\(\int \frac {15-8 \log (x^2)+\log ^2(x^2)+(-64+128 x-64 x^2+(16-32 x+16 x^2) \log (x^2)) \log ^3(-15+8 \log (x^2)-\log ^2(x^2))+(-30 x+30 x^2+(16 x-16 x^2) \log (x^2)+(-2 x+2 x^2) \log ^2(x^2)) \log ^4(-15+8 \log (x^2)-\log ^2(x^2))}{15 x-8 x \log (x^2)+x \log ^2(x^2)} \, dx\) [45]

Optimal result

Integrand size = 140, antiderivative size = 24 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\log (x)+(-1+x)^2 \log ^4\left (1-\left (-4+\log \left (x^2\right )\right )^2\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\log (x)+(-1+x)^2 \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right ) \] Input:

Integrate[(15 - 8*Log[x^2] + Log[x^2]^2 + (-64 + 128*x - 64*x^2 + (16 - 32 
*x + 16*x^2)*Log[x^2])*Log[-15 + 8*Log[x^2] - Log[x^2]^2]^3 + (-30*x + 30* 
x^2 + (16*x - 16*x^2)*Log[x^2] + (-2*x + 2*x^2)*Log[x^2]^2)*Log[-15 + 8*Lo 
g[x^2] - Log[x^2]^2]^4)/(15*x - 8*x*Log[x^2] + x*Log[x^2]^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\log ^2\left (x^2\right )+\left (30 x^2+\left (2 x^2-2 x\right ) \log ^2\left (x^2\right )+\left (16 x-16 x^2\right ) \log \left (x^2\right )-30 x\right ) \log ^4\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )+\left (-64 x^2+\left (16 x^2-32 x+16\right ) \log \left (x^2\right )+128 x-64\right ) \log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )-8 \log \left (x^2\right )+15}{x \log ^2\left (x^2\right )-8 x \log \left (x^2\right )+15 x} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (2 (x-1) \log ^4\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )+\frac {16 (x-1)^2 \left (\log \left (x^2\right )-4\right ) \log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )}{x \left (\log \left (x^2\right )-5\right ) \left (\log \left (x^2\right )-3\right )}+\frac {1}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -12 \text {Subst}\left (\int \frac {\log (x-5) \log ^2\left (-x^2+8 x-15\right )}{x-5}dx,x,\log \left (x^2\right )\right )-12 \text {Subst}\left (\int \frac {\log (x-5) \log ^2\left (-x^2+8 x-15\right )}{x-3}dx,x,\log \left (x^2\right )\right )-12 \text {Subst}\left (\int \frac {\log (x-3) \log ^2\left (-x^2+8 x-15\right )}{x-5}dx,x,\log \left (x^2\right )\right )-12 \text {Subst}\left (\int \frac {\log (x-3) \log ^2\left (-x^2+8 x-15\right )}{x-3}dx,x,\log \left (x^2\right )\right )+\text {Subst}\left (\int \log ^4\left (-\log ^2(x)+8 \log (x)-15\right )dx,x,x^2\right )-32 \text {Subst}\left (\int \frac {\log ^3\left (-\log ^2(x)+8 \log (x)-15\right )}{\log ^2(x)-8 \log (x)+15}dx,x,x^2\right )+8 \text {Subst}\left (\int \frac {\log (x) \log ^3\left (-\log ^2(x)+8 \log (x)-15\right )}{\log ^2(x)-8 \log (x)+15}dx,x,x^2\right )-2 \int \log ^4\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )dx+128 \int \frac {\log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+15}dx-32 \int \frac {\log \left (x^2\right ) \log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )}{\log ^2\left (x^2\right )-8 \log \left (x^2\right )+15}dx+4 \log \left (\log \left (x^2\right )-5\right ) \log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )+4 \log \left (\log \left (x^2\right )-3\right ) \log ^3\left (-\log ^2\left (x^2\right )+8 \log \left (x^2\right )-15\right )+\log (x)\)

Input:

Int[(15 - 8*Log[x^2] + Log[x^2]^2 + (-64 + 128*x - 64*x^2 + (16 - 32*x + 1 
6*x^2)*Log[x^2])*Log[-15 + 8*Log[x^2] - Log[x^2]^2]^3 + (-30*x + 30*x^2 + 
(16*x - 16*x^2)*Log[x^2] + (-2*x + 2*x^2)*Log[x^2]^2)*Log[-15 + 8*Log[x^2] 
 - Log[x^2]^2]^4)/(15*x - 8*x*Log[x^2] + x*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(24)=48\).

