ^2(x) \log ^2(-2 (\log (x)-8)) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{x (\log (x)-8) \left (x^3-\log ^2(-2 (\log (x)-8)) x-\log (x) \log ^2(-2 (\log (x)-8))\right )}dx+32 \int \frac {x^2 \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{(\log (x)-8) \left (-x^3+\log ^2(-2 (\log (x)-8)) x+\log (x) \log ^2(-2 (\log (x)-8))\right )}dx-4 \int \frac {x^2 \log (x) \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{(\log (x)-8) \left (-x^3+\log ^2(-2 (\log (x)-8)) x+\log (x) \log ^2(-2 (\log (x)-8))\right )}dx+4 \int \frac {x^2 \log \left (\frac {x^2}{\log ^2(-2 (\log (x)-8))}-1-\frac {\log (x)}{x}\right )}{(\log (x)-8) \log (-2 (\log (x)-8)) \left (-x^3+\log ^2(-2 (\log (x)-8)) x+\log (x) \log ^2(-2 (\log (x)-8))\right )}dx\)

input
Int[((4*x^3 + (32*x^3 - 4*x^3*Log[x])*Log[16 - 2*Log[x]] + (-16 + 18*Log[x 
] - 2*Log[x]^2)*Log[16 - 2*Log[x]]^3)*Log[(x^3 + (-x - Log[x])*Log[16 - 2* 
Log[x]]^2)/(x*Log[16 - 2*Log[x]]^2)])/((8*x^4 - x^4*Log[x])*Log[16 - 2*Log 
[x]] + (-8*x^2 + (-8*x + x^2)*Log[x] + x*Log[x]^2)*Log[16 - 2*Log[x]]^3),x 
]
 
output
$Aborted
 

3.15.72.3.1 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] 
]
 
3.15.72.4 Maple [F]

\[\int \frac {\left (\left (-2 \ln \left (x \right )^{2}+18 \ln \left (x \right )-16\right ) \ln \left (-2 \ln \left (x \right )+16\right )^{3}+\left (-4 x^{3} \ln \left (x \right )+32 x^{3}\right ) \ln \left (-2 \ln \left (x \right )+16\right )+4 x^{3}\right ) \ln \left (\frac {\left (-x -\ln \left (x \right )\right ) \ln \left (-2 \ln \left (x \right )+16\right )^{2}+x^{3}}{x \ln \left (-2 \ln \left (x \right )+16\right )^{2}}\right )}{\left (x \ln \left (x \right )^{2}+\left (x^{2}-8 x \right ) \ln \left (x \right )-8 x^{2}\right ) \ln \left (-2 \ln \left (x \right )+16\right )^{3}+\left (-x^{4} \ln \left (x \right )+8 x^{4}\right ) \ln \left (-2 \ln \left (x \right )+16\right )}d x\]

input
int(((-2*ln(x)^2+18*ln(x)-16)*ln(-2*ln(x)+16)^3+(-4*x^3*ln(x)+32*x^3)*ln(- 
2*ln(x)+16)+4*x^3)*ln(((-x-ln(x))*ln(-2*ln(x)+16)^2+x^3)/x/ln(-2*ln(x)+16) 
^2)/((x*ln(x)^2+(x^2-8*x)*ln(x)-8*x^2)*ln(-2*ln(x)+16)^3+(-x^4*ln(x)+8*x^4 
)*ln(-2*ln(x)+16)),x)
 
output
int(((-2*ln(x)^2+18*ln(x)-16)*ln(-2*ln(x)+16)^3+(-4*x^3*ln(x)+32*x^3)*ln(- 
2*ln(x)+16)+4*x^3)*ln(((-x-ln(x))*ln(-2*ln(x)+16)^2+x^3)/x/ln(-2*ln(x)+16) 
^2)/((x*ln(x)^2+(x^2-8*x)*ln(x)-8*x^2)*ln(-2*ln(x)+16)^3+(-x^4*ln(x)+8*x^4 
)*ln(-2*ln(x)+16)),x)
 
3.15.72.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\log \left (\frac {x^{3} - {\left (x + \log \left (x\right )\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )^{2}}{x \log \left (-2 \, \log \left (x\right ) + 16\right )^{2}}\right )^{2} \]

input
integrate(((-2*log(x)^2+18*log(x)-16)*log(-2*log(x)+16)^3+(-4*x^3*log(x)+3 
2*x^3)*log(-2*log(x)+16)+4*x^3)*log(((-x-log(x))*log(-2*log(x)+16)^2+x^3)/ 
x/log(-2*log(x)+16)^2)/((x*log(x)^2+(x^2-8*x)*log(x)-8*x^2)*log(-2*log(x)+ 
16)^3+(-x^4*log(x)+8*x^4)*log(-2*log(x)+16)),x, algorithm=\
 
output
log((x^3 - (x + log(x))*log(-2*log(x) + 16)^2)/(x*log(-2*log(x) + 16)^2))^ 
2
 
