\(\int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+((-x^2-x^3) \log ^2(2)+(1+x) \log ^4(2)) \log (\frac {x}{1+x}) \log (\log (\frac {x}{1+x})) \log (\log (\log (\frac {x}{1+x})))}{(10 x^2+10 x^3) \log (\frac {x}{1+x}) \log (\log (\frac {x}{1+x}))} \, dx\) [513]

Optimal result

Integrand size = 112, antiderivative size = 33 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\left (x-(x-\log (2))^2 \log ^2(2)\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \] Output:

1/10*(x-ln(2)^2*(x-ln(2))^2)/x*ln(ln(ln(x/(1+x))))
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {\left (x^2 \log ^2(2)+\log ^4(2)-x \left (1+2 \log ^3(2)\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{10 x} \] Input:

Integrate[(x - x^2*Log[2]^2 + 2*x*Log[2]^3 - Log[2]^4 + ((-x^2 - x^3)*Log[ 
2]^2 + (1 + x)*Log[2]^4)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]*Log[Log[Log[x/ 
(1 + x)]]])/((10*x^2 + 10*x^3)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]),x]
 

Output:

-1/10*((x^2*Log[2]^2 + Log[2]^4 - x*(1 + 2*Log[2]^3))*Log[Log[Log[x/(1 + x 
)]]])/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.

\(\displaystyle \int \frac {x^2 \left (-\log ^2(2)\right )+\left (\left (-x^3-x^2\right ) \log ^2(2)+(x+1) \log ^4(2)\right ) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )+x+2 x \log ^3(2)-\log ^4(2)}{\left (10 x^3+10 x^2\right ) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {x^2 \left (-\log ^2(2)\right )+\left (\left (-x^3-x^2\right ) \log ^2(2)+(x+1) \log ^4(2)\right ) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )+x \left (1+2 \log ^3(2)\right )-\log ^4(2)}{\left (10 x^3+10 x^2\right ) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {x^2 \left (-\log ^2(2)\right )+\left (\left (-x^3-x^2\right ) \log ^2(2)+(x+1) \log ^4(2)\right ) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )+x \left (1+2 \log ^3(2)\right )-\log ^4(2)}{x^2 (10 x+10) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^2 \left (-\log ^2(2)\right )+x \left (1+2 \log ^3(2)\right )-\log ^4(2)}{10 x^2 (x+1) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}-\frac {\log ^2(2) (x-\log (2)) (x+\log (2)) \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )}{10 x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{10} \log ^4(2) \int \frac {1}{x^2 \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx+\frac {1}{10} \log ^4(2) \int \frac {\log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )}{x^2}dx-\frac {1}{10} \log ^2(2) \int \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )dx+\frac {1}{10} \left (1+\log ^4(2)+2 \log ^3(2)\right ) \int \frac {1}{x \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx-\frac {1}{10} \left (1+\log ^4(2)+2 \log ^3(2)+\log ^2(2)\right ) \int \frac {1}{(x+1) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx\)

Input:

Int[(x - x^2*Log[2]^2 + 2*x*Log[2]^3 - Log[2]^4 + ((-x^2 - x^3)*Log[2]^2 + 
 (1 + x)*Log[2]^4)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]*Log[Log[Log[x/(1 + x 
)]]])/((10*x^2 + 10*x^3)*Log[x/(1 + x)]*Log[Log[x/(1 + x)]]),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(31)=62\).

Time = 42.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12

method	result	size
parallelrisch	\(-\frac {\ln \left (2\right )^{4} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-2 \ln \left (2\right )^{3} x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )+\ln \left (2\right )^{2} x^{2} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )}{10 x}\)	\(70\)

Input:

int((((1+x)*ln(2)^4+(-x^3-x^2)*ln(2)^2)*ln(x/(1+x))*ln(ln(x/(1+x)))*ln(ln( 
ln(x/(1+x))))-ln(2)^4+2*x*ln(2)^3-x^2*ln(2)^2+x)/(10*x^3+10*x^2)/ln(x/(1+x 
))/ln(ln(x/(1+x))),x,method=_RETURNVERBOSE)
 

Output:

-1/10*(ln(2)^4*ln(ln(ln(x/(1+x))))-2*ln(2)^3*x*ln(ln(ln(x/(1+x))))+ln(2)^2 
*x^2*ln(ln(ln(x/(1+x))))-x*ln(ln(ln(x/(1+x)))))/x
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {{\left (x^{2} \log \left (2\right )^{2} - 2 \, x \log \left (2\right )^{3} + \log \left (2\right )^{4} - x\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right )}{10 \, x} \] Input:

integrate((((1+x)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(1+x))*log(log(x/(1+ 
x)))*log(log(log(x/(1+x))))-log(2)^4+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+ 
10*x^2)/log(x/(1+x))/log(log(x/(1+x))),x, algorithm="fricas")
 

