\(\int \frac {9 x-3 x^4+(3-x^3) \log (16)+(-9+3 x^3) \log (3-x^3)+(-9 x-9 x^3+3 x^4) \log (x) \log (\frac {1}{\log (x)})}{(-9 x+3 x^4) \log (x)} \, dx\) [2151]

Optimal result

Integrand size = 73, antiderivative size = 24 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x+\frac {\log (16)}{3}-\log \left (3-x^3\right )\right ) \log \left (\frac {1}{\log (x)}\right ) \] Output:

ln(1/ln(x))*(4/3*ln(2)+x-ln(-x^3+3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x-\log \left (3-x^3\right )\right ) \log \left (\frac {1}{\log (x)}\right )-\frac {1}{3} \log (16) \log (\log (x)) \] Input:

Integrate[(9*x - 3*x^4 + (3 - x^3)*Log[16] + (-9 + 3*x^3)*Log[3 - x^3] + ( 
-9*x - 9*x^3 + 3*x^4)*Log[x]*Log[Log[x]^(-1)])/((-9*x + 3*x^4)*Log[x]),x]
 

Output:

(x - Log[3 - x^3])*Log[Log[x]^(-1)] - (Log[16]*Log[Log[x]])/3
 

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.

\(\displaystyle \int \frac {-3 x^4+\left (3 x^3-9\right ) \log \left (3-x^3\right )+\left (3-x^3\right ) \log (16)+\left (3 x^4-9 x^3-9 x\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+9 x}{\left (3 x^4-9 x\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-3 x^4+\left (3 x^3-9\right ) \log \left (3-x^3\right )+\left (3-x^3\right ) \log (16)+\left (3 x^4-9 x^3-9 x\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+9 x}{x \left (3 x^3-9\right ) \log (x)}dx\)

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\log \left (3-x^3\right )}{x \log (x)}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{-x-\sqrt [3]{-3}}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{\sqrt [3]{3}-x}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-1)^{2/3} \sqrt [3]{3}-x}dx+x \log \left (\frac {1}{\log (x)}\right )-\frac {1}{3} \log (16) \log (\log (x))\)

Input:

Int[(9*x - 3*x^4 + (3 - x^3)*Log[16] + (-9 + 3*x^3)*Log[3 - x^3] + (-9*x - 
 9*x^3 + 3*x^4)*Log[x]*Log[Log[x]^(-1)])/((-9*x + 3*x^4)*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 68.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
risch	\(\left (-x +\ln \left (-x^{3}+3\right )\right ) \ln \left (\ln \left (x \right )\right )-\frac {4 \ln \left (2\right ) \ln \left (\ln \left (x \right )\right )}{3}\)	\(25\)
parallelrisch	\(\frac {4 \ln \left (\frac {1}{\ln \left (x \right )}\right ) \ln \left (2\right )}{3}+\ln \left (\frac {1}{\ln \left (x \right )}\right ) x -\ln \left (-x^{3}+3\right ) \ln \left (\frac {1}{\ln \left (x \right )}\right )\)	\(33\)

Input:

int(((3*x^4-9*x^3-9*x)*ln(x)*ln(1/ln(x))+(3*x^3-9)*ln(-x^3+3)+4*(-x^3+3)*l 
n(2)-3*x^4+9*x)/(3*x^4-9*x)/ln(x),x,method=_RETURNVERBOSE)
 

Output:

(-x+ln(-x^3+3))*ln(ln(x))-4/3*ln(2)*ln(ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\frac {1}{3} \, {\left (3 \, x + 4 \, \log \left (2\right ) - 3 \, \log \left (-x^{3} + 3\right )\right )} \log \left (\frac {1}{\log \left (x\right )}\right ) \] Input:

integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* 
(-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x - \log {\left (3 - x^{3} \right )}\right ) \log {\left (\frac {1}{\log {\left (x \right )}} \right )} - \frac {4 \log {\left (2 \right )} \log {\left (\log {\left (x \right )} \right )}}{3} \] Input:

integrate(((3*x**4-9*x**3-9*x)*ln(x)*ln(1/ln(x))+(3*x**3-9)*ln(-x**3+3)+4* 
(-x**3+3)*ln(2)-3*x**4+9*x)/(3*x**4-9*x)/ln(x),x)
 

