\(\int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+(-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)) \log (\frac {1}{4} (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)))}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx\) [1362]

Optimal result

Integrand size = 115, antiderivative size = 22 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=x \log \left (-1-x-\frac {\log (25)}{4}+(x+\log (5 x))^2\right ) \] Output:

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

Mathematica [F]

\[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=\int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx \] Input:

Integrate[(4*x + 8*x^2 + (8 + 8*x)*Log[5*x] + (-4 - 4*x + 4*x^2 - Log[25] 
+ 8*x*Log[5*x] + 4*Log[5*x]^2)*Log[(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5 
*x] + 4*Log[5*x]^2)/4])/(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5*x] + 4*Log 
[5*x]^2),x]
 

Output:

Integrate[(4*x + 8*x^2 + (8 + 8*x)*Log[5*x] + (-4 - 4*x + 4*x^2 - Log[25] 
+ 8*x*Log[5*x] + 4*Log[5*x]^2)*Log[(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5 
*x] + 4*Log[5*x]^2)/4])/(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5*x] + 4*Log 
[5*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.

\(\displaystyle \int \frac {8 x^2+\left (4 x^2-4 x+4 \log ^2(5 x)+8 x \log (5 x)-4-\log (25)\right ) \log \left (\frac {1}{4} \left (4 x^2-4 x+4 \log ^2(5 x)+8 x \log (5 x)-4-\log (25)\right )\right )+4 x+(8 x+8) \log (5 x)}{4 x^2-4 x+4 \log ^2(5 x)+8 x \log (5 x)-4-\log (25)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-8 x^2-\left (4 x^2-4 x+4 \log ^2(5 x)+8 x \log (5 x)-4-\log (25)\right ) \log \left (\frac {1}{4} \left (4 x^2-4 x+4 \log ^2(5 x)+8 x \log (5 x)-4-\log (25)\right )\right )-4 x-(8 x+8) \log (5 x)}{-4 x^2+4 x-4 \log ^2(5 x)-8 x \log (5 x)+4 \left (1+\frac {\log (5)}{2}\right )}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(4*x + 8*x^2 + (8 + 8*x)*Log[5*x] + (-4 - 4*x + 4*x^2 - Log[25] + 8*x* 
Log[5*x] + 4*Log[5*x]^2)*Log[(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5*x] + 
4*Log[5*x]^2)/4])/(-4 - 4*x + 4*x^2 - Log[25] + 8*x*Log[5*x] + 4*Log[5*x]^ 
2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32

method	result	size
risch	\(\ln \left (\ln \left (5 x \right )^{2}+2 x \ln \left (5 x \right )-\frac {\ln \left (5\right )}{2}+x^{2}-x -1\right ) x\)	\(29\)
parallelrisch	\(\ln \left (\ln \left (5 x \right )^{2}+2 x \ln \left (5 x \right )-\frac {\ln \left (5\right )}{2}+x^{2}-x -1\right ) x\)	\(29\)
default	\(-x \ln \left (2\right )+x \ln \left (2 \ln \left (5 x \right )^{2}+4 x \ln \left (5 x \right )+2 x^{2}-\ln \left (5\right )-2 x -2\right )\)	\(39\)

Input:

int(((4*ln(5*x)^2+8*x*ln(5*x)-2*ln(5)+4*x^2-4*x-4)*ln(ln(5*x)^2+2*x*ln(5*x 
)-1/2*ln(5)+x^2-x-1)+(8*x+8)*ln(5*x)+8*x^2+4*x)/(4*ln(5*x)^2+8*x*ln(5*x)-2 
*ln(5)+4*x^2-4*x-4),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=x \log \left (x^{2} + 2 \, x \log \left (5 \, x\right ) + \log \left (5 \, x\right )^{2} - x - \frac {1}{2} \, \log \left (5\right ) - 1\right ) \] Input:

