\(\int \frac {75+90 x+3 x^2+(120 x-24 x^2+(75-30 x+3 x^2) \log (\frac {x}{e})) \log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x}) \log (-\log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x}))}{(40 x-8 x^2+(25-10 x+x^2) \log (\frac {x}{e})) \log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x})} \, dx\) [2435]

Optimal result

Integrand size = 132, antiderivative size = 24 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 x \log \left (-\log \left (\frac {8 x}{5-x}+\log \left (\frac {x}{e}\right )\right )\right ) \] Output:

3*x*ln(-ln(ln(x/exp(1))+8*x/(5-x)))
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 x \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \] Input:

Integrate[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x 
/E])*Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]*Log[-Log[(-8*x + (-5 + x)*Lo 
g[x/E])/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8*x 
 + (-5 + x)*Log[x/E])/(-5 + x)]),x]
 

Output:

3*x*Log[-Log[(5 - 9*x + (-5 + x)*Log[x])/(-5 + x)]]
 

Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7292, 7293, 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.

\(\displaystyle \int \frac {3 x^2+\left (-24 x^2+\left (3 x^2-30 x+75\right ) \log \left (\frac {x}{e}\right )+120 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right ) \log \left (-\log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )\right )+90 x+75}{\left (-8 x^2+\left (x^2-10 x+25\right ) \log \left (\frac {x}{e}\right )+40 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-3 x^2-\left (-24 x^2+\left (3 x^2-30 x+75\right ) \log \left (\frac {x}{e}\right )+120 x\right ) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right ) \log \left (-\log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )\right )-90 x-75}{(5-x) (-9 x+x \log (x)-5 \log (x)+5) \log \left (\frac {(x-5) \log \left (\frac {x}{e}\right )-8 x}{x-5}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {3 \left (x^2+30 x+25\right )}{(x-5) (-9 x+x \log (x)-5 \log (x)+5) \log \left (\frac {-9 x+(x-5) \log (x)+5}{x-5}\right )}+3 \log \left (-\log \left (\frac {-9 x+(x-5) \log (x)+5}{x-5}\right )\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 x \log \left (-\log \left (-\frac {-9 x-(5-x) \log (x)+5}{5-x}\right )\right )\)

Input:

Int[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x/E])*L 
og[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]*Log[-Log[(-8*x + (-5 + x)*Log[x/E] 
)/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8*x + (-5 
 + x)*Log[x/E])/(-5 + x)]),x]
 

Output:

3*x*Log[-Log[-((5 - 9*x - (5 - x)*Log[x])/(5 - x))]]
 

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] 
]
 

Maple [A] (verified)

Time = 17.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(3 x \ln \left (-\ln \left (\frac {\left (-5+x \right ) \ln \left (x \,{\mathrm e}^{-1}\right )-8 x}{-5+x}\right )\right )\)	\(29\)
risch	\(3 \ln \left (\ln \left (-5+x \right )-\ln \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )+\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right )+\operatorname {csgn}\left (\frac {i}{-5+x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )}{-5+x}\right )+\operatorname {csgn}\left (i \left (\left (\ln \left (x \,{\mathrm e}^{-1}\right )-8\right ) x -5 \ln \left (x \,{\mathrm e}^{-1}\right )\right )\right )\right )}{2}\right ) x\)	\(148\)

Input:

int((((3*x^2-30*x+75)*ln(x/exp(1))-24*x^2+120*x)*ln(((-5+x)*ln(x/exp(1))-8 
*x)/(-5+x))*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))+3*x^2+90*x+75)/((x^2 
-10*x+25)*ln(x/exp(1))-8*x^2+40*x)/ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)),x, 
method=_RETURNVERBOSE)
 

Output:

3*x*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left (\frac {{\left (x - 5\right )} \log \left (x e^{\left (-1\right )}\right ) - 8 \, x}{x - 5}\right )\right ) \] Input:

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ 
exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 
*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 
8*x)/(-5+x)),x, algorithm="fricas")
 

Output:

3*x*log(-log(((x - 5)*log(x*e^(-1)) - 8*x)/(x - 5)))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((3*x**2-30*x+75)*ln(x/exp(1))-24*x**2+120*x)*ln(((-5+x)*ln(x/e 
xp(1))-8*x)/(-5+x))*ln(-ln(((-5+x)*ln(x/exp(1))-8*x)/(-5+x)))+3*x**2+90*x+ 
75)/((x**2-10*x+25)*ln(x/exp(1))-8*x**2+40*x)/ln(((-5+x)*ln(x/exp(1))-8*x) 
/(-5+x)),x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left ({\left (x - 5\right )} \log \left (x\right ) - 9 \, x + 5\right ) + \log \left (x - 5\right )\right ) \] Input:

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ 
exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 
*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 
8*x)/(-5+x)),x, algorithm="maxima")
 

Output:

3*x*log(-log((x - 5)*log(x) - 9*x + 5) + log(x - 5))
 

Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3 \, x \log \left (-\log \left (x \log \left (x\right ) - 9 \, x - 5 \, \log \left (x\right ) + 5\right ) + \log \left (x - 5\right )\right ) \] Input:

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/ 
exp(1))-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90 
*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))- 
8*x)/(-5+x)),x, algorithm="giac")
 

Output:

3*x*log(-log(x*log(x) - 9*x - 5*log(x) + 5) + log(x - 5))
 

Mupad [B] (verification not implemented)

Time = 5.81 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=3\,x\,\ln \left (-\ln \left (-\frac {8\,x-\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (x-5\right )}{x-5}\right )\right ) \] Input:

int((90*x + 3*x^2 + log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5))*log(-log( 
-(8*x - log(x*exp(-1))*(x - 5))/(x - 5)))*(120*x + log(x*exp(-1))*(3*x^2 - 
 30*x + 75) - 24*x^2) + 75)/(log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5))* 
(40*x + log(x*exp(-1))*(x^2 - 10*x + 25) - 8*x^2)),x)
 

Output:

3*x*log(-log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 4.00 \[ \int \frac {75+90 x+3 x^2+\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{\left (40 x-8 x^2+\left (25-10 x+x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx=90 \,\mathrm {log}\left (\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right )+3 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right ) x -90 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {\mathrm {log}\left (\frac {x}{e}\right ) x -5 \,\mathrm {log}\left (\frac {x}{e}\right )-8 x}{-5+x}\right )\right ) \] Input:

int((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((-5+x)*log(x/exp(1) 
)-8*x)/(-5+x))*log(-log(((-5+x)*log(x/exp(1))-8*x)/(-5+x)))+3*x^2+90*x+75) 
/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((-5+x)*log(x/exp(1))-8*x)/( 
-5+x)),x)
 

Output:

3*(30*log(log((log(x/e)*x - 5*log(x/e) - 8*x)/(x - 5))) + log( - log((log( 
x/e)*x - 5*log(x/e) - 8*x)/(x - 5)))*x - 30*log( - log((log(x/e)*x - 5*log 
(x/e) - 8*x)/(x - 5))))