Output:

(x - log(3 - x**3))*log(1/log(x)) - 4*log(2)*log(log(x))/3
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-\frac {1}{3} \, {\left (3 \, x + 4 \, \log \left (2\right )\right )} \log \left (\log \left (x\right )\right ) + \log \left (-x^{3} + 3\right ) \log \left (\log \left (x\right )\right ) \] Input:

integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* 
(-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-x \log \left (\log \left (x\right )\right ) - \frac {4}{3} \, \log \left (2\right ) \log \left (\log \left (x\right )\right ) + \log \left (-x^{3} + 3\right ) \log \left (\log \left (x\right )\right ) \] Input:

integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* 
(-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm="giac")
 

Output:

-x*log(log(x)) - 4/3*log(2)*log(log(x)) + log(-x^3 + 3)*log(log(x))
 

Mupad [B] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-\ln \left (\frac {1}{\ln \left (x\right )}\right )\,\left (\ln \left (3-x^3\right )+\frac {3\,x^2-x^5}{x\,\left (x^3-3\right )}\right )-\frac {4\,\ln \left (\ln \left (x\right )\right )\,\ln \left (2\right )}{3} \] Input:

int((4*log(2)*(x^3 - 3) - 9*x + 3*x^4 - log(3 - x^3)*(3*x^3 - 9) + log(1/l 
og(x))*log(x)*(9*x + 9*x^3 - 3*x^4))/(log(x)*(9*x - 3*x^4)),x)
 

Output:

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

Reduce [F]

\[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\frac {\left (27 \,3^{\frac {5}{6}} \left (\int \frac {\mathrm {log}\left (\mathrm {log}\left (x \right )\right )}{x^{4}-3 x}d x \right )-9 \,3^{\frac {5}{6}} \left (\int \frac {\mathrm {log}\left (-x^{3}+3\right )}{\mathrm {log}\left (x \right ) x^{4}-3 \,\mathrm {log}\left (x \right ) x}d x \right )+3 \,3^{\frac {5}{6}} \left (\int \frac {\mathrm {log}\left (-x^{3}+3\right ) x^{2}}{\mathrm {log}\left (x \right ) x^{3}-3 \,\mathrm {log}\left (x \right )}d x \right )+9 \,3^{\frac {5}{6}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (x \right )-4 \,3^{\frac {5}{6}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) \mathrm {log}\left (2\right )-3 \,3^{\frac {5}{6}} \mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x -9 \,3^{\frac {5}{6}} \mathrm {log}\left (x \right )\right ) 3^{\frac {1}{6}}}{9} \] Input:

int(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4*(-x^3+ 
3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x)
 

Output:

( - 3*3**(7/9)*3**(1/18)*int(x**3/(log(x)*x**3 - 3*log(x)),x) + 27*3**(7/9 
)*3**(1/18)*int(log(log(x))/(x**4 - 3*x),x) + 9*3**(7/9)*3**(1/18)*int(1/( 
log(x)*x**3 - 3*log(x)),x) + 3*3**(2/3)*3**(1/6)*int(x**3/(log(x)*x**3 - 3 
*log(x)),x) - 9*3**(2/3)*3**(1/6)*int(log( - x**3 + 3)/(log(x)*x**4 - 3*lo 
g(x)*x),x) + 3*3**(2/3)*3**(1/6)*int((log( - x**3 + 3)*x**2)/(log(x)*x**3 
- 3*log(x)),x) - 9*3**(2/3)*3**(1/6)*int(1/(log(x)*x**3 - 3*log(x)),x) + 9 
*3**(2/3)*3**(1/6)*log(log(x))*log(x) - 4*3**(2/3)*3**(1/6)*log(log(x))*lo 
g(2) - 3*3**(2/3)*3**(1/6)*log(log(x))*x - 9*3**(2/3)*3**(1/6)*log(x))/(3* 
3**(2/3)*3**(1/6))