Integrand size = 156, antiderivative size = 25 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {x}{3 x+\frac {1}{2} x \log \left (\log \left (\log \left (9+\frac {4 x}{5}+\log (2)\right )\right )\right )} \]
Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{6+\log \left (\log \left (\log \left (9+\frac {4 x}{5}+\log (2)\right )\right )\right )} \]
Integrate[-8/((1620 + 144*x + 180*Log[2])*Log[(45 + 4*x + 5*Log[2])/5]*Log [Log[(45 + 4*x + 5*Log[2])/5]] + (540 + 48*x + 60*Log[2])*Log[(45 + 4*x + 5*Log[2])/5]*Log[Log[(45 + 4*x + 5*Log[2])/5]]*Log[Log[Log[(45 + 4*x + 5*L og[2])/5]]] + (45 + 4*x + 5*Log[2])*Log[(45 + 4*x + 5*Log[2])/5]*Log[Log[( 45 + 4*x + 5*Log[2])/5]]*Log[Log[Log[(45 + 4*x + 5*Log[2])/5]]]^2),x]
Time = 0.46 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {27, 7239, 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.
| \(\displaystyle \int -\frac {8}{(4 x+45+5 \log (2)) \log \left (\frac {1}{5} (4 x+45+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (4 x+45+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (4 x+45+5 \log (2))\right )\right )\right )+(48 x+540+60 \log (2)) \log \left (\frac {1}{5} (4 x+45+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (4 x+45+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (4 x+45+5 \log (2))\right )\right )\right )+(144 x+1620+180 \log (2)) \log \left (\frac {1}{5} (4 x+45+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (4 x+45+5 \log (2))\right )\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -8 \int \frac {1}{(4 x+\log (32)+45) \log \left (\frac {1}{5} (4 x+\log (32)+45)\right ) \log \left (\log \left (\frac {1}{5} (4 x+\log (32)+45)\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (4 x+\log (32)+45)\right )\right )\right )+12 (4 x+5 (9+\log (2))) \log \left (\frac {1}{5} (4 x+\log (32)+45)\right ) \log \left (\log \left (\frac {1}{5} (4 x+\log (32)+45)\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (4 x+\log (32)+45)\right )\right )\right )+36 (4 x+5 (9+\log (2))) \log \left (\frac {1}{5} (4 x+\log (32)+45)\right ) \log \left (\log \left (\frac {1}{5} (4 x+\log (32)+45)\right )\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -8 \int \frac {1}{(4 x+\log (32)+45) \log \left (\frac {4 x}{5}+\log (2)+9\right ) \log \left (\log \left (\frac {4 x}{5}+\log (2)+9\right )\right ) \left (\log \left (\log \left (\log \left (\frac {4 x}{5}+\log (2)+9\right )\right )\right )+6\right )^2}dx\) |
| \(\Big \downarrow \) 7237 |
| \(\displaystyle \frac {2}{\log \left (\log \left (\log \left (\frac {4 x}{5}+9+\log (2)\right )\right )\right )+6}\) |
Int[-8/((1620 + 144*x + 180*Log[2])*Log[(45 + 4*x + 5*Log[2])/5]*Log[Log[( 45 + 4*x + 5*Log[2])/5]] + (540 + 48*x + 60*Log[2])*Log[(45 + 4*x + 5*Log[ 2])/5]*Log[Log[(45 + 4*x + 5*Log[2])/5]]*Log[Log[Log[(45 + 4*x + 5*Log[2]) /5]]] + (45 + 4*x + 5*Log[2])*Log[(45 + 4*x + 5*Log[2])/5]*Log[Log[(45 + 4 *x + 5*Log[2])/5]]*Log[Log[Log[(45 + 4*x + 5*Log[2])/5]]]^2),x]
3.15.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 10.91 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
| method | result | size |
| risch | \(\frac {2}{\ln \left (\ln \left (\ln \left (\ln \left (2\right )+\frac {4 x}{5}+9\right )\right )\right )+6}\) | \(17\) |
| parallelrisch | \(\frac {2}{\ln \left (\ln \left (\ln \left (\ln \left (2\right )+\frac {4 x}{5}+9\right )\right )\right )+6}\) | \(17\) |