3.15.72.6 Sympy [A] (verification not implemented)

Time = 2.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\log {\left (\frac {x^{3} + \left (- x - \log {\left (x \right )}\right ) \log {\left (16 - 2 \log {\left (x \right )} \right )}^{2}}{x \log {\left (16 - 2 \log {\left (x \right )} \right )}^{2}} \right )}^{2} \]

input
integrate(((-2*ln(x)**2+18*ln(x)-16)*ln(-2*ln(x)+16)**3+(-4*x**3*ln(x)+32* 
x**3)*ln(-2*ln(x)+16)+4*x**3)*ln(((-x-ln(x))*ln(-2*ln(x)+16)**2+x**3)/x/ln 
(-2*ln(x)+16)**2)/((x*ln(x)**2+(x**2-8*x)*ln(x)-8*x**2)*ln(-2*ln(x)+16)**3 
+(-x**4*ln(x)+8*x**4)*ln(-2*ln(x)+16)),x)
 
output
log((x**3 + (-x - log(x))*log(16 - 2*log(x))**2)/(x*log(16 - 2*log(x))**2) 
)**2
 
3.15.72.7 Maxima [F]

\[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\int { -\frac {2 \, {\left ({\left (\log \left (x\right )^{2} - 9 \, \log \left (x\right ) + 8\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )^{3} - 2 \, x^{3} + 2 \, {\left (x^{3} \log \left (x\right ) - 8 \, x^{3}\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )\right )} \log \left (\frac {x^{3} - {\left (x + \log \left (x\right )\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )^{2}}{x \log \left (-2 \, \log \left (x\right ) + 16\right )^{2}}\right )}{{\left (x \log \left (x\right )^{2} - 8 \, x^{2} + {\left (x^{2} - 8 \, x\right )} \log \left (x\right )\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )^{3} - {\left (x^{4} \log \left (x\right ) - 8 \, x^{4}\right )} \log \left (-2 \, \log \left (x\right ) + 16\right )} \,d x } \]

input
integrate(((-2*log(x)^2+18*log(x)-16)*log(-2*log(x)+16)^3+(-4*x^3*log(x)+3 
2*x^3)*log(-2*log(x)+16)+4*x^3)*log(((-x-log(x))*log(-2*log(x)+16)^2+x^3)/ 
x/log(-2*log(x)+16)^2)/((x*log(x)^2+(x^2-8*x)*log(x)-8*x^2)*log(-2*log(x)+ 
16)^3+(-x^4*log(x)+8*x^4)*log(-2*log(x)+16)),x, algorithm=\
 
output
-2*integrate(((log(x)^2 - 9*log(x) + 8)*log(-2*log(x) + 16)^3 - 2*x^3 + 2* 
(x^3*log(x) - 8*x^3)*log(-2*log(x) + 16))*log((x^3 - (x + log(x))*log(-2*l 
og(x) + 16)^2)/(x*log(-2*log(x) + 16)^2))/((x*log(x)^2 - 8*x^2 + (x^2 - 8* 
x)*log(x))*log(-2*log(x) + 16)^3 - (x^4*log(x) - 8*x^4)*log(-2*log(x) + 16 
)), x)
 
3.15.72.8 Giac [F(-1)]

Timed out. \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx=\text {Timed out} \]

input
integrate(((-2*log(x)^2+18*log(x)-16)*log(-2*log(x)+16)^3+(-4*x^3*log(x)+3 
2*x^3)*log(-2*log(x)+16)+4*x^3)*log(((-x-log(x))*log(-2*log(x)+16)^2+x^3)/ 
x/log(-2*log(x)+16)^2)/((x*log(x)^2+(x^2-8*x)*log(x)-8*x^2)*log(-2*log(x)+ 
16)^3+(-x^4*log(x)+8*x^4)*log(-2*log(x)+16)),x, algorithm=\
 
output
Timed out
 
3.15.72.9 Mupad [B] (verification not implemented)

Time = 12.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {\left (4 x^3+\left (32 x^3-4 x^3 \log (x)\right ) \log (16-2 \log (x))+\left (-16+18 \log (x)-2 \log ^2(x)\right ) \log ^3(16-2 \log (x))\right ) \log \left (\frac {x^3+(-x-\log (x)) \log ^2(16-2 \log (x))}{x \log ^2(16-2 \log (x))}\right )}{\left (8 x^4-x^4 \log (x)\right ) \log (16-2 \log (x))+\left (-8 x^2+\left (-8 x+x^2\right ) \log (x)+x \log ^2(x)\right ) \log ^3(16-2 \log (x))} \, dx={\ln \left (-\frac {{\ln \left (16-2\,\ln \left (x\right )\right )}^2\,\left (x+\ln \left (x\right )\right )-x^3}{x\,{\ln \left (16-2\,\ln \left (x\right )\right )}^2}\right )}^2 \]

input
int((log(-(log(16 - 2*log(x))^2*(x + log(x)) - x^3)/(x*log(16 - 2*log(x))^ 
2))*(log(16 - 2*log(x))^3*(2*log(x)^2 - 18*log(x) + 16) + log(16 - 2*log(x 
))*(4*x^3*log(x) - 32*x^3) - 4*x^3))/(log(16 - 2*log(x))*(x^4*log(x) - 8*x 
^4) + log(16 - 2*log(x))^3*(log(x)*(8*x - x^2) - x*log(x)^2 + 8*x^2)),x)
 
output
log(-(log(16 - 2*log(x))^2*(x + log(x)) - x^3)/(x*log(16 - 2*log(x))^2))^2