Integrand size = 95, antiderivative size = 24 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=3+\log \left (2 \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )\right ) \end {dmath*}
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (\log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (e^2+x\right )^2}\right )\right )\right ) \end {dmath*}
Integrate[(-4*x*Log[x] + (E^2 + x)*Log[3/(E^4 + 2*E^2*x + x^2)])/((E^2*x + x^2)*Log[x]*Log[3/(E^4 + 2*E^2*x + x^2)]*Log[(Log[x]*Log[3/(E^4 + 2*E^2*x + x^2)]^2)/5]),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.
\(\displaystyle \int \frac {\left (x+e^2\right ) \log \left (\frac {3}{x^2+2 e^2 x+e^4}\right )-4 x \log (x)}{\left (x^2+e^2 x\right ) \log (x) \log \left (\frac {3}{x^2+2 e^2 x+e^4}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{x^2+2 e^2 x+e^4}\right )\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x+e^2\right ) \log \left (\frac {3}{x^2+2 e^2 x+e^4}\right )-4 x \log (x)}{x \left (x+e^2\right ) \log (x) \log \left (\frac {3}{x^2+2 e^2 x+e^4}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{x^2+2 e^2 x+e^4}\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 x \log (x)-x \log \left (\frac {3}{\left (x+e^2\right )^2}\right )-e^2 \log \left (\frac {3}{\left (x+e^2\right )^2}\right )}{e^2 \left (x+e^2\right ) \log (x) \log \left (\frac {3}{\left (x+e^2\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (x+e^2\right )^2}\right )\right )}+\frac {-4 x \log (x)+x \log \left (\frac {3}{\left (x+e^2\right )^2}\right )+e^2 \log \left (\frac {3}{\left (x+e^2\right )^2}\right )}{e^2 x \log (x) \log \left (\frac {3}{\left (x+e^2\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (x+e^2\right )^2}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x \log (x) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (x+e^2\right )^2}\right )\right )}dx-4 \int \frac {1}{\left (x+e^2\right ) \log \left (\frac {3}{\left (x+e^2\right )^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{\left (x+e^2\right )^2}\right )\right )}dx\) |
Int[(-4*x*Log[x] + (E^2 + x)*Log[3/(E^4 + 2*E^2*x + x^2)])/((E^2*x + x^2)* Log[x]*Log[3/(E^4 + 2*E^2*x + x^2)]*Log[(Log[x]*Log[3/(E^4 + 2*E^2*x + x^2 )]^2)/5]),x]
3.10.80.3.1 Defintions of rubi rules used
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])
Time = 94.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(\ln \left (\ln \left (\frac {\ln \left (\frac {3}{2 \,{\mathrm e}^{2} x +x^{2}+{\mathrm e}^{4}}\right )^{2} \ln \left (x \right )}{5}\right )\right )\) | \(27\) |
risch | \(\text {Expression too large to display}\) | \(1123\) |
int((-4*x*ln(x)+(x+exp(2))*ln(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2)*x+x^2) /ln(3/(exp(2)^2+2*exp(2)*x+x^2))/ln(x)/ln(1/5*ln(3/(exp(2)^2+2*exp(2)*x+x^ 2))^2*ln(x)),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (\log \left (\frac {1}{5} \, \log \left (x\right ) \log \left (\frac {3}{x^{2} + 2 \, x e^{2} + e^{4}}\right )^{2}\right )\right ) \end {dmath*}
integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2 )*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2 *exp(2)*x+x^2))^2*log(x)),x, algorithm=\
Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log {\left (\log {\left (\frac {\log {\left (x \right )} \log {\left (\frac {3}{x^{2} + 2 x e^{2} + e^{4}} \right )}^{2}}{5} \right )} \right )} \end {dmath*}
integrate((-4*x*ln(x)+(x+exp(2))*ln(3/(exp(2)**2+2*exp(2)*x+x**2)))/(exp(2 )*x+x**2)/ln(3/(exp(2)**2+2*exp(2)*x+x**2))/ln(x)/ln(1/5*ln(3/(exp(2)**2+2 *exp(2)*x+x**2))**2*ln(x)),x)
Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (-\frac {1}{2} \, \log \left (5\right ) + \log \left (-\log \left (3\right ) + 2 \, \log \left (x + e^{2}\right )\right ) + \frac {1}{2} \, \log \left (\log \left (x\right )\right )\right ) \end {dmath*}
integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2 )*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2 *exp(2)*x+x^2))^2*log(x)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (21) = 42\).
Time = 1.89 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\log \left (-\log \left (5\right ) + \log \left (\log \left (3\right )^{2} \log \left (x\right ) - 2 \, \log \left (3\right ) \log \left (x^{2} + 2 \, x e^{2} + e^{4}\right ) \log \left (x\right ) + \log \left (x^{2} + 2 \, x e^{2} + e^{4}\right )^{2} \log \left (x\right )\right )\right ) \end {dmath*}
integrate((-4*x*log(x)+(x+exp(2))*log(3/(exp(2)^2+2*exp(2)*x+x^2)))/(exp(2 )*x+x^2)/log(3/(exp(2)^2+2*exp(2)*x+x^2))/log(x)/log(1/5*log(3/(exp(2)^2+2 *exp(2)*x+x^2))^2*log(x)),x, algorithm=\
log(-log(5) + log(log(3)^2*log(x) - 2*log(3)*log(x^2 + 2*x*e^2 + e^4)*log( x) + log(x^2 + 2*x*e^2 + e^4)^2*log(x)))
Time = 19.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {-4 x \log (x)+\left (e^2+x\right ) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right )}{\left (e^2 x+x^2\right ) \log (x) \log \left (\frac {3}{e^4+2 e^2 x+x^2}\right ) \log \left (\frac {1}{5} \log (x) \log ^2\left (\frac {3}{e^4+2 e^2 x+x^2}\right )\right )} \, dx=\ln \left (\ln \left (\frac {{\ln \left (\frac {3}{x^2+2\,{\mathrm {e}}^2\,x+{\mathrm {e}}^4}\right )}^2\,\ln \left (x\right )}{5}\right )\right ) \end {dmath*}