| default | \(\frac {2}{\ln \left (\ln \left (-\ln \left (5\right )+\ln \left (5 \ln \left (2\right )+4 x +45\right )\right )\right )+6}\) | \(24\) |
int(-8/((5*ln(2)+4*x+45)*ln(ln(2)+4/5*x+9)*ln(ln(ln(2)+4/5*x+9))*ln(ln(ln( ln(2)+4/5*x+9)))^2+(60*ln(2)+48*x+540)*ln(ln(2)+4/5*x+9)*ln(ln(ln(2)+4/5*x +9))*ln(ln(ln(ln(2)+4/5*x+9)))+(180*ln(2)+144*x+1620)*ln(ln(2)+4/5*x+9)*ln (ln(ln(2)+4/5*x+9))),x,method=_RETURNVERBOSE)
Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (\log \left (\frac {4}{5} \, x + \log \left (2\right ) + 9\right )\right )\right ) + 6} \]
integrate(-8/((5*log(2)+4*x+45)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9 ))*log(log(log(log(2)+4/5*x+9)))^2+(60*log(2)+48*x+540)*log(log(2)+4/5*x+9 )*log(log(log(2)+4/5*x+9))*log(log(log(log(2)+4/5*x+9)))+(180*log(2)+144*x +1620)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9))),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log {\left (\log {\left (\log {\left (\frac {4 x}{5} + \log {\left (2 \right )} + 9 \right )} \right )} \right )} + 6} \]
integrate(-8/((5*ln(2)+4*x+45)*ln(ln(2)+4/5*x+9)*ln(ln(ln(2)+4/5*x+9))*ln( ln(ln(ln(2)+4/5*x+9)))**2+(60*ln(2)+48*x+540)*ln(ln(2)+4/5*x+9)*ln(ln(ln(2 )+4/5*x+9))*ln(ln(ln(ln(2)+4/5*x+9)))+(180*ln(2)+144*x+1620)*ln(ln(2)+4/5* x+9)*ln(ln(ln(2)+4/5*x+9))),x)
Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (-\log \left (5\right ) + \log \left (4 \, x + 5 \, \log \left (2\right ) + 45\right )\right )\right ) + 6} \]
integrate(-8/((5*log(2)+4*x+45)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9 ))*log(log(log(log(2)+4/5*x+9)))^2+(60*log(2)+48*x+540)*log(log(2)+4/5*x+9 )*log(log(log(2)+4/5*x+9))*log(log(log(log(2)+4/5*x+9)))+(180*log(2)+144*x +1620)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9))),x, algorithm=\
Time = 0.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (-\log \left (5\right ) + \log \left (4 \, x + 5 \, \log \left (2\right ) + 45\right )\right )\right ) + 6} \]
integrate(-8/((5*log(2)+4*x+45)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9 ))*log(log(log(log(2)+4/5*x+9)))^2+(60*log(2)+48*x+540)*log(log(2)+4/5*x+9 )*log(log(log(2)+4/5*x+9))*log(log(log(log(2)+4/5*x+9)))+(180*log(2)+144*x +1620)*log(log(2)+4/5*x+9)*log(log(log(2)+4/5*x+9))),x, algorithm=\
Timed out. \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\int -\frac {8}{\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (4\,x+5\,\ln \left (2\right )+45\right )\,{\ln \left (\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\right )}^2+\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (48\,x+60\,\ln \left (2\right )+540\right )\,\ln \left (\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\right )+\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (144\,x+180\,\ln \left (2\right )+1620\right )} \,d x \]
int(-8/(log((4*x)/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*(144*x + 180*log(2) + 1620) + log((4*x)/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*log(log(log((4*x)/5 + log(2) + 9)))^2*(4*x + 5*log(2) + 45) + log((4*x )/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*log(log(log((4*x)/5 + log (2) + 9)))*(48*x + 60*log(2) + 540)),x)
int(-8/(log((4*x)/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*(144*x + 180*log(2) + 1620) + log((4*x)/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*log(log(log((4*x)/5 + log(2) + 9)))^2*(4*x + 5*log(2) + 45) + log((4*x )/5 + log(2) + 9)*log(log((4*x)/5 + log(2) + 9))*log(log(log((4*x)/5 + log (2) + 9)))*(48*x + 60*log(2) + 540)), x)