\(\int \frac {4-x+8 x \log (8 e^{-x/4} x)+(-1-2 x^2+2 x^3+2 x \log (x)) \log ^2(8 e^{-x/4} x)}{4 x \log (8 e^{-x/4} x)+(x^3+x \log (x)) \log ^2(8 e^{-x/4} x)} \, dx\) [554]

Optimal result

Integrand size = 90, antiderivative size = 29 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=3+2 x-\log \left (x^2+\log (x)+\frac {4}{\log \left (8 e^{-x/4} x\right )}\right ) \] Output:

2*x-ln(ln(x)+4/ln(8*x/exp(1/4*x))+x^2)+3
 

Mathematica [F]

\[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=\int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx \] Input:

Integrate[(4 - x + 8*x*Log[(8*x)/E^(x/4)] + (-1 - 2*x^2 + 2*x^3 + 2*x*Log[ 
x])*Log[(8*x)/E^(x/4)]^2)/(4*x*Log[(8*x)/E^(x/4)] + (x^3 + x*Log[x])*Log[( 
8*x)/E^(x/4)]^2),x]
 

Output:

Integrate[(4 - x + 8*x*Log[(8*x)/E^(x/4)] + (-1 - 2*x^2 + 2*x^3 + 2*x*Log[ 
x])*Log[(8*x)/E^(x/4)]^2)/(4*x*Log[(8*x)/E^(x/4)] + (x^3 + x*Log[x])*Log[( 
8*x)/E^(x/4)]^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.

\(\displaystyle \int \frac {\left (2 x^3-2 x^2+2 x \log (x)-1\right ) \log ^2\left (8 e^{-x/4} x\right )-x+8 x \log \left (8 e^{-x/4} x\right )+4}{\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )+4 x \log \left (8 e^{-x/4} x\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 x^3-2 x^2+2 x \log (x)-1}{x \left (x^2+\log (x)\right )}+\frac {x^5-4 x^4+2 x^3 \log (x)+32 x^2-8 x^2 \log (x)+x \log ^2(x)-4 \log ^2(x)+16}{4 x \left (x^2+\log (x)\right ) \left (x^2 \log \left (8 e^{-x/4} x\right )+\log (x) \log \left (8 e^{-x/4} x\right )+4\right )}+\frac {4-x}{4 x \log \left (8 e^{-x/4} x\right )}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(4 - x + 8*x*Log[(8*x)/E^(x/4)] + (-1 - 2*x^2 + 2*x^3 + 2*x*Log[x])*Lo 
g[(8*x)/E^(x/4)]^2)/(4*x*Log[(8*x)/E^(x/4)] + (x^3 + x*Log[x])*Log[(8*x)/E 
^(x/4)]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66

method	result	size
parallelrisch	\(2 x +\ln \left (\ln \left (8 x \,{\mathrm e}^{-\frac {x}{4}}\right )\right )-\ln \left (\ln \left (8 x \,{\mathrm e}^{-\frac {x}{4}}\right ) x^{2}+\ln \left (8 x \,{\mathrm e}^{-\frac {x}{4}}\right ) \ln \left (x \right )+4\right )\)	\(48\)
risch	\(2 x -\ln \left (x^{2}+\ln \left (x \right )\right )+\ln \left (\ln \left ({\mathrm e}^{\frac {x}{4}}\right )+\frac {i \left (-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \right )+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}-\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2} \operatorname {csgn}\left (i x \right )+6 i \ln \left (2\right )+2 i \ln \left (x \right )\right )}{2}\right )-\ln \left (\ln \left ({\mathrm e}^{\frac {x}{4}}\right )+\frac {i \left (-\pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \right )+\pi \,x^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}-\pi \,x^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2} \operatorname {csgn}\left (i x \right )-\ln \left (x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2}+\ln \left (x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right ) \operatorname {csgn}\left (i x \right )+\ln \left (x \right ) \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{3}-\ln \left (x \right ) \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-\frac {x}{4}}\right )^{2} \operatorname {csgn}\left (i x \right )+6 i \ln \left (2\right ) \ln \left (x \right )+6 i \ln \left (2\right ) x^{2}+2 i \ln \left (x \right )^{2}+2 i x^{2} \ln \left (x \right )+8 i\right )}{2 x^{2}+2 \ln \left (x \right )}\right )\)	\(342\)

