\(\int \frac {-2+(-4 x-2 x^2) \log (5)+((-1+4 x+x^2) \log (5)+\log (x^2)) \log ((-1+4 x+x^2) \log (5)+\log (x^2))}{((-x+4 x^2+x^3) \log (5)+x \log (x^2)) \log ((-1+4 x+x^2) \log (5)+\log (x^2))} \, dx\) [1189]

Optimal result

Integrand size = 92, antiderivative size = 22 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\log \left (\frac {x}{\log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}\right ) \] Output:

ln(x/ln(ln(x^2)+(x^2+4*x-1)*ln(5)))
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx \] Input:

Integrate[(-2 + (-4*x - 2*x^2)*Log[5] + ((-1 + 4*x + x^2)*Log[5] + Log[x^2 
])*Log[(-1 + 4*x + x^2)*Log[5] + Log[x^2]])/(((-x + 4*x^2 + x^3)*Log[5] + 
x*Log[x^2])*Log[(-1 + 4*x + x^2)*Log[5] + Log[x^2]]),x]
 

Output:

Integrate[(-2 + (-4*x - 2*x^2)*Log[5] + ((-1 + 4*x + x^2)*Log[5] + Log[x^2 
])*Log[(-1 + 4*x + x^2)*Log[5] + Log[x^2]])/(((-x + 4*x^2 + x^3)*Log[5] + 
x*Log[x^2])*Log[(-1 + 4*x + x^2)*Log[5] + Log[x^2]]), x]
 

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.

\(\displaystyle \int \frac {\left (-2 x^2-4 x\right ) \log (5)+\left (\left (x^2+4 x-1\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (x^2+4 x-1\right ) \log (5)+\log \left (x^2\right )\right )-2}{\left (x \log \left (x^2\right )+\left (x^3+4 x^2-x\right ) \log (5)\right ) \log \left (\left (x^2+4 x-1\right ) \log (5)+\log \left (x^2\right )\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {1}{x}-\frac {2 \left (x^2 \log (5)+x \log (25)+1\right )}{x \left (x^2 \log (5)+\log \left (x^2\right )+4 x \log (5)-\log (5)\right ) \log \left (\left (x^2+4 x-1\right ) \log (5)+\log \left (x^2\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \log (25) \int \frac {1}{\left (\log (5) x^2+4 \log (5) x+\log \left (x^2\right )-\log (5)\right ) \log \left (\log (5) \left (x^2+4 x-1\right )+\log \left (x^2\right )\right )}dx-2 \int \frac {1}{x \left (\log (5) x^2+4 \log (5) x+\log \left (x^2\right )-\log (5)\right ) \log \left (\log (5) \left (x^2+4 x-1\right )+\log \left (x^2\right )\right )}dx-2 \log (5) \int \frac {x}{\left (\log (5) x^2+4 \log (5) x+\log \left (x^2\right )-\log (5)\right ) \log \left (\log (5) \left (x^2+4 x-1\right )+\log \left (x^2\right )\right )}dx+\log (x)\)

Input:

Int[(-2 + (-4*x - 2*x^2)*Log[5] + ((-1 + 4*x + x^2)*Log[5] + Log[x^2])*Log 
[(-1 + 4*x + x^2)*Log[5] + Log[x^2]])/(((-x + 4*x^2 + x^3)*Log[5] + x*Log[ 
x^2])*Log[(-1 + 4*x + x^2)*Log[5] + Log[x^2]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\ln \left (5\right )-\ln \left (\ln \left (\ln \left (x^{2}\right )+\left (x^{2}+4 x -1\right ) \ln \left (5\right )\right )\right )+\frac {\ln \left (x^{2}\right )}{2}\)	\(30\)

Input:

int(((ln(x^2)+(x^2+4*x-1)*ln(5))*ln(ln(x^2)+(x^2+4*x-1)*ln(5))+(-2*x^2-4*x 
)*ln(5)-2)/(x*ln(x^2)+(x^3+4*x^2-x)*ln(5))/ln(ln(x^2)+(x^2+4*x-1)*ln(5)),x 
,method=_RETURNVERBOSE)
 