Time = 1.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.83

method	result	size
parallelrisch	\({\ln \left (-\ln \left (x^{2}\right )^{2}+8 \ln \left (x^{2}\right )-15\right )}^{4} x^{2}-2 x {\ln \left (-\ln \left (x^{2}\right )^{2}+8 \ln \left (x^{2}\right )-15\right )}^{4}+{\ln \left (-\ln \left (x^{2}\right )^{2}+8 \ln \left (x^{2}\right )-15\right )}^{4}+\ln \left (x \right )\)	\(68\)

Input:

int((((2*x^2-2*x)*ln(x^2)^2+(-16*x^2+16*x)*ln(x^2)+30*x^2-30*x)*ln(-ln(x^2 
)^2+8*ln(x^2)-15)^4+((16*x^2-32*x+16)*ln(x^2)-64*x^2+128*x-64)*ln(-ln(x^2) 
^2+8*ln(x^2)-15)^3+ln(x^2)^2-8*ln(x^2)+15)/(x*ln(x^2)^2-8*x*ln(x^2)+15*x), 
x,method=_RETURNVERBOSE)
 

Output:

ln(-ln(x^2)^2+8*ln(x^2)-15)^4*x^2-2*x*ln(-ln(x^2)^2+8*ln(x^2)-15)^4+ln(-ln 
(x^2)^2+8*ln(x^2)-15)^4+ln(x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx={\left (x^{2} - 2 \, x + 1\right )} \log \left (-\log \left (x^{2}\right )^{2} + 8 \, \log \left (x^{2}\right ) - 15\right )^{4} + \frac {1}{2} \, \log \left (x^{2}\right ) \] Input:

integrate((((2*x^2-2*x)*log(x^2)^2+(-16*x^2+16*x)*log(x^2)+30*x^2-30*x)*lo 
g(-log(x^2)^2+8*log(x^2)-15)^4+((16*x^2-32*x+16)*log(x^2)-64*x^2+128*x-64) 
*log(-log(x^2)^2+8*log(x^2)-15)^3+log(x^2)^2-8*log(x^2)+15)/(x*log(x^2)^2- 
8*x*log(x^2)+15*x),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\left (x^{2} - 2 x + 1\right ) \log {\left (- \log {\left (x^{2} \right )}^{2} + 8 \log {\left (x^{2} \right )} - 15 \right )}^{4} + \log {\left (x \right )} \] Input:

integrate((((2*x**2-2*x)*ln(x**2)**2+(-16*x**2+16*x)*ln(x**2)+30*x**2-30*x 
)*ln(-ln(x**2)**2+8*ln(x**2)-15)**4+((16*x**2-32*x+16)*ln(x**2)-64*x**2+12 
8*x-64)*ln(-ln(x**2)**2+8*ln(x**2)-15)**3+ln(x**2)**2-8*ln(x**2)+15)/(x*ln 
(x**2)**2-8*x*ln(x**2)+15*x),x)
 

Output:

(x**2 - 2*x + 1)*log(-log(x**2)**2 + 8*log(x**2) - 15)**4 + log(x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (24) = 48\).