Input:

int(((2*x*ln(x)+2*x^3-2*x^2-1)*ln(8*x/exp(1/4*x))^2+8*x*ln(8*x/exp(1/4*x)) 
-x+4)/((x*ln(x)+x^3)*ln(8*x/exp(1/4*x))^2+4*x*ln(8*x/exp(1/4*x))),x,method 
=_RETURNVERBOSE)
 

Output:

2*x+ln(ln(8*x/exp(1/4*x)))-ln(ln(8*x/exp(1/4*x))*x^2+ln(x)*ln(8*x/exp(1/4* 
x))+4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=2 \, x - \log \left (-x^{3} + 12 \, x^{2} \log \left (2\right ) + {\left (4 \, x^{2} - x + 12 \, \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2} + 16\right ) + \log \left (-x + 12 \, \log \left (2\right ) + 4 \, \log \left (x\right )\right ) \] Input:

integrate(((2*x*log(x)+2*x^3-2*x^2-1)*log(8*x/exp(1/4*x))^2+8*x*log(8*x/ex 
p(1/4*x))-x+4)/((x*log(x)+x^3)*log(8*x/exp(1/4*x))^2+4*x*log(8*x/exp(1/4*x 
))),x, algorithm="fricas")
 

Output:

2*x - log(-x^3 + 12*x^2*log(2) + (4*x^2 - x + 12*log(2))*log(x) + 4*log(x) 
^2 + 16) + log(-x + 12*log(2) + 4*log(x))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).

Time = 0.61 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=2 x + \log {\left (- \frac {x}{4} + \log {\left (x \right )} + 3 \log {\left (2 \right )} \right )} - \log {\left (- \frac {x^{3}}{4} + 3 x^{2} \log {\left (2 \right )} + \left (x^{2} - \frac {x}{4} + 3 \log {\left (2 \right )}\right ) \log {\left (x \right )} + \log {\left (x \right )}^{2} + 4 \right )} \] Input:

integrate(((2*x*ln(x)+2*x**3-2*x**2-1)*ln(8*x/exp(1/4*x))**2+8*x*ln(8*x/ex 
p(1/4*x))-x+4)/((x*ln(x)+x**3)*ln(8*x/exp(1/4*x))**2+4*x*ln(8*x/exp(1/4*x) 
)),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=2 \, x - \log \left (-\frac {1}{4} \, x^{3} + 3 \, x^{2} \log \left (2\right ) + \frac {1}{4} \, {\left (4 \, x^{2} - x + 12 \, \log \left (2\right )\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 4\right ) + \log \left (-\frac {1}{4} \, x + 3 \, \log \left (2\right ) + \log \left (x\right )\right ) \] Input:

integrate(((2*x*log(x)+2*x^3-2*x^2-1)*log(8*x/exp(1/4*x))^2+8*x*log(8*x/ex 
p(1/4*x))-x+4)/((x*log(x)+x^3)*log(8*x/exp(1/4*x))^2+4*x*log(8*x/exp(1/4*x 
))),x, algorithm="maxima")
 

Output:

2*x - log(-1/4*x^3 + 3*x^2*log(2) + 1/4*(4*x^2 - x + 12*log(2))*log(x) + l 
og(x)^2 + 4) + log(-1/4*x + 3*log(2) + log(x))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).

Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=2 \, x - \log \left (-x^{3} + 12 \, x^{2} \log \left (2\right ) + 4 \, x^{2} \log \left (x\right ) - x \log \left (x\right ) + 12 \, \log \left (2\right ) \log \left (x\right ) + 4 \, \log \left (x\right )^{2} + 16\right ) + \log \left (-x + 12 \, \log \left (2\right ) + 4 \, \log \left (x\right )\right ) \] Input:

integrate(((2*x*log(x)+2*x^3-2*x^2-1)*log(8*x/exp(1/4*x))^2+8*x*log(8*x/ex 
p(1/4*x))-x+4)/((x*log(x)+x^3)*log(8*x/exp(1/4*x))^2+4*x*log(8*x/exp(1/4*x 
))),x, algorithm="giac")
 

Output:

2*x - log(-x^3 + 12*x^2*log(2) + 4*x^2*log(x) - x*log(x) + 12*log(2)*log(x 
) + 4*log(x)^2 + 16) + log(-x + 12*log(2) + 4*log(x))
 