integrate(((4*log(5*x)^2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4)*log(log(5*x)^2 
+2*x*log(5*x)-1/2*log(5)+x^2-x-1)+(8*x+8)*log(5*x)+8*x^2+4*x)/(4*log(5*x)^ 
2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=x \log {\left (x^{2} + 2 x \log {\left (5 x \right )} - x + \log {\left (5 x \right )}^{2} - 1 - \frac {\log {\left (5 \right )}}{2} \right )} \] Input:

integrate(((4*ln(5*x)**2+8*x*ln(5*x)-2*ln(5)+4*x**2-4*x-4)*ln(ln(5*x)**2+2 
*x*ln(5*x)-1/2*ln(5)+x**2-x-1)+(8*x+8)*ln(5*x)+8*x**2+4*x)/(4*ln(5*x)**2+8 
*x*ln(5*x)-2*ln(5)+4*x**2-4*x-4),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).

Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.23 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=-x \log \left (2\right ) + x \log \left (2 \, x^{2} + 2 \, x {\left (2 \, \log \left (5\right ) - 1\right )} + 2 \, \log \left (5\right )^{2} + 4 \, {\left (x + \log \left (5\right )\right )} \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - \log \left (5\right ) - 2\right ) \] Input:

integrate(((4*log(5*x)^2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4)*log(log(5*x)^2 
+2*x*log(5*x)-1/2*log(5)+x^2-x-1)+(8*x+8)*log(5*x)+8*x^2+4*x)/(4*log(5*x)^ 
2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=-x \log \left (2\right ) + x \log \left (2 \, x^{2} + 4 \, x \log \left (5\right ) + 2 \, \log \left (5\right )^{2} + 4 \, x \log \left (x\right ) + 4 \, \log \left (5\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 2 \, x - \log \left (5\right ) - 2\right ) \] Input:

integrate(((4*log(5*x)^2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4)*log(log(5*x)^2 
+2*x*log(5*x)-1/2*log(5)+x^2-x-1)+(8*x+8)*log(5*x)+8*x^2+4*x)/(4*log(5*x)^ 
2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx=x\,\ln \left (x^2+2\,x\,\ln \left (5\,x\right )-x+{\ln \left (5\,x\right )}^2-\frac {\ln \left (5\right )}{2}-1\right ) \] Input:

int(-(4*x - log(2*x*log(5*x) - log(5)/2 - x + log(5*x)^2 + x^2 - 1)*(4*x + 
 2*log(5) - 8*x*log(5*x) - 4*log(5*x)^2 - 4*x^2 + 4) + 8*x^2 + log(5*x)*(8 
*x + 8))/(4*x + 2*log(5) - 8*x*log(5*x) - 4*log(5*x)^2 - 4*x^2 + 4),x)
 

Output:

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

Reduce [F]

\[ \int \frac {4 x+8 x^2+(8+8 x) \log (5 x)+\left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right ) \log \left (\frac {1}{4} \left (-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)\right )\right )}{-4-4 x+4 x^2-\log (25)+8 x \log (5 x)+4 \log ^2(5 x)} \, dx =\text {Too large to display} \] Input:

int(((4*log(5*x)^2+8*x*log(5*x)-2*log(5)+4*x^2-4*x-4)*log(log(5*x)^2+2*x*l 
og(5*x)-1/2*log(5)+x^2-x-1)+(8*x+8)*log(5*x)+8*x^2+4*x)/(4*log(5*x)^2+8*x* 
log(5*x)-2*log(5)+4*x^2-4*x-4),x)
 

Output:

4*int(x**2/(2*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) - 
 int(log((2*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)/(2* 
log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x)*log(5) - 2*int( 
log((2*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)/(2*log(5 
*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 4*int(log(5*x)/(2* 
log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 2*int((log((2 
*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)*log(5*x)**2)/( 
2*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 2*int((log( 
(2*log(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)*x**2)/(2*log 
(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 4*int((log((2*lo 
g(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)*log(5*x)*x)/(2*lo 
g(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) - 2*int((log((2*l 
og(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2)/2)*x)/(2*log(5*x)** 
2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 4*int((log(5*x)*x)/(2*l 
og(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x) + 2*int(x/(2*log 
(5*x)**2 + 4*log(5*x)*x - log(5) + 2*x**2 - 2*x - 2),x)