Time = 0.21 (sec) , antiderivative size = 284, normalized size of antiderivative = 11.83 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx={\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, \log \left (x\right ) - 3\right )^{4} + 4 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, \log \left (x\right ) - 3\right )^{3} \log \left (-2 \, \log \left (x\right ) + 5\right ) + 6 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, \log \left (x\right ) - 3\right )^{2} \log \left (-2 \, \log \left (x\right ) + 5\right )^{2} + 4 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2 \, \log \left (x\right ) - 3\right ) \log \left (-2 \, \log \left (x\right ) + 5\right )^{3} + {\left (x^{2} - 2 \, x + 1\right )} \log \left (-2 \, \log \left (x\right ) + 5\right )^{4} - \frac {1}{4} \, {\left (\log \left (\log \left (x\right ) - \frac {3}{2}\right ) - \log \left (\log \left (x\right ) - \frac {5}{2}\right )\right )} \log \left (x^{2}\right )^{2} + \frac {1}{2} \, {\left ({\left (2 \, \log \left (x\right ) - 3\right )} \log \left (\log \left (x\right ) - \frac {3}{2}\right ) - {\left (2 \, \log \left (x\right ) - 5\right )} \log \left (\log \left (x\right ) - \frac {5}{2}\right ) - 2\right )} \log \left (x^{2}\right ) + 2 \, {\left (\log \left (\log \left (x\right ) - \frac {3}{2}\right ) - \log \left (\log \left (x\right ) - \frac {5}{2}\right )\right )} \log \left (x^{2}\right ) - {\left (\log \left (x\right )^{2} - 3 \, \log \left (x\right )\right )} \log \left (\log \left (x\right ) - \frac {3}{2}\right ) - 2 \, {\left (2 \, \log \left (x\right ) - 3\right )} \log \left (\log \left (x\right ) - \frac {3}{2}\right ) + {\left (\log \left (x\right )^{2} - 5 \, \log \left (x\right )\right )} \log \left (\log \left (x\right ) - \frac {5}{2}\right ) + 2 \, {\left (2 \, \log \left (x\right ) - 5\right )} \log \left (\log \left (x\right ) - \frac {5}{2}\right ) + 3 \, \log \left (x\right ) - \frac {9}{4} \, \log \left (2 \, \log \left (x\right ) - 3\right ) + \frac {25}{4} \, \log \left (2 \, \log \left (x\right ) - 5\right ) - \frac {15}{4} \, \log \left (\log \left (x\right ) - \frac {3}{2}\right ) + \frac {15}{4} \, \log \left (\log \left (x\right ) - \frac {5}{2}\right ) + 4 \] Input:

integrate((((2*x^2-2*x)*log(x^2)^2+(-16*x^2+16*x)*log(x^2)+30*x^2-30*x)*lo 
g(-log(x^2)^2+8*log(x^2)-15)^4+((16*x^2-32*x+16)*log(x^2)-64*x^2+128*x-64) 
*log(-log(x^2)^2+8*log(x^2)-15)^3+log(x^2)^2-8*log(x^2)+15)/(x*log(x^2)^2- 
8*x*log(x^2)+15*x),x, algorithm="maxima")
 

Output:

(x^2 - 2*x + 1)*log(2*log(x) - 3)^4 + 4*(x^2 - 2*x + 1)*log(2*log(x) - 3)^ 
3*log(-2*log(x) + 5) + 6*(x^2 - 2*x + 1)*log(2*log(x) - 3)^2*log(-2*log(x) 
 + 5)^2 + 4*(x^2 - 2*x + 1)*log(2*log(x) - 3)*log(-2*log(x) + 5)^3 + (x^2 
- 2*x + 1)*log(-2*log(x) + 5)^4 - 1/4*(log(log(x) - 3/2) - log(log(x) - 5/ 
2))*log(x^2)^2 + 1/2*((2*log(x) - 3)*log(log(x) - 3/2) - (2*log(x) - 5)*lo 
g(log(x) - 5/2) - 2)*log(x^2) + 2*(log(log(x) - 3/2) - log(log(x) - 5/2))* 
log(x^2) - (log(x)^2 - 3*log(x))*log(log(x) - 3/2) - 2*(2*log(x) - 3)*log( 
log(x) - 3/2) + (log(x)^2 - 5*log(x))*log(log(x) - 5/2) + 2*(2*log(x) - 5) 
*log(log(x) - 5/2) + 3*log(x) - 9/4*log(2*log(x) - 3) + 25/4*log(2*log(x) 
- 5) - 15/4*log(log(x) - 3/2) + 15/4*log(log(x) - 5/2) + 4
 

