Integrand size = 76, antiderivative size = 23 \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \]
Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \]
Integrate[((x/Log[(3*Log[x/2])/4])^(-1 + E^2)^(-1)*(-1 + Log[x/2]*Log[(3*L og[x/2])/4]))/((-x + E^2*x)*Log[x/2]*Log[(3*Log[x/2])/4]),x]
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.
| \(\displaystyle \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{e^2-1}} \left (\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )-1\right )}{\left (e^2 x-x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{e^2-1}} \left (\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )-1\right )}{\left (e^2-1\right ) x \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int -\frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (1-\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{x \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}dx}{1-e^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (1-\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{x \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}dx}{1-e^2}\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {x^{\frac {1}{1-e^2}} \left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{e^2-1}} \sqrt [e^2-1]{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \int \frac {x^{-1+\frac {1}{-1+e^2}} \log ^{-1+\frac {1}{1-e^2}}\left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right ) \left (1-\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\log \left (\frac {x}{2}\right )}dx}{1-e^2}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {x^{\frac {1}{1-e^2}} \left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{e^2-1}} \sqrt [e^2-1]{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \int \left (\frac {x^{-1+\frac {1}{-1+e^2}} \log ^{-1+\frac {1}{1-e^2}}\left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}{\log \left (\frac {x}{2}\right )}-x^{-1+\frac {1}{-1+e^2}} \sqrt [1-e^2]{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )dx}{1-e^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{\frac {1}{1-e^2}} \left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{e^2-1}} \sqrt [e^2-1]{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \left (\int \frac {x^{-1+\frac {1}{-1+e^2}} \log ^{-1+\frac {1}{1-e^2}}\left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}{\log \left (\frac {x}{2}\right )}dx-\int x^{-1+\frac {1}{-1+e^2}} \sqrt [1-e^2]{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}dx\right )}{1-e^2}\) |
Int[((x/Log[(3*Log[x/2])/4])^(-1 + E^2)^(-1)*(-1 + Log[x/2]*Log[(3*Log[x/2 ])/4]))/((-x + E^2*x)*Log[x/2]*Log[(3*Log[x/2])/4]),x]
3.15.8.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
Leaf count of result is larger than twice the leaf count of optimal. \(53\) vs. \(2(20)=40\).
Time = 9.56 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35
| method | result | size |
| parallelrisch | \(\frac {{\mathrm e}^{\frac {\ln \left (\frac {x}{\ln \left (\frac {3 \ln \left (\frac {x}{2}\right )}{4}\right )}\right )}{{\mathrm e}^{2}-1}} {\mathrm e}^{2}-{\mathrm e}^{\frac {\ln \left (\frac {x}{\ln \left (\frac {3 \ln \left (\frac {x}{2}\right )}{4}\right )}\right )}{{\mathrm e}^{2}-1}}}{{\mathrm e}^{2}-1}\) | \(54\) |
int((ln(1/2*x)*ln(3/4*ln(1/2*x))-1)*exp(ln(x/ln(3/4*ln(1/2*x)))/(exp(2)-1) )/(exp(2)*x-x)/ln(1/2*x)/ln(3/4*ln(1/2*x)),x,method=_RETURNVERBOSE)
(exp(ln(x/ln(3/4*ln(1/2*x)))/(exp(2)-1))*exp(2)-exp(ln(x/ln(3/4*ln(1/2*x)) )/(exp(2)-1)))/(exp(2)-1)
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )} \]
integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)) )/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x)/log(3/4*log(1/2*x)),x, algorithm=\
Timed out. \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\text {Timed out} \]
integrate((ln(1/2*x)*ln(3/4*ln(1/2*x))-1)*exp(ln(x/ln(3/4*ln(1/2*x)))/(exp (2)-1))/(exp(2)*x-x)/ln(1/2*x)/ln(3/4*ln(1/2*x)),x)
\[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\int { \frac {{\left (\log \left (\frac {1}{2} \, x\right ) \log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right ) - 1\right )} \left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )}}{{\left (x e^{2} - x\right )} \log \left (\frac {1}{2} \, x\right ) \log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )} \,d x } \]
integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)) )/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x)/log(3/4*log(1/2*x)),x, algorithm=\
integrate((log(1/2*x)*log(3/4*log(1/2*x)) - 1)*(x/log(3/4*log(1/2*x)))^(1/ (e^2 - 1))/((x*e^2 - x)*log(1/2*x)*log(3/4*log(1/2*x))), x)
Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx=\left (\frac {x}{\log \left (\frac {3}{4} \, \log \left (\frac {1}{2} \, x\right )\right )}\right )^{\left (\frac {1}{e^{2} - 1}\right )} \]
integrate((log(1/2*x)*log(3/4*log(1/2*x))-1)*exp(log(x/log(3/4*log(1/2*x)) )/(exp(2)-1))/(exp(2)*x-x)/log(1/2*x)/log(3/4*log(1/2*x)),x, algorithm=\
Time = 11.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {\left (\frac {x}{\log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )}\right )^{\frac {1}{-1+e^2}} \left (-1+\log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )\right )}{\left (-x+e^2 x\right ) \log \left (\frac {x}{2}\right ) \log \left (\frac {3}{4} \log \left (\frac {x}{2}\right )\right )} \, dx={\left (\frac {x}{\ln \left (\frac {3\,\ln \left (\frac {x}{2}\right )}{4}\right )}\right )}^{\frac {1}{{\mathrm {e}}^2-1}} \]