Integrand size = 69, antiderivative size = 21 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\log ^2\left (2+\frac {1}{2} (1+\log (4))-\log (\log (-1+\log (x)))\right ) \] Output:
ln(-ln(ln(ln(x)-1))+ln(2)+5/2)^2
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\log ^2\left (\frac {5}{2}+\log (2)-\log (\log (-1+\log (x)))\right ) \] Input:
Integrate[(4*Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2])/((5*x + x*Log[ 4] + (-5*x - x*Log[4])*Log[x])*Log[-1 + Log[x]] + (-2*x + 2*x*Log[x])*Log[ -1 + Log[x]]*Log[Log[-1 + Log[x]]]),x]
Output:
Log[5/2 + Log[2] - Log[Log[-1 + Log[x]]]]^2
Time = 0.89 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {27, 3039, 7237}
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.
\(\displaystyle \int \frac {4 \log \left (\frac {1}{2} (-2 \log (\log (\log (x)-1))+5+\log (4))\right )}{(5 x+x \log (4)+(x (-\log (4))-5 x) \log (x)) \log (\log (x)-1)+(2 x \log (x)-2 x) \log (\log (\log (x)-1)) \log (\log (x)-1)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {\log \left (\frac {1}{2} (-2 \log (\log (\log (x)-1))+\log (4)+5)\right )}{(-((5+\log (4)) \log (x) x)+\log (4) x+5 x) \log (\log (x)-1)-2 (x-x \log (x)) \log (\log (x)-1) \log (\log (\log (x)-1))}dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle 4 \int \frac {\log \left (\frac {1}{2} (-2 \log (\log (\log (x)-1))+\log (4)+5)\right )}{(1-\log (x)) \log (\log (x)-1) (-2 \log (\log (\log (x)-1))+\log (4)+5)}d\log (x)\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \log ^2\left (\frac {1}{2} (-2 \log (\log (\log (x)-1))+5+\log (4))\right )\) |
Input:
Int[(4*Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2])/((5*x + x*Log[4] + ( -5*x - x*Log[4])*Log[x])*Log[-1 + Log[x]] + (-2*x + 2*x*Log[x])*Log[-1 + L og[x]]*Log[Log[-1 + Log[x]]]),x]
Output:
Log[(5 + Log[4] - 2*Log[Log[-1 + Log[x]]])/2]^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 180.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\ln \left (-\ln \left (\ln \left (\ln \left (x \right )-1\right )\right )+\ln \left (2\right )+\frac {5}{2}\right )^{2}\) | \(16\) |
default | \(\ln \left (-2 \ln \left (\ln \left (\ln \left (x \right )-1\right )\right )+2 \ln \left (2\right )+5\right )^{2}-2 \ln \left (-2 \ln \left (\ln \left (\ln \left (x \right )-1\right )\right )+2 \ln \left (2\right )+5\right ) \ln \left (2\right )\) | \(38\) |
Input:
int(4*ln(-ln(ln(ln(x)-1))+ln(2)+5/2)/((2*x*ln(x)-2*x)*ln(ln(x)-1)*ln(ln(ln (x)-1))+((-2*x*ln(2)-5*x)*ln(x)+2*x*ln(2)+5*x)*ln(ln(x)-1)),x,method=_RETU RNVERBOSE)
Output:
ln(-ln(ln(ln(x)-1))+ln(2)+5/2)^2
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\log \left (\log \left (2\right ) - \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) + \frac {5}{2}\right )^{2} \] Input:
integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log( x)-1)*log(log(log(x)-1))+((-2*x*log(2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log (x)-1)),x, algorithm="fricas")
Output:
log(log(2) - log(log(log(x) - 1)) + 5/2)^2
Time = 2.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\log {\left (- \log {\left (\log {\left (\log {\left (x \right )} - 1 \right )} \right )} + \log {\left (2 \right )} + \frac {5}{2} \right )}^{2} \] Input:
integrate(4*ln(-ln(ln(ln(x)-1))+ln(2)+5/2)/((2*x*ln(x)-2*x)*ln(ln(x)-1)*ln (ln(ln(x)-1))+((-2*x*ln(2)-5*x)*ln(x)+2*x*ln(2)+5*x)*ln(ln(x)-1)),x)
Output:
log(-log(log(log(x) - 1)) + log(2) + 5/2)**2
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (15) = 30\).
Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.38 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=2 \, \log \left (\log \left (2\right ) - \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) + \frac {5}{2}\right ) \log \left (-2 \, \log \left (2\right ) + 2 \, \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) - 5\right ) - \log \left (-2 \, \log \left (2\right ) + 2 \, \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) - 5\right )^{2} \] Input:
integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log( x)-1)*log(log(log(x)-1))+((-2*x*log(2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log (x)-1)),x, algorithm="maxima")
Output:
2*log(log(2) - log(log(log(x) - 1)) + 5/2)*log(-2*log(2) + 2*log(log(log(x ) - 1)) - 5) - log(-2*log(2) + 2*log(log(log(x) - 1)) - 5)^2
Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).
Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=-2 \, \log \left (2\right ) \log \left (2 \, \log \left (2\right ) - 2 \, \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) + 5\right ) + \log \left (2 \, \log \left (2\right ) - 2 \, \log \left (\log \left (\log \left (x\right ) - 1\right )\right ) + 5\right )^{2} \] Input:
integrate(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log( x)-1)*log(log(log(x)-1))+((-2*x*log(2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log (x)-1)),x, algorithm="giac")
Output:
-2*log(2)*log(2*log(2) - 2*log(log(log(x) - 1)) + 5) + log(2*log(2) - 2*lo g(log(log(x) - 1)) + 5)^2
Timed out. \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\int \frac {4\,\ln \left (\ln \left (2\right )-\ln \left (\ln \left (\ln \left (x\right )-1\right )\right )+\frac {5}{2}\right )}{\ln \left (\ln \left (x\right )-1\right )\,\left (5\,x+2\,x\,\ln \left (2\right )-\ln \left (x\right )\,\left (5\,x+2\,x\,\ln \left (2\right )\right )\right )-\ln \left (\ln \left (\ln \left (x\right )-1\right )\right )\,\ln \left (\ln \left (x\right )-1\right )\,\left (2\,x-2\,x\,\ln \left (x\right )\right )} \,d x \] Input:
int((4*log(log(2) - log(log(log(x) - 1)) + 5/2))/(log(log(x) - 1)*(5*x + 2 *x*log(2) - log(x)*(5*x + 2*x*log(2))) - log(log(log(x) - 1))*log(log(x) - 1)*(2*x - 2*x*log(x))),x)
Output:
int((4*log(log(2) - log(log(log(x) - 1)) + 5/2))/(log(log(x) - 1)*(5*x + 2 *x*log(2) - log(x)*(5*x + 2*x*log(2))) - log(log(log(x) - 1))*log(log(x) - 1)*(2*x - 2*x*log(x))), x)
Time = 0.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {4 \log \left (\frac {1}{2} (5+\log (4)-2 \log (\log (-1+\log (x))))\right )}{(5 x+x \log (4)+(-5 x-x \log (4)) \log (x)) \log (-1+\log (x))+(-2 x+2 x \log (x)) \log (-1+\log (x)) \log (\log (-1+\log (x)))} \, dx=\mathrm {log}\left (-\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )-1\right )\right )+\mathrm {log}\left (2\right )+\frac {5}{2}\right )^{2} \] Input:
int(4*log(-log(log(log(x)-1))+log(2)+5/2)/((2*x*log(x)-2*x)*log(log(x)-1)* log(log(log(x)-1))+((-2*x*log(2)-5*x)*log(x)+2*x*log(2)+5*x)*log(log(x)-1) ),x)
Output:
log(( - 2*log(log(log(x) - 1)) + 2*log(2) + 5)/2)**2