\(\int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+(4 \log (2) \log (x)-8 x \log ^2(x)) \log (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)})}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx\) [2799]

Optimal result

Integrand size = 71, antiderivative size = 23 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=x-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right ) \] Input:

Integrate[(-4*Log[2] + 7*Log[2]*Log[x] - 22*x*Log[x]^2 + (4*Log[2]*Log[x] 
- 8*x*Log[x]^2)*Log[(-8*x^2*Log[2] + 16*x^3*Log[x])/Log[x]])/(-(Log[2]*Log 
[x]) + 2*x*Log[x]^2),x]
 

Output:

x - 4*x*Log[8*x^2*(2*x - Log[2]/Log[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.

\(\displaystyle \int \frac {\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {16 x^3 \log (x)-8 x^2 \log (2)}{\log (x)}\right )-22 x \log ^2(x)+7 \log (2) \log (x)-4 \log (2)}{2 x \log ^2(x)-\log (2) \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {16 x^3 \log (x)-8 x^2 \log (2)}{\log (x)}\right )+22 x \log ^2(x)-7 \log (2) \log (x)+4 \log (2)}{\log (x) (\log (2)-2 x \log (x))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {22 x \log ^2(x)-\log (128) \log (x)+\log (16)}{\log (x) (\log (2)-2 x \log (x))}-4 \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \log (2) \log (16) \int \frac {1}{\log ^2(2)-x \log (4) \log (x)}dx-\log (256) \int \frac {x}{x \log (4) \log (x)-\log ^2(2)}dx-4 \log (2) \int \frac {1}{\log (2)-2 x \log (x)}dx+8 \int \frac {x}{2 x \log (x)-\log (2)}dx-4 \operatorname {LogIntegral}(x)+\frac {\log (16) \operatorname {LogIntegral}(x)}{\log (2)}-4 x \log \left (8 x^2 \left (2 x-\frac {\log (2)}{\log (x)}\right )\right )+x\)

Input:

Int[(-4*Log[2] + 7*Log[2]*Log[x] - 22*x*Log[x]^2 + (4*Log[2]*Log[x] - 8*x* 
Log[x]^2)*Log[(-8*x^2*Log[2] + 16*x^3*Log[x])/Log[x]])/(-(Log[2]*Log[x]) + 
 2*x*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17

method	result	size
norman	\(x -4 x \ln \left (\frac {16 x^{3} \ln \left (x \right )-8 x^{2} \ln \left (2\right )}{\ln \left (x \right )}\right )\)	\(27\)
parallelrisch	\(x -4 x \ln \left (\frac {16 x^{3} \ln \left (x \right )-8 x^{2} \ln \left (2\right )}{\ln \left (x \right )}\right )\)	\(27\)
default	\(-12 x \ln \left (2\right )+x -4 \ln \left (\frac {x^{2} \left (2 x \ln \left (x \right )-\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) x\)	\(30\)
risch	\(-4 x \ln \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )+4 x \ln \left (\ln \left (x \right )\right )-2 i \pi x \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+2 i \pi x \,\operatorname {csgn}\left (i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )+2 i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi x \,\operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x^{2}\right )-2 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \pi x \,\operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}-2 i \pi x \,\operatorname {csgn}\left (i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )\right ) \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}-2 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{3}-4 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+2 i \pi x \operatorname {csgn}\left (\frac {i \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{3}-4 i \pi x +4 i \pi x \operatorname {csgn}\left (\frac {i x^{2} \left (-2 x \ln \left (x \right )+\ln \left (2\right )\right )}{\ln \left (x \right )}\right )^{2}+2 i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}-12 x \ln \left (2\right )-8 x \ln \left (x \right )+x\)	\(392\)

Input:

int(((-8*x*ln(x)^2+4*ln(2)*ln(x))*ln((16*x^3*ln(x)-8*x^2*ln(2))/ln(x))-22* 
x*ln(x)^2+7*ln(2)*ln(x)-4*ln(2))/(2*x*ln(x)^2-ln(2)*ln(x)),x,method=_RETUR 
NVERBOSE)
 

Output:

x-4*x*ln((16*x^3*ln(x)-8*x^2*ln(2))/ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-4 \, x \log \left (\frac {8 \, {\left (2 \, x^{3} \log \left (x\right ) - x^{2} \log \left (2\right )\right )}}{\log \left (x\right )}\right ) + x \] Input:

