Integrand size = 54, antiderivative size = 22 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=16-\frac {1}{4} \log \left (\log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )\right ) \]
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=-\frac {1}{4} \log \left (\log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )\right ) \]
Integrate[1/(((-8*x - 4*E^5*x)*Log[7/x] + 4*x*Log[7/x]*Log[Log[7/x]])*Log[ 2 + E^5 - Log[Log[7/x]]]),x]
Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3039, 27, 3039, 2837, 2739, 14}
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.
\(\displaystyle \int \frac {1}{\left (\left (-4 e^5 x-8 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\log \left (\frac {7}{x}\right )\right ) \log \left (\frac {7}{x}\right )\right ) \log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )} \, dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -\int -\frac {1}{4 \log \left (\frac {7}{x}\right ) \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right ) \log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )}d\log \left (\frac {7}{x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\log \left (\frac {7}{x}\right ) \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right ) \log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )}d\log \left (\frac {7}{x}\right )\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle \frac {1}{4} \int \frac {1}{\left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right ) \log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )}d\log \left (\log \left (\frac {7}{x}\right )\right )\) |
\(\Big \downarrow \) 2837 |
\(\displaystyle -\frac {1}{4} \int \frac {1}{\left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right ) \log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )}d\left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )\) |
\(\Big \downarrow \) 2739 |
\(\displaystyle -\frac {1}{4} \int \frac {1}{\log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )}d\log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )\) |
\(\Big \downarrow \) 14 |
\(\displaystyle -\frac {1}{4} \log \left (\log \left (-\log \left (\log \left (\frac {7}{x}\right )\right )+e^5+2\right )\right )\) |
Int[1/(((-8*x - 4*E^5*x)*Log[7/x] + 4*x*Log[7/x]*Log[Log[7/x]])*Log[2 + E^ 5 - Log[Log[7/x]]]),x]
3.6.67.3.1 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[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1/( b*n) Subst[Int[x^p, x], x, a + b*Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p} , x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[1/e Subst[Int[(f*(x/d))^q*(a + b*Log[c*x ^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x] && EqQ[e*f - d*g, 0]
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]
\[\int \frac {1}{\left (4 x \ln \left (\frac {7}{x}\right ) \ln \left (\ln \left (\frac {7}{x}\right )\right )+\left (-4 x \,{\mathrm e}^{5}-8 x \right ) \ln \left (\frac {7}{x}\right )\right ) \ln \left (-\ln \left (\ln \left (\frac {7}{x}\right )\right )+{\mathrm e}^{5}+2\right )}d x\]
Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=-\frac {1}{4} \, \log \left (\log \left (e^{5} - \log \left (\log \left (\frac {7}{x}\right )\right ) + 2\right )\right ) \]
integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-l og(log(7/x))+exp(5)+2),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=- \frac {\log {\left (\log {\left (- \log {\left (\log {\left (\frac {7}{x} \right )} \right )} + 2 + e^{5} \right )} \right )}}{4} \]
Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=-\frac {1}{4} \, \log \left (\log \left (e^{5} - \log \left (\log \left (7\right ) - \log \left (x\right )\right ) + 2\right )\right ) \]
integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-l og(log(7/x))+exp(5)+2),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=-\frac {1}{4} \, \log \left ({\left | \log \left (e^{5} - \log \left (\log \left (\frac {7}{x}\right )\right ) + 2\right ) \right |}\right ) \]
integrate(1/(4*x*log(7/x)*log(log(7/x))+(-4*x*exp(5)-8*x)*log(7/x))/log(-l og(log(7/x))+exp(5)+2),x, algorithm=\
Time = 10.88 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (\left (-8 x-4 e^5 x\right ) \log \left (\frac {7}{x}\right )+4 x \log \left (\frac {7}{x}\right ) \log \left (\log \left (\frac {7}{x}\right )\right )\right ) \log \left (2+e^5-\log \left (\log \left (\frac {7}{x}\right )\right )\right )} \, dx=-\frac {\ln \left (\ln \left ({\mathrm {e}}^5-\ln \left (\ln \left (\frac {7}{x}\right )\right )+2\right )\right )}{4} \]