Integrand size = 90, antiderivative size = 27 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\log \left (\frac {8 \left (e^4+\left (-x+\frac {\log \left (\log ^2(x)\right )}{x}\right )^2\right )^4}{x^2}\right ) \end {dmath*}
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=-2 \left (5 \log (x)-2 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )\right ) \end {dmath*}
Integrate[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log [Log[x]^2] - 10*Log[x]*Log[Log[x]^2]^2)/((E^4*x^3 + x^5)*Log[x] - 2*x^3*Lo g[x]*Log[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]
Time = 0.96 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {-16 x^2+\left (4 x^2 \log (x)+16\right ) \log \left (\log ^2(x)\right )+\left (6 x^4-2 e^4 x^2\right ) \log (x)-10 \log (x) \log ^2\left (\log ^2(x)\right )}{-2 x^3 \log (x) \log \left (\log ^2(x)\right )+\left (x^5+e^4 x^3\right ) \log (x)+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-16 x^2+\left (4 x^2 \log (x)+16\right ) \log \left (\log ^2(x)\right )+\left (6 x^4-2 e^4 x^2\right ) \log (x)-10 \log (x) \log ^2\left (\log ^2(x)\right )}{x \log (x) \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {8 \left (2 x^4 \log (x)-2 x^2-2 x^2 \log (x) \log \left (\log ^2(x)\right )+e^4 x^2 \log (x)+2 \log \left (\log ^2(x)\right )\right )}{x \log (x) \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}-\frac {10}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \log \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )-10 \log (x)\) |
Int[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log[Log[x ]^2] - 10*Log[x]*Log[Log[x]^2]^2)/((E^4*x^3 + x^5)*Log[x] - 2*x^3*Log[x]*L og[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]
3.8.32.3.1 Defintions of rubi rules used
Time = 8.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(-10 \ln \left (x \right )+4 \ln \left (x^{4}+x^{2} {\mathrm e}^{4}-2 x^{2} \ln \left (\ln \left (x \right )^{2}\right )+\ln \left (\ln \left (x \right )^{2}\right )^{2}\right )\) | \(36\) |
risch | \(-10 \ln \left (x \right )+4 \ln \left (-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{4} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{16}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{3} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}-\frac {3 \pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{4}}{8}+\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{5}}{4}-\frac {\pi ^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{6}}{16}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{4}+\frac {x^{4}}{4}+\frac {x^{2} {\mathrm e}^{4}}{4}+\left (-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )}{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}-\frac {i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}}{2}-x^{2}\right ) \ln \left (\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )^{2}\right )\) | \(259\) |
int((-10*ln(x)*ln(ln(x)^2)^2+(4*x^2*ln(x)+16)*ln(ln(x)^2)+(-2*x^2*exp(4)+6 *x^4)*ln(x)-16*x^2)/(x*ln(x)*ln(ln(x)^2)^2-2*x^3*ln(x)*ln(ln(x)^2)+(x^3*ex p(4)+x^5)*ln(x)),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \end {dmath*}
integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=- 10 \log {\left (x \right )} + 4 \log {\left (x^{4} - 2 x^{2} \log {\left (\log {\left (x \right )}^{2} \right )} + x^{2} e^{4} + \log {\left (\log {\left (x \right )}^{2} \right )}^{2} \right )} \end {dmath*}
integrate((-10*ln(x)*ln(ln(x)**2)**2+(4*x**2*ln(x)+16)*ln(ln(x)**2)+(-2*x* *2*exp(4)+6*x**4)*ln(x)-16*x**2)/(x*ln(x)*ln(ln(x)**2)**2-2*x**3*ln(x)*ln( ln(x)**2)+(x**3*exp(4)+x**5)*ln(x)),x)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (\frac {1}{4} \, x^{4} + \frac {1}{4} \, x^{2} e^{4} - x^{2} \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right ) - 10 \, \log \left (x\right ) \end {dmath*}
integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm=\
Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \left (x\right )^{2}\right ) + \log \left (\log \left (x\right )^{2}\right )^{2}\right ) - 10 \, \log \left (x\right ) \end {dmath*}
integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2* x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*log(x)*log(log(x)^2)^2-2*x^3*log(x)*lo g(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm=\
Timed out. \begin {dmath*} \int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{\left (e^4 x^3+x^5\right ) \log (x)-2 x^3 \log (x) \log \left (\log ^2(x)\right )+x \log (x) \log ^2\left (\log ^2(x)\right )} \, dx=\int -\frac {10\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )+\ln \left (x\right )\,\left (2\,x^2\,{\mathrm {e}}^4-6\,x^4\right )-\ln \left ({\ln \left (x\right )}^2\right )\,\left (4\,x^2\,\ln \left (x\right )+16\right )+16\,x^2}{\ln \left (x\right )\,\left (x^5+{\mathrm {e}}^4\,x^3\right )+x\,{\ln \left ({\ln \left (x\right )}^2\right )}^2\,\ln \left (x\right )-2\,x^3\,\ln \left ({\ln \left (x\right )}^2\right )\,\ln \left (x\right )} \,d x \end {dmath*}
int(-(10*log(log(x)^2)^2*log(x) + log(x)*(2*x^2*exp(4) - 6*x^4) - log(log( x)^2)*(4*x^2*log(x) + 16) + 16*x^2)/(log(x)*(x^3*exp(4) + x^5) + x*log(log (x)^2)^2*log(x) - 2*x^3*log(log(x)^2)*log(x)),x)