Integrand size = 115, antiderivative size = 20 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \]
Time = 0.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {43}{2}+x^2\right )\right )\right )} \]
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]
Time = 0.69 (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 |
\(\displaystyle \frac {81}{e^3 x^4 \log ^4\left (\log \left (\log \left (\frac {1}{2} \left (2 x^2+43\right )\right )\right )\right )}\) |
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]
3.12.77.3.1 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[(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])
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]
Time = 70.92 (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 | \(\text {Expression too large to display}\) | \(5536\) |
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)
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \]
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=\
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81}{x^{4} e^{3} \log {\left (\log {\left (\log {\left (x^{2} + \frac {43}{2} \right )} \right )} \right )}^{4}} \]
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)
Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81 \, e^{\left (-3\right )}}{x^{4} \log \left (\log \left (-\log \left (2\right ) + \log \left (2 \, x^{2} + 43\right )\right )\right )^{4}} \]
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=\
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).
Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {324 \, e^{\left (-3\right )}}{{\left (2 \, x^{2} + 43\right )}^{2} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} - 86 \, {\left (2 \, x^{2} + 43\right )} \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4} + 1849 \, \log \left (\log \left (\log \left (x^{2} + \frac {43}{2}\right )\right )\right )^{4}} \]
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=\
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)
Time = 14.80 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1296 x^2+\left (-13932-648 x^2\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )}{e^3 \left (43 x^5+2 x^7\right ) \log \left (\frac {1}{2} \left (43+2 x^2\right )\right ) \log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right ) \log ^5\left (\log \left (\log \left (\frac {1}{2} \left (43+2 x^2\right )\right )\right )\right )} \, dx=\frac {81\,{\mathrm {e}}^{-3}}{x^4\,{\ln \left (\ln \left (\ln \left (x^2+\frac {43}{2}\right )\right )\right )}^4} \]