\(\int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+(-15 x^3+(-120 x^2+15 x^3) \log (-8+x)+(8 x^2-x^3) \log ^2(-8+x)+(15 x^2+(120 x-15 x^2) \log (-8+x)+(-8 x+x^2) \log ^2(-8+x)) \log (2 x)) \log ^2(\frac {x-\log (2 x)}{x})}{((8 x^2-x^3) \log ^2(-8+x)+(-8 x+x^2) \log ^2(-8+x) \log (2 x)) \log ^2(\frac {x-\log (2 x)}{x})} \, dx\) [1282]

Optimal result

Integrand size = 179, antiderivative size = 27 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \] Output:

x+4/ln(1-ln(2*x)/x)-15/ln(-8+x)*x
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \] Input:

Integrate[((32 - 4*x)*Log[-8 + x]^2 + (-32 + 4*x)*Log[-8 + x]^2*Log[2*x] + 
 (-15*x^3 + (-120*x^2 + 15*x^3)*Log[-8 + x] + (8*x^2 - x^3)*Log[-8 + x]^2 
+ (15*x^2 + (120*x - 15*x^2)*Log[-8 + x] + (-8*x + x^2)*Log[-8 + x]^2)*Log 
[2*x])*Log[(x - Log[2*x])/x]^2)/(((8*x^2 - x^3)*Log[-8 + x]^2 + (-8*x + x^ 
2)*Log[-8 + x]^2*Log[2*x])*Log[(x - Log[2*x])/x]^2),x]
 

Output:

x - (15*x)/Log[-8 + x] + 4/Log[1 - Log[2*x]/x]
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {7239, 2009}

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.

\(\displaystyle \int \frac {\left (-15 x^3+\left (15 x^2+\left (x^2-8 x\right ) \log ^2(x-8)+\left (120 x-15 x^2\right ) \log (x-8)\right ) \log (2 x)+\left (8 x^2-x^3\right ) \log ^2(x-8)+\left (15 x^3-120 x^2\right ) \log (x-8)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )+(32-4 x) \log ^2(x-8)+(4 x-32) \log (2 x) \log ^2(x-8)}{\left (\left (x^2-8 x\right ) \log (2 x) \log ^2(x-8)+\left (8 x^2-x^3\right ) \log ^2(x-8)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \left (\frac {x^2 \log ^2\left (1-\frac {\log (2 x)}{x}\right )-\log (2 x) \left (x \log ^2\left (1-\frac {\log (2 x)}{x}\right )+4\right )+4}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )}+\frac {15 x}{(x-8) \log ^2(x-8)}-\frac {15}{\log (x-8)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle x+\frac {15 (8-x)}{\log (x-8)}-\frac {120}{\log (x-8)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}\)

Input:

Int[((32 - 4*x)*Log[-8 + x]^2 + (-32 + 4*x)*Log[-8 + x]^2*Log[2*x] + (-15* 
x^3 + (-120*x^2 + 15*x^3)*Log[-8 + x] + (8*x^2 - x^3)*Log[-8 + x]^2 + (15* 
x^2 + (120*x - 15*x^2)*Log[-8 + x] + (-8*x + x^2)*Log[-8 + x]^2)*Log[2*x]) 
*Log[(x - Log[2*x])/x]^2)/(((8*x^2 - x^3)*Log[-8 + x]^2 + (-8*x + x^2)*Log 
[-8 + x]^2*Log[2*x])*Log[(x - Log[2*x])/x]^2),x]
 

Output:

x - 120/Log[-8 + x] + (15*(8 - x))/Log[-8 + x] + 4/Log[1 - Log[2*x]/x]
 

Defintions of rubi rules used

rule 2009

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

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

Maple [A] (verified)

Time = 16.63 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52

method	result	size
default	\(x -\frac {15 \left (-8+x \right )}{\ln \left (-8+x \right )}-\frac {120}{\ln \left (-8+x \right )}+\frac {4}{\ln \left (\frac {x -\ln \left (2\right )-\ln \left (x \right )}{x}\right )}\)	\(41\)
parts	\(x -\frac {15 \left (-8+x \right )}{\ln \left (-8+x \right )}-\frac {120}{\ln \left (-8+x \right )}+\frac {4}{\ln \left (\frac {x -\ln \left (2\right )-\ln \left (x \right )}{x}\right )}\)	\(41\)
parallelrisch	\(\frac {16 x \ln \left (-8+x \right ) \ln \left (-\frac {-x +\ln \left (2 x \right )}{x}\right )-240 \ln \left (-\frac {-x +\ln \left (2 x \right )}{x}\right ) x +64 \ln \left (-\frac {-x +\ln \left (2 x \right )}{x}\right ) \ln \left (-8+x \right )+64 \ln \left (-8+x \right )}{16 \ln \left (-8+x \right ) \ln \left (-\frac {-x +\ln \left (2 x \right )}{x}\right )}\)	\(90\)