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2) 
)/log(x))-22*x*log(x)^2+7*log(2)*log(x)-4*log(2))/(2*x*log(x)^2-log(2)*log 
(x)),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=- 4 x \log {\left (\frac {16 x^{3} \log {\left (x \right )} - 8 x^{2} \log {\left (2 \right )}}{\log {\left (x \right )}} \right )} + x \] Input:

integrate(((-8*x*ln(x)**2+4*ln(2)*ln(x))*ln((16*x**3*ln(x)-8*x**2*ln(2))/l 
n(x))-22*x*ln(x)**2+7*ln(2)*ln(x)-4*ln(2))/(2*x*ln(x)**2-ln(2)*ln(x)),x)
 

Output:

-4*x*log((16*x**3*log(x) - 8*x**2*log(2))/log(x)) + x
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-x {\left (12 \, \log \left (2\right ) - 1\right )} - 4 \, x \log \left (2 \, x \log \left (x\right ) - \log \left (2\right )\right ) - 8 \, x \log \left (x\right ) + 4 \, x \log \left (\log \left (x\right )\right ) \] Input:

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2) 
)/log(x))-22*x*log(x)^2+7*log(2)*log(x)-4*log(2))/(2*x*log(x)^2-log(2)*log 
(x)),x, algorithm="maxima")
 

Output:

-x*(12*log(2) - 1) - 4*x*log(2*x*log(x) - log(2)) - 8*x*log(x) + 4*x*log(l 
og(x))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=-x {\left (12 \, \log \left (2\right ) - 1\right )} - 4 \, x \log \left (2 \, x \log \left (x\right ) - \log \left (2\right )\right ) - 8 \, x \log \left (x\right ) + 4 \, x \log \left (\log \left (x\right )\right ) \] Input:

integrate(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2) 
)/log(x))-22*x*log(x)^2+7*log(2)*log(x)-4*log(2))/(2*x*log(x)^2-log(2)*log 
(x)),x, algorithm="giac")
 

Output:

-x*(12*log(2) - 1) - 4*x*log(2*x*log(x) - log(2)) - 8*x*log(x) + 4*x*log(l 
og(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=\int -\frac {4\,\ln \left (2\right )+22\,x\,{\ln \left (x\right )}^2-7\,\ln \left (2\right )\,\ln \left (x\right )+\ln \left (\frac {16\,x^3\,\ln \left (x\right )-8\,x^2\,\ln \left (2\right )}{\ln \left (x\right )}\right )\,\left (8\,x\,{\ln \left (x\right )}^2-4\,\ln \left (2\right )\,\ln \left (x\right )\right )}{2\,x\,{\ln \left (x\right )}^2-\ln \left (2\right )\,\ln \left (x\right )} \,d x \] Input:

int(-(4*log(2) + 22*x*log(x)^2 - 7*log(2)*log(x) + log((16*x^3*log(x) - 8* 
x^2*log(2))/log(x))*(8*x*log(x)^2 - 4*log(2)*log(x)))/(2*x*log(x)^2 - log( 
2)*log(x)),x)
 

Output:

int(-(4*log(2) + 22*x*log(x)^2 - 7*log(2)*log(x) + log((16*x^3*log(x) - 8* 
x^2*log(2))/log(x))*(8*x*log(x)^2 - 4*log(2)*log(x)))/(2*x*log(x)^2 - log( 
2)*log(x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-4 \log (2)+7 \log (2) \log (x)-22 x \log ^2(x)+\left (4 \log (2) \log (x)-8 x \log ^2(x)\right ) \log \left (\frac {-8 x^2 \log (2)+16 x^3 \log (x)}{\log (x)}\right )}{-\log (2) \log (x)+2 x \log ^2(x)} \, dx=x \left (-4 \,\mathrm {log}\left (\frac {16 \,\mathrm {log}\left (x \right ) x^{3}-8 \,\mathrm {log}\left (2\right ) x^{2}}{\mathrm {log}\left (x \right )}\right )+1\right ) \] Input:

int(((-8*x*log(x)^2+4*log(2)*log(x))*log((16*x^3*log(x)-8*x^2*log(2))/log( 
x))-22*x*log(x)^2+7*log(2)*log(x)-4*log(2))/(2*x*log(x)^2-log(2)*log(x)),x 
)
 

Output:

x*( - 4*log((16*log(x)*x**3 - 8*log(2)*x**2)/log(x)) + 1)