Giac [A] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx={\left (x^{2} - 2 \, x + 1\right )} \log \left (-\log \left (x^{2}\right )^{2} + 8 \, \log \left (x^{2}\right ) - 15\right )^{4} + \log \left (x\right ) \] Input:

integrate((((2*x^2-2*x)*log(x^2)^2+(-16*x^2+16*x)*log(x^2)+30*x^2-30*x)*lo 
g(-log(x^2)^2+8*log(x^2)-15)^4+((16*x^2-32*x+16)*log(x^2)-64*x^2+128*x-64) 
*log(-log(x^2)^2+8*log(x^2)-15)^3+log(x^2)^2-8*log(x^2)+15)/(x*log(x^2)^2- 
8*x*log(x^2)+15*x),x, algorithm="giac")
 

Output:

(x^2 - 2*x + 1)*log(-log(x^2)^2 + 8*log(x^2) - 15)^4 + log(x)
 

Mupad [B] (verification not implemented)

Time = 2.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\left (x^2-2\,x+1\right )\,{\ln \left (\ln \left (x^{16}\right )-{\ln \left (x^2\right )}^2-15\right )}^4+\ln \left (x\right ) \] Input:

int((log(x^2)^2 - log(8*log(x^2) - log(x^2)^2 - 15)^4*(30*x - log(x^2)*(16 
*x - 16*x^2) + log(x^2)^2*(2*x - 2*x^2) - 30*x^2) - 8*log(x^2) + log(8*log 
(x^2) - log(x^2)^2 - 15)^3*(128*x + log(x^2)*(16*x^2 - 32*x + 16) - 64*x^2 
 - 64) + 15)/(15*x - 8*x*log(x^2) + x*log(x^2)^2),x)
 

Output:

log(x) + log(log(x^16) - log(x^2)^2 - 15)^4*(x^2 - 2*x + 1)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.33 \[ \int \frac {15-8 \log \left (x^2\right )+\log ^2\left (x^2\right )+\left (-64+128 x-64 x^2+\left (16-32 x+16 x^2\right ) \log \left (x^2\right )\right ) \log ^3\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )+\left (-30 x+30 x^2+\left (16 x-16 x^2\right ) \log \left (x^2\right )+\left (-2 x+2 x^2\right ) \log ^2\left (x^2\right )\right ) \log ^4\left (-15+8 \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{15 x-8 x \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=-\frac {15 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )-5\right )}{4}-\frac {15 \,\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )-3\right )}{4}+{\mathrm {log}\left (-\mathrm {log}\left (x^{2}\right )^{2}+8 \,\mathrm {log}\left (x^{2}\right )-15\right )}^{4} x^{2}-2 {\mathrm {log}\left (-\mathrm {log}\left (x^{2}\right )^{2}+8 \,\mathrm {log}\left (x^{2}\right )-15\right )}^{4} x +{\mathrm {log}\left (-\mathrm {log}\left (x^{2}\right )^{2}+8 \,\mathrm {log}\left (x^{2}\right )-15\right )}^{4}+\frac {15 \,\mathrm {log}\left (-\mathrm {log}\left (x^{2}\right )^{2}+8 \,\mathrm {log}\left (x^{2}\right )-15\right )}{4}+\mathrm {log}\left (x \right ) \] Input:

int((((2*x^2-2*x)*log(x^2)^2+(-16*x^2+16*x)*log(x^2)+30*x^2-30*x)*log(-log 
(x^2)^2+8*log(x^2)-15)^4+((16*x^2-32*x+16)*log(x^2)-64*x^2+128*x-64)*log(- 
log(x^2)^2+8*log(x^2)-15)^3+log(x^2)^2-8*log(x^2)+15)/(x*log(x^2)^2-8*x*lo 
g(x^2)+15*x),x)
 

Output:

( - 15*log(log(x**2) - 5) - 15*log(log(x**2) - 3) + 4*log( - log(x**2)**2 
+ 8*log(x**2) - 15)**4*x**2 - 8*log( - log(x**2)**2 + 8*log(x**2) - 15)**4 
*x + 4*log( - log(x**2)**2 + 8*log(x**2) - 15)**4 + 15*log( - log(x**2)**2 
 + 8*log(x**2) - 15) + 4*log(x))/4