Mupad [B] (verification not implemented)

Time = 6.25 (sec) , antiderivative size = 491, normalized size of antiderivative = 16.93 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx =\text {Too large to display} \] Input:

int((8*x*log(8*x*exp(-x/4)) - x + log(8*x*exp(-x/4))^2*(2*x*log(x) - 2*x^2 
 + 2*x^3 - 1) + 4)/(log(8*x*exp(-x/4))^2*(x*log(x) + x^3) + 4*x*log(8*x*ex 
p(-x/4))),x)
 

Output:

2*x - log((16*x + 4*x*log(x)^2 - 17*x^2*log(x) + 4*x^3*log(x) - 16*log(x)^ 
2 - 48*x^2*log(2) + 12*x^3*log(2) - 48*log(2)*log(x) + 4*x*log(x) + 4*x^3 
- x^4 + 12*x*log(2)*log(x) - 64)/x) + log(x*(x - 4)) - log(9*x^3*log(2)^2 
- 36*x^2*log(2)^2 - 8*x + 3*x*log(2) + (9*x*log(2)^2)/2 + (45*x^2*log(2))/ 
4 + 3*x^3*log(2) + (45*x^4*log(2))/2 - 6*x^5*log(2) - 18*log(2)^2 + (515*x 
^2)/8 - (287*x^3)/32 + 30*x^4 - (23*x^5)/16 - (7*x^6)/2 + x^7 + 32) + log( 
1/x^2) + log(32*x - 384*log(2) - 128*log(x) - (261*x^2*log(2)^2)/2 + 432*x 
^2*log(2)^3 - 72*x^3*log(2)^2 - 108*x^3*log(2)^3 - 261*x^4*log(2)^2 + 72*x 
^5*log(2)^2 + 72*log(2)^2*log(x) - (515*x^2*log(x))/2 + (287*x^3*log(x))/8 
 - 120*x^4*log(x) + (23*x^5*log(x))/4 + 14*x^6*log(x) - 4*x^7*log(x) + 96* 
x*log(2) - 54*x*log(2)^2 - (1539*x^2*log(2))/2 - 54*x*log(2)^3 + (951*x^3* 
log(2))/8 - 357*x^4*log(2) + (159*x^5*log(2))/4 + 36*x^6*log(2) - 12*x^7*l 
og(2) + 32*x*log(x) + 216*log(2)^3 - 8*x^2 + (515*x^3)/8 - (287*x^4)/32 + 
30*x^5 - (23*x^6)/16 - (7*x^7)/2 + x^8 - 12*x*log(2)*log(x) - 18*x*log(2)^ 
2*log(x) - 45*x^2*log(2)*log(x) - 12*x^3*log(2)*log(x) - 90*x^4*log(2)*log 
(x) + 24*x^5*log(2)*log(x) + 144*x^2*log(2)^2*log(x) - 36*x^3*log(2)^2*log 
(x))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {4-x+8 x \log \left (8 e^{-x/4} x\right )+\left (-1-2 x^2+2 x^3+2 x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )}{4 x \log \left (8 e^{-x/4} x\right )+\left (x^3+x \log (x)\right ) \log ^2\left (8 e^{-x/4} x\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {8 x}{e^{\frac {x}{4}}}\right )\right )-\mathrm {log}\left (\mathrm {log}\left (\frac {8 x}{e^{\frac {x}{4}}}\right ) \mathrm {log}\left (x \right )+\mathrm {log}\left (\frac {8 x}{e^{\frac {x}{4}}}\right ) x^{2}+4\right )-8 \,\mathrm {log}\left (\frac {8 x}{e^{\frac {x}{4}}}\right )+8 \,\mathrm {log}\left (x \right ) \] Input:

int(((2*x*log(x)+2*x^3-2*x^2-1)*log(8*x/exp(1/4*x))^2+8*x*log(8*x/exp(1/4* 
x))-x+4)/((x*log(x)+x^3)*log(8*x/exp(1/4*x))^2+4*x*log(8*x/exp(1/4*x))),x)
 

Output:

log(log((8*x)/e**(x/4))) - log(log((8*x)/e**(x/4))*log(x) + log((8*x)/e**( 
x/4))*x**2 + 4) - 8*log((8*x)/e**(x/4)) + 8*log(x)