Output:

-1/10*(x^2*log(2)^2 - 2*x*log(2)^3 + log(2)^4 - x)*log(log(log(x/(x + 1))) 
)/x
 

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\left (2 \log {\left (2 \right )}^{3} + 1\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10} + \frac {\left (- x^{2} \log {\left (2 \right )}^{2} - \log {\left (2 \right )}^{4}\right ) \log {\left (\log {\left (\log {\left (\frac {x}{x + 1} \right )} \right )} \right )}}{10 x} \] Input:

integrate((((1+x)*ln(2)**4+(-x**3-x**2)*ln(2)**2)*ln(x/(1+x))*ln(ln(x/(1+x 
)))*ln(ln(ln(x/(1+x))))-ln(2)**4+2*x*ln(2)**3-x**2*ln(2)**2+x)/(10*x**3+10 
*x**2)/ln(x/(1+x))/ln(ln(x/(1+x))),x)
 

Output:

(2*log(2)**3 + 1)*log(log(log(x/(x + 1))))/10 + (-x**2*log(2)**2 - log(2)* 
*4)*log(log(log(x/(x + 1))))/(10*x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=-\frac {{\left (x^{2} \log \left (2\right )^{2} + \log \left (2\right )^{4} - {\left (2 \, \log \left (2\right )^{3} + 1\right )} x\right )} \log \left (\log \left (-\log \left (x + 1\right ) + \log \left (x\right )\right )\right )}{10 \, x} \] Input:

integrate((((1+x)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(1+x))*log(log(x/(1+ 
x)))*log(log(log(x/(1+x))))-log(2)^4+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+ 
10*x^2)/log(x/(1+x))/log(log(x/(1+x))),x, algorithm="maxima")
 

Output:

-1/10*(x^2*log(2)^2 + log(2)^4 - (2*log(2)^3 + 1)*x)*log(log(-log(x + 1) + 
 log(x)))/x
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {1}{10} \, {\left (2 \, \log \left (2\right )^{3} + 1\right )} \log \left ({\left | \log \left (-\log \left (x + 1\right ) + \log \left (x\right )\right ) \right |}\right ) - \frac {1}{10} \, {\left (x \log \left (2\right )^{2} + \frac {\log \left (2\right )^{4}}{x}\right )} \log \left (\log \left (\log \left (\frac {x}{x + 1}\right )\right )\right ) \] Input:

integrate((((1+x)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(1+x))*log(log(x/(1+ 
x)))*log(log(log(x/(1+x))))-log(2)^4+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+ 
10*x^2)/log(x/(1+x))/log(log(x/(1+x))),x, algorithm="giac")
 

Output:

1/10*(2*log(2)^3 + 1)*log(abs(log(-log(x + 1) + log(x)))) - 1/10*(x*log(2) 
^2 + log(2)^4/x)*log(log(log(x/(x + 1))))
 

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\ln \left (\ln \left (\ln \left (\frac {x}{x+1}\right )\right )\right )\,\left (x-x^2\,{\ln \left (2\right )}^2+2\,x\,{\ln \left (2\right )}^3-{\ln \left (2\right )}^4\right )}{10\,x} \] Input:

int((x - x^2*log(2)^2 + 2*x*log(2)^3 - log(2)^4 + log(log(log(x/(x + 1)))) 
*log(x/(x + 1))*log(log(x/(x + 1)))*(log(2)^4*(x + 1) - log(2)^2*(x^2 + x^ 
3)))/(log(x/(x + 1))*log(log(x/(x + 1)))*(10*x^2 + 10*x^3)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {x-x^2 \log ^2(2)+2 x \log ^3(2)-\log ^4(2)+\left (\left (-x^2-x^3\right ) \log ^2(2)+(1+x) \log ^4(2)\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right ) \log \left (\log \left (\log \left (\frac {x}{1+x}\right )\right )\right )}{\left (10 x^2+10 x^3\right ) \log \left (\frac {x}{1+x}\right ) \log \left (\log \left (\frac {x}{1+x}\right )\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {x}{x +1}\right )\right )\right ) \left (-\mathrm {log}\left (2\right )^{4}+2 \mathrm {log}\left (2\right )^{3} x -\mathrm {log}\left (2\right )^{2} x^{2}+x \right )}{10 x} \] Input:

int((((1+x)*log(2)^4+(-x^3-x^2)*log(2)^2)*log(x/(1+x))*log(log(x/(1+x)))*l 
og(log(log(x/(1+x))))-log(2)^4+2*x*log(2)^3-x^2*log(2)^2+x)/(10*x^3+10*x^2 
)/log(x/(1+x))/log(log(x/(1+x))),x)
 

Output:

(log(log(log(x/(x + 1))))*( - log(2)**4 + 2*log(2)**3*x - log(2)**2*x**2 + 
 x))/(10*x)