Input:

int(((((x^2-8*x)*ln(-8+x)^2+(-15*x^2+120*x)*ln(-8+x)+15*x^2)*ln(2*x)+(-x^3 
+8*x^2)*ln(-8+x)^2+(15*x^3-120*x^2)*ln(-8+x)-15*x^3)*ln((x-ln(2*x))/x)^2+( 
4*x-32)*ln(-8+x)^2*ln(2*x)+(-4*x+32)*ln(-8+x)^2)/((x^2-8*x)*ln(-8+x)^2*ln( 
2*x)+(-x^3+8*x^2)*ln(-8+x)^2)/ln((x-ln(2*x))/x)^2,x,method=_RETURNVERBOSE)
 

Output:

x-15/ln(-8+x)*(-8+x)-120/ln(-8+x)+4/ln((x-ln(2)-ln(x))/x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right ) + 4 \, \log \left (x - 8\right )}{\log \left (x - 8\right ) \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right )} \] Input:

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2 
*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3-120*x^2)*log(-8+x)-15*x^3)*log((x-log 
(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8*x 
)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, 
algorithm="fricas")
 

Output:

((x*log(x - 8) - 15*x)*log((x - log(2*x))/x) + 4*log(x - 8))/(log(x - 8)*l 
og((x - log(2*x))/x))
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x - \frac {15 x}{\log {\left (x - 8 \right )}} + \frac {4}{\log {\left (\frac {x - \log {\left (2 x \right )}}{x} \right )}} \] Input:

integrate(((((x**2-8*x)*ln(-8+x)**2+(-15*x**2+120*x)*ln(-8+x)+15*x**2)*ln( 
2*x)+(-x**3+8*x**2)*ln(-8+x)**2+(15*x**3-120*x**2)*ln(-8+x)-15*x**3)*ln((x 
-ln(2*x))/x)**2+(4*x-32)*ln(-8+x)**2*ln(2*x)+(-4*x+32)*ln(-8+x)**2)/((x**2 
-8*x)*ln(-8+x)**2*ln(2*x)+(-x**3+8*x**2)*ln(-8+x)**2)/ln((x-ln(2*x))/x)**2 
,x)
                                                                                    
                                                                                    
 

Output:

x - 15*x/log(x - 8) + 4/log((x - log(2*x))/x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (27) = 54\).

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (x - \log \left (2\right ) - \log \left (x\right )\right ) - {\left (x \log \left (x\right ) - 4\right )} \log \left (x - 8\right ) + 15 \, x \log \left (x\right )}{\log \left (x - \log \left (2\right ) - \log \left (x\right )\right ) \log \left (x - 8\right ) - \log \left (x - 8\right ) \log \left (x\right )} \] Input:

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2 
*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3-120*x^2)*log(-8+x)-15*x^3)*log((x-log 
(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8*x 
)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, 
algorithm="maxima")
 

Output:

((x*log(x - 8) - 15*x)*log(x - log(2) - log(x)) - (x*log(x) - 4)*log(x - 8 
) + 15*x*log(x))/(log(x - log(2) - log(x))*log(x - 8) - log(x - 8)*log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (27) = 54\).

Time = 0.40 (sec) , antiderivative size = 235, normalized size of antiderivative = 8.70 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=x + \frac {4 \, {\left ({\left (x - 8\right )} \log \left (2\right ) - \log \left (2\right ) \log \left (2 \, x\right ) + {\left (x - 8\right )} \log \left (x\right ) - \log \left (2 \, x\right ) \log \left (x\right ) - x + 8 \, \log \left (2\right ) + \log \left (2 \, x\right ) + 8 \, \log \left (x\right )\right )}}{{\left (x - 8\right )} \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - \log \left (2\right ) \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - {\left (x - 8\right )} \log \left (2 \, x\right ) \log \left (x\right ) + \log \left (2\right ) \log \left (2 \, x\right ) \log \left (x\right ) - \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) \log \left (x\right ) + \log \left (2 \, x\right ) \log \left (x\right )^{2} - {\left (x - 8\right )} \log \left (x - \log \left (2 \, x\right )\right ) + \log \left (2\right ) \log \left (x - \log \left (2 \, x\right )\right ) + 8 \, \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) + {\left (x - 8\right )} \log \left (x\right ) - \log \left (2\right ) \log \left (x\right ) - 8 \, \log \left (2 \, x\right ) \log \left (x\right ) + \log \left (x - \log \left (2 \, x\right )\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 8 \, \log \left (x - \log \left (2 \, x\right )\right ) + 8 \, \log \left (x\right )} - \frac {15 \, x}{\log \left (x - 8\right )} - 8 \] Input:

