\(\int \frac {-1296 x^2+(-13932-648 x^2) \log (\frac {1}{2} (43+2 x^2)) \log (\log (\frac {1}{2} (43+2 x^2))) \log (\log (\log (\frac {1}{2} (43+2 x^2))))}{e^3 (43 x^5+2 x^7) \log (\frac {1}{2} (43+2 x^2)) \log (\log (\frac {1}{2} (43+2 x^2))) \log ^5(\log (\log (\frac {1}{2} (43+2 x^2))))} \, dx\) [1177]

Optimal result

Integrand size = 115, antiderivative size = 20 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \] Output:

81/ln(ln(ln(x^2+43/2)))^4/x^4/exp(3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \] Input:

Integrate[(-1296*x^2 + (-13932 - 648*x^2)*Log[(43 + 2*x^2)/2]*Log[Log[(43 
+ 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]])/(E^3*(43*x^5 + 2*x^7)*Log[(43 
+ 2*x^2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]]^5),x]
 

Output:

81/(E^3*x^4*Log[Log[Log[43/2 + x^2]]]^4)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {27, 27, 2026, 7238}

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.

\(\displaystyle \int \frac {\left (-648 x^2-13932\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )-1296 x^2}{e^3 \left (2 x^7+43 x^5\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int -\frac {324 \left (4 x^2+\left (2 x^2+43\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )\right )}{\left (2 x^7+43 x^5\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}dx}{e^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {324 \int \frac {4 x^2+\left (2 x^2+43\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}{\left (2 x^7+43 x^5\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}dx}{e^3}\)

\(\Big \downarrow \) 2026

\(\displaystyle -\frac {324 \int \frac {4 x^2+\left (2 x^2+43\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}{x^5 \left (2 x^2+43\right ) \log \left (\frac {1}{2} \left (2 x^2+43\right )\right ) \log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}dx}{e^3}\)

\(\Big \downarrow \) 7238

\(\displaystyle \frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}\)

Input:

Int[(-1296*x^2 + (-13932 - 648*x^2)*Log[(43 + 2*x^2)/2]*Log[Log[(43 + 2*x^ 
2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]])/(E^3*(43*x^5 + 2*x^7)*Log[(43 + 2*x^ 
2)/2]*Log[Log[(43 + 2*x^2)/2]]*Log[Log[Log[(43 + 2*x^2)/2]]]^5),x]
 

Output:

81/(E^3*x^4*Log[Log[Log[(43 + 2*x^2)/2]]]^4)
 

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 2026

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])
 

rule 7238

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y* 
z, u*z^(n - m), x]}, Simp[q*y^(m + 1)*(z^(m + 1)/(m + 1)), x] /;  !FalseQ[q 
]] /; FreeQ[{m, n}, x] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 107.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {81 \,{\mathrm e}^{-3}}{{\ln \left (\ln \left (\ln \left (x^{2}+\frac {43}{2}\right )\right )\right )}^{4} x^{4}}\)	\(18\)
parallelrisch	\(\frac {81 \,{\mathrm e}^{-3}}{{\ln \left (\ln \left (\ln \left (x^{2}+\frac {43}{2}\right )\right )\right )}^{4} x^{4}}\)	\(20\)
default	\(\text {Expression too large to display}\)	\(5536\)

Input:

int(((-648*x^2-13932)*ln(x^2+43/2)*ln(ln(x^2+43/2))*ln(ln(ln(x^2+43/2)))-1 
296*x^2)/(2*x^7+43*x^5)/exp(3)/ln(x^2+43/2)/ln(ln(x^2+43/2))/ln(ln(ln(x^2+ 
43/2)))^5,x,method=_RETURNVERBOSE)
 

Output:

81/ln(ln(ln(x^2+43/2)))^4/x^4*exp(-3)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \] Input:

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x 
^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/exp(3)/log(x^2+43/2)/log(log(x^2+43/2) 
)/log(log(log(x^2+43/2)))^5,x, algorithm="fricas")
 

Output:

81*e^(-3)/(x^4*log(log(log(x^2 + 43/2)))^4)
 

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{x^{4} e^{3} \log {\left (\log {\left (\log {\left (x^{2} + \frac {43}{2} \right )} \right )} \right )}^{4}} \] Input:

integrate(((-648*x**2-13932)*ln(x**2+43/2)*ln(ln(x**2+43/2))*ln(ln(ln(x**2 
+43/2)))-1296*x**2)/(2*x**7+43*x**5)/exp(3)/ln(x**2+43/2)/ln(ln(x**2+43/2) 
)/ln(ln(ln(x**2+43/2)))**5,x)
 

Output:

81*exp(-3)/(x**4*log(log(log(x**2 + 43/2)))**4)
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (-\log \left (2\right ) + \log \left (2 \, x^{2} + 43\right )\right )\right )^{4}} \] Input:

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x 
^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/exp(3)/log(x^2+43/2)/log(log(x^2+43/2) 
)/log(log(log(x^2+43/2)))^5,x, algorithm="maxima")
 

Output:

81*e^(-3)/(x^4*log(log(-log(2) + log(2*x^2 + 43)))^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).

Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {324 \, e^{\left (-3\right )}}{{\left (2 \, x^{2} + 43\right )}^{2} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} - 86 \, {\left (2 \, x^{2} + 43\right )} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} + 1849 \, \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \] Input:

integrate(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x 
^2+43/2)))-1296*x^2)/(2*x^7+43*x^5)/exp(3)/log(x^2+43/2)/log(log(x^2+43/2) 
)/log(log(log(x^2+43/2)))^5,x, algorithm="giac")
 

Output:

324*e^(-3)/((2*x^2 + 43)^2*log(log(log(x^2 + 43/2)))^4 - 86*(2*x^2 + 43)*l 
og(log(log(x^2 + 43/2)))^4 + 1849*log(log(log(x^2 + 43/2)))^4)
 

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81\,{\mathrm {e}}^{-3}}{x^4\,{\ln \left (\ln \left (\ln \left (x^2+\frac {43}{2}\right )\right )\right )}^4} \] Input:

int(-(exp(-3)*(1296*x^2 + log(log(x^2 + 43/2))*log(log(log(x^2 + 43/2)))*l 
og(x^2 + 43/2)*(648*x^2 + 13932)))/(log(log(x^2 + 43/2))*log(log(log(x^2 + 
 43/2)))^5*log(x^2 + 43/2)*(43*x^5 + 2*x^7)),x)
 

Output:

(81*exp(-3))/(x^4*log(log(log(x^2 + 43/2)))^4)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{{\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x^{2}+\frac {43}{2}\right )\right )\right )}^{4} e^{3} x^{4}} \] Input:

int(((-648*x^2-13932)*log(x^2+43/2)*log(log(x^2+43/2))*log(log(log(x^2+43/ 
2)))-1296*x^2)/(2*x^7+43*x^5)/exp(3)/log(x^2+43/2)/log(log(x^2+43/2))/log( 
log(log(x^2+43/2)))^5,x)
 

Output:

81/(log(log(log((2*x**2 + 43)/2)))**4*e**3*x**4)