Output:

ln(5)-ln(ln(ln(x^2)+(x^2+4*x-1)*ln(5)))+1/2*ln(x^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\frac {1}{2} \, \log \left (x^{2}\right ) - \log \left (\log \left ({\left (x^{2} + 4 \, x - 1\right )} \log \left (5\right ) + \log \left (x^{2}\right )\right )\right ) \] Input:

integrate(((log(x^2)+(x^2+4*x-1)*log(5))*log(log(x^2)+(x^2+4*x-1)*log(5))+ 
(-2*x^2-4*x)*log(5)-2)/(x*log(x^2)+(x^3+4*x^2-x)*log(5))/log(log(x^2)+(x^2 
+4*x-1)*log(5)),x, algorithm="fricas")
 

Output:

1/2*log(x^2) - log(log((x^2 + 4*x - 1)*log(5) + log(x^2)))
 

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\log {\left (x \right )} - \log {\left (\log {\left (\left (x^{2} + 4 x - 1\right ) \log {\left (5 \right )} + \log {\left (x^{2} \right )} \right )} \right )} \] Input:

integrate(((ln(x**2)+(x**2+4*x-1)*ln(5))*ln(ln(x**2)+(x**2+4*x-1)*ln(5))+( 
-2*x**2-4*x)*ln(5)-2)/(x*ln(x**2)+(x**3+4*x**2-x)*ln(5))/ln(ln(x**2)+(x**2 
+4*x-1)*ln(5)),x)
 

Output:

log(x) - log(log((x**2 + 4*x - 1)*log(5) + log(x**2)))
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - \log \left (5\right ) + 2 \, \log \left (x\right )\right )\right ) \] Input:

integrate(((log(x^2)+(x^2+4*x-1)*log(5))*log(log(x^2)+(x^2+4*x-1)*log(5))+ 
(-2*x^2-4*x)*log(5)-2)/(x*log(x^2)+(x^3+4*x^2-x)*log(5))/log(log(x^2)+(x^2 
+4*x-1)*log(5)),x, algorithm="maxima")
 

Output:

log(x) - log(log(x^2*log(5) + 4*x*log(5) - log(5) + 2*log(x)))
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (x^{2} \log \left (5\right ) + 4 \, x \log \left (5\right ) - \log \left (5\right ) + \log \left (x^{2}\right )\right )\right ) \] Input:

integrate(((log(x^2)+(x^2+4*x-1)*log(5))*log(log(x^2)+(x^2+4*x-1)*log(5))+ 
(-2*x^2-4*x)*log(5)-2)/(x*log(x^2)+(x^3+4*x^2-x)*log(5))/log(log(x^2)+(x^2 
+4*x-1)*log(5)),x, algorithm="giac")
 

Output:

log(x) - log(log(x^2*log(5) + 4*x*log(5) - log(5) + log(x^2)))
 

Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\ln \left (x\right )-\ln \left (\ln \left (\ln \left (x^2\right )+\ln \left (5\right )\,\left (x^2+4\,x-1\right )\right )\right ) \] Input:

int(-(log(5)*(4*x + 2*x^2) - log(log(x^2) + log(5)*(4*x + x^2 - 1))*(log(x 
^2) + log(5)*(4*x + x^2 - 1)) + 2)/(log(log(x^2) + log(5)*(4*x + x^2 - 1)) 
*(log(5)*(4*x^2 - x + x^3) + x*log(x^2))),x)
 

Output:

log(x) - log(log(log(x^2) + log(5)*(4*x + x^2 - 1)))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {-2+\left (-4 x-2 x^2\right ) \log (5)+\left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )}{\left (\left (-x+4 x^2+x^3\right ) \log (5)+x \log \left (x^2\right )\right ) \log \left (\left (-1+4 x+x^2\right ) \log (5)+\log \left (x^2\right )\right )} \, dx=\frac {\mathrm {log}\left (x^{2}\right )}{2}-\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )+\mathrm {log}\left (5\right ) x^{2}+4 \,\mathrm {log}\left (5\right ) x -\mathrm {log}\left (5\right )\right )\right ) \] Input:

int(((log(x^2)+(x^2+4*x-1)*log(5))*log(log(x^2)+(x^2+4*x-1)*log(5))+(-2*x^ 
2-4*x)*log(5)-2)/(x*log(x^2)+(x^3+4*x^2-x)*log(5))/log(log(x^2)+(x^2+4*x-1 
)*log(5)),x)
 

Output:

(log(x**2) - 2*log(log(log(x**2) + log(5)*x**2 + 4*log(5)*x - log(5))))/2