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2 
*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3-120*x^2)*log(-8+x)-15*x^3)*log((x-log 
(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8*x 
)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, 
algorithm="giac")
 

Output:

x + 4*((x - 8)*log(2) - log(2)*log(2*x) + (x - 8)*log(x) - log(2*x)*log(x) 
 - x + 8*log(2) + log(2*x) + 8*log(x))/((x - 8)*log(2*x)*log(x - log(2*x)) 
 - log(2)*log(2*x)*log(x - log(2*x)) - (x - 8)*log(2*x)*log(x) + log(2)*lo 
g(2*x)*log(x) - log(2*x)*log(x - log(2*x))*log(x) + log(2*x)*log(x)^2 - (x 
 - 8)*log(x - log(2*x)) + log(2)*log(x - log(2*x)) + 8*log(2*x)*log(x - lo 
g(2*x)) + (x - 8)*log(x) - log(2)*log(x) - 8*log(2*x)*log(x) + log(x - log 
(2*x))*log(x) - log(x)^2 - 8*log(x - log(2*x)) + 8*log(x)) - 15*x/log(x - 
8) - 8
 

Mupad [B] (verification not implemented)

Time = 7.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {4}{\ln \left (\frac {x-\ln \left (2\,x\right )}{x}\right )}-\frac {15\,x-\ln \left (x-8\right )\,\left (15\,x-120\right )}{\ln \left (x-8\right )}-14\,x \] Input:

int(-(log(x - 8)^2*(4*x - 32) + log((x - log(2*x))/x)^2*(log(x - 8)*(120*x 
^2 - 15*x^3) - log(2*x)*(log(x - 8)*(120*x - 15*x^2) - log(x - 8)^2*(8*x - 
 x^2) + 15*x^2) - log(x - 8)^2*(8*x^2 - x^3) + 15*x^3) - log(2*x)*log(x - 
8)^2*(4*x - 32))/(log((x - log(2*x))/x)^2*(log(x - 8)^2*(8*x^2 - x^3) - lo 
g(2*x)*log(x - 8)^2*(8*x - x^2))),x)
 

Output:

4/log((x - log(2*x))/x) - (15*x - log(x - 8)*(15*x - 120))/log(x - 8) - 14 
*x
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.37 \[ \int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+\left (-15 x^3+\left (-120 x^2+15 x^3\right ) \log (-8+x)+\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (15 x^2+\left (120 x-15 x^2\right ) \log (-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x)\right ) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )}{\left (\left (8 x^2-x^3\right ) \log ^2(-8+x)+\left (-8 x+x^2\right ) \log ^2(-8+x) \log (2 x)\right ) \log ^2\left (\frac {x-\log (2 x)}{x}\right )} \, dx=\frac {\mathrm {log}\left (x -8\right ) \mathrm {log}\left (\frac {-\mathrm {log}\left (2 x \right )+x}{x}\right ) x +4 \,\mathrm {log}\left (x -8\right )-15 \,\mathrm {log}\left (\frac {-\mathrm {log}\left (2 x \right )+x}{x}\right ) x}{\mathrm {log}\left (x -8\right ) \mathrm {log}\left (\frac {-\mathrm {log}\left (2 x \right )+x}{x}\right )} \] Input:

int(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2*x)+(- 
x^3+8*x^2)*log(-8+x)^2+(15*x^3-120*x^2)*log(-8+x)-15*x^3)*log((x-log(2*x)) 
/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8*x)*log( 
-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x)
 

Output:

(log(x - 8)*log(( - log(2*x) + x)/x)*x + 4*log(x - 8) - 15*log(( - log(2*x 
) + x)/x)*x)/(log(x - 8)*log(( - log(2*x) + x)/x))