Integrand size = 68, antiderivative size = 17 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\frac {4}{x^4}}}{\log (\log (6+2 x))} \] Output:
exp(2/x^4)^2/ln(ln(2*x+6))
Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\frac {4}{x^4}}}{\log (\log (2 (3+x)))} \] Input:
Integrate[(-(E^(4/x^4)*x^5) + E^(4/x^4)*(-48 - 16*x)*Log[6 + 2*x]*Log[Log[ 6 + 2*x]])/((3*x^5 + x^6)*Log[6 + 2*x]*Log[Log[6 + 2*x]]^2),x]
Output:
E^(4/x^4)/Log[Log[2*(3 + 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.
\(\displaystyle \int \frac {e^{\frac {4}{x^4}} (-16 x-48) \log (2 x+6) \log (\log (2 x+6))-e^{\frac {4}{x^4}} x^5}{\left (x^6+3 x^5\right ) \log (2 x+6) \log ^2(\log (2 x+6))} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {4}{x^4}} (-16 x-48) \log (2 x+6) \log (\log (2 x+6))-e^{\frac {4}{x^4}} x^5}{x^5 (x+3) \log (2 x+6) \log ^2(\log (2 x+6))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {4}{x^4}} \left (-x^5-16 x \log (2 (x+3)) \log (\log (2 (x+3)))-48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{243 x \log (2 x+6) \log ^2(\log (2 x+6))}+\frac {e^{\frac {4}{x^4}} \left (-x^5-16 x \log (2 (x+3)) \log (\log (2 (x+3)))-48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{3 x^5 \log (2 x+6) \log ^2(\log (2 x+6))}+\frac {e^{\frac {4}{x^4}} \left (x^5+16 x \log (2 (x+3)) \log (\log (2 (x+3)))+48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{243 (x+3) \log (2 x+6) \log ^2(\log (2 x+6))}+\frac {e^{\frac {4}{x^4}} \left (x^5+16 x \log (2 (x+3)) \log (\log (2 (x+3)))+48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{9 x^4 \log (2 x+6) \log ^2(\log (2 x+6))}+\frac {e^{\frac {4}{x^4}} \left (-x^5-16 x \log (2 (x+3)) \log (\log (2 (x+3)))-48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{27 x^3 \log (2 x+6) \log ^2(\log (2 x+6))}+\frac {e^{\frac {4}{x^4}} \left (x^5+16 x \log (2 (x+3)) \log (\log (2 (x+3)))+48 \log (2 (x+3)) \log (\log (2 (x+3)))\right )}{81 x^2 \log (2 x+6) \log ^2(\log (2 x+6))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{\frac {4}{x^4}}}{(-x-3) \log (2 x+6) \log ^2(\log (2 x+6))}dx+\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(-x-3) \log (\log (2 x+6))}dx+\frac {16}{81} \int \frac {e^{\frac {4}{x^4}}}{(x+3) \log (\log (2 x+6))}dx-16 \int \frac {e^{\frac {4}{x^4}}}{x^5 \log (\log (2 x+6))}dx\) |
Input:
Int[(-(E^(4/x^4)*x^5) + E^(4/x^4)*(-48 - 16*x)*Log[6 + 2*x]*Log[Log[6 + 2* x]])/((3*x^5 + x^6)*Log[6 + 2*x]*Log[Log[6 + 2*x]]^2),x]
Output:
$Aborted
Time = 22.70 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {{\mathrm e}^{\frac {4}{x^{4}}}}{\ln \left (\ln \left (2 x +6\right )\right )}\) | \(17\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {4}{x^{4}}}}{\ln \left (\ln \left (2 x +6\right )\right )}\) | \(19\) |
Input:
int(((-16*x-48)*exp(2/x^4)^2*ln(2*x+6)*ln(ln(2*x+6))-x^5*exp(2/x^4)^2)/(x^ 6+3*x^5)/ln(2*x+6)/ln(ln(2*x+6))^2,x,method=_RETURNVERBOSE)
Output:
exp(4/x^4)/ln(ln(2*x+6))
Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \left (2 \, x + 6\right )\right )} \] Input:
integrate(((-16*x-48)*exp(2/x^4)^2*log(2*x+6)*log(log(2*x+6))-x^5*exp(2/x^ 4)^2)/(x^6+3*x^5)/log(2*x+6)/log(log(2*x+6))^2,x, algorithm="fricas")
Output:
e^(4/x^4)/log(log(2*x + 6))
Time = 0.17 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\frac {4}{x^{4}}}}{\log {\left (\log {\left (2 x + 6 \right )} \right )}} \] Input:
integrate(((-16*x-48)*exp(2/x**4)**2*ln(2*x+6)*ln(ln(2*x+6))-x**5*exp(2/x* *4)**2)/(x**6+3*x**5)/ln(2*x+6)/ln(ln(2*x+6))**2,x)
Output:
exp(4/x**4)/log(log(2*x + 6))
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \left (2\right ) + \log \left (x + 3\right )\right )} \] Input:
integrate(((-16*x-48)*exp(2/x^4)^2*log(2*x+6)*log(log(2*x+6))-x^5*exp(2/x^ 4)^2)/(x^6+3*x^5)/log(2*x+6)/log(log(2*x+6))^2,x, algorithm="maxima")
Output:
e^(4/x^4)/log(log(2) + log(x + 3))
Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\left (\frac {4}{x^{4}}\right )}}{\log \left (\log \left (2 \, x + 6\right )\right )} \] Input:
integrate(((-16*x-48)*exp(2/x^4)^2*log(2*x+6)*log(log(2*x+6))-x^5*exp(2/x^ 4)^2)/(x^6+3*x^5)/log(2*x+6)/log(log(2*x+6))^2,x, algorithm="giac")
Output:
e^(4/x^4)/log(log(2*x + 6))
Timed out. \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\int -\frac {x^5\,{\mathrm {e}}^{\frac {4}{x^4}}+\ln \left (\ln \left (2\,x+6\right )\right )\,{\mathrm {e}}^{\frac {4}{x^4}}\,\ln \left (2\,x+6\right )\,\left (16\,x+48\right )}{{\ln \left (\ln \left (2\,x+6\right )\right )}^2\,\ln \left (2\,x+6\right )\,\left (x^6+3\,x^5\right )} \,d x \] Input:
int(-(x^5*exp(4/x^4) + log(log(2*x + 6))*exp(4/x^4)*log(2*x + 6)*(16*x + 4 8))/(log(log(2*x + 6))^2*log(2*x + 6)*(3*x^5 + x^6)),x)
Output:
int(-(x^5*exp(4/x^4) + log(log(2*x + 6))*exp(4/x^4)*log(2*x + 6)*(16*x + 4 8))/(log(log(2*x + 6))^2*log(2*x + 6)*(3*x^5 + x^6)), x)
Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{\frac {4}{x^4}} x^5+e^{\frac {4}{x^4}} (-48-16 x) \log (6+2 x) \log (\log (6+2 x))}{\left (3 x^5+x^6\right ) \log (6+2 x) \log ^2(\log (6+2 x))} \, dx=\frac {e^{\frac {4}{x^{4}}}}{\mathrm {log}\left (\mathrm {log}\left (2 x +6\right )\right )} \] Input:
int(((-16*x-48)*exp(2/x^4)^2*log(2*x+6)*log(log(2*x+6))-x^5*exp(2/x^4)^2)/ (x^6+3*x^5)/log(2*x+6)/log(log(2*x+6))^2,x)
Output:
e**(4/x**4)/log(log(2*x + 6))