Integrand size = 112, antiderivative size = 33 \begin {dmath*} \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} \end {dmath*}
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \begin {dmath*} \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} \end {dmath*}
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]
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 |
\(\displaystyle -\frac {1}{10} \log ^4(2) \int \frac {1}{x^2 \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx+\frac {1}{10} \log ^4(2) \int \frac {\log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )}{x^2}dx-\frac {1}{10} \log ^2(2) \int \log \left (\log \left (\log \left (\frac {x}{x+1}\right )\right )\right )dx+\frac {1}{10} \left (1+\log ^4(2)+2 \log ^3(2)\right ) \int \frac {1}{x \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx-\frac {1}{10} \left (1+\log ^4(2)+2 \log ^3(2)+\log ^2(2)\right ) \int \frac {1}{(x+1) \log \left (\frac {x}{x+1}\right ) \log \left (\log \left (\frac {x}{x+1}\right )\right )}dx\) |
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]
3.6.13.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(31)=62\).
Time = 46.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.12
method | result | size |
parallelrisch | \(-\frac {\ln \left (2\right )^{4} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-2 \ln \left (2\right )^{3} x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )+\ln \left (2\right )^{2} x^{2} \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )-x \ln \left (\ln \left (\ln \left (\frac {x}{1+x}\right )\right )\right )}{10 x}\) | \(70\) |
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)
-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
Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \begin {dmath*} \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} \end {dmath*}
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=\
Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \begin {dmath*} \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} \end {dmath*}
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)
(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)
Time = 0.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \begin {dmath*} \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} \end {dmath*}
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=\
Time = 0.51 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \begin {dmath*} \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 (\log \left (-\log \left (x + 1\right ) + \log \left (x\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 ) \end {dmath*}
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=\
1/10*(2*log(2)^3 + 1)*log(log(-log(x + 1) + log(x))) - 1/10*(x*log(2)^2 + log(2)^4/x)*log(log(log(x/(x + 1))))
Time = 16.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \begin {dmath*} \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} \end {dmath*}