\(\int -\frac {8}{(1620+144 x+180 \log (2)) \log (\frac {1}{5} (45+4 x+5 \log (2))) \log (\log (\frac {1}{5} (45+4 x+5 \log (2))))+(540+48 x+60 \log (2)) \log (\frac {1}{5} (45+4 x+5 \log (2))) \log (\log (\frac {1}{5} (45+4 x+5 \log (2)))) \log (\log (\log (\frac {1}{5} (45+4 x+5 \log (2)))))+(45+4 x+5 \log (2)) \log (\frac {1}{5} (45+4 x+5 \log (2))) \log (\log (\frac {1}{5} (45+4 x+5 \log (2)))) \log ^2(\log (\log (\frac {1}{5} (45+4 x+5 \log (2)))))} \, dx\) [1474]

Optimal result

Integrand size = 156, antiderivative size = 25 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {x}{3 x+\frac {1}{2} x \log \left (\log \left (\log \left (9+\frac {4 x}{5}+\log (2)\right )\right )\right )} \] Output:

x/(3*x+1/2*ln(ln(ln(ln(2)+4/5*x+9)))*x)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{6+\log \left (\log \left (\log \left (9+\frac {4 x}{5}+\log (2)\right )\right )\right )} \] Input:

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]
 

Output:

2/(6 + Log[Log[Log[9 + (4*x)/5 + Log[2]]]])
 

Rubi [A] (verified)

Time = 0.47 (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

\(\displaystyle \frac {2}{\log \left (\log \left (\log \left (\frac {4 x}{5}+9+\log (2)\right )\right )\right )+6}\)

Input:

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]
 

Output:

2/(6 + Log[Log[Log[9 + (4*x)/5 + Log[2]]]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7237

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]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

Maple [A] (verified)

Time = 10.70 (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	\(\frac {2}{\ln \left (\ln \left (-\ln \left (5\right )+\ln \left (5 \ln \left (2\right )+4 x +45\right )\right )\right )+6}\)	\(24\)

Input:

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)
 

Output:

2/(ln(ln(ln(ln(2)+4/5*x+9)))+6)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (\log \left (\frac {4}{5} \, x + \log \left (2\right ) + 9\right )\right )\right ) + 6} \] Input:

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="fricas" 
)
 

Output:

2/(log(log(log(4/5*x + log(2) + 9))) + 6)
 

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log {\left (\log {\left (\log {\left (\frac {4 x}{5} + \log {\left (2 \right )} + 9 \right )} \right )} \right )} + 6} \] Input:

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)
 

Output:

2/(log(log(log(4*x/5 + log(2) + 9))) + 6)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (-\log \left (5\right ) + \log \left (4 \, x + 5 \, \log \left (2\right ) + 45\right )\right )\right ) + 6} \] Input:

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="maxima" 
)
 

Output:

2/(log(log(-log(5) + log(4*x + 5*log(2) + 45))) + 6)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\frac {2}{\log \left (\log \left (-\log \left (5\right ) + \log \left (4 \, x + 5 \, \log \left (2\right ) + 45\right )\right )\right ) + 6} \] Input:

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="giac")
 

Output:

2/(log(log(-log(5) + log(4*x + 5*log(2) + 45))) + 6)
 

Mupad [F(-1)]

Timed out. \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=\int -\frac {8}{\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (4\,x+5\,\ln \left (2\right )+45\right )\,{\ln \left (\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\right )}^2+\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (48\,x+60\,\ln \left (2\right )+540\right )\,\ln \left (\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\right )+\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\,\ln \left (\ln \left (\frac {4\,x}{5}+\ln \left (2\right )+9\right )\right )\,\left (144\,x+180\,\ln \left (2\right )+1620\right )} \,d x \] Input:

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)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int -\frac {8}{(1620+144 x+180 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )+(540+48 x+60 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log \left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )+(45+4 x+5 \log (2)) \log \left (\frac {1}{5} (45+4 x+5 \log (2))\right ) \log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{5} (45+4 x+5 \log (2))\right )\right )\right )} \, dx=-\frac {\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )+\frac {4 x}{5}+9\right )\right )\right )}{3 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )+\frac {4 x}{5}+9\right )\right )\right )+18} \] Input:

int(-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)
 

Output:

( - log(log(log((5*log(2) + 4*x + 45)/5))))/(3*(log(log(log((5*log(2) + 4* 
x + 45)/5))) + 6))