Integrand size = 78, antiderivative size = 24 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \]
Time = 3.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \]
Integrate[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x )*(3 + x^2))*Log[x^4/16]*Log[Log[x^4/16]])/(x^2*Log[x^4/16]),x]
Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(24)=48\).
Time = 1.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {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.
\(\displaystyle \int \frac {-8 x^2+4 e^{\frac {x^2-3}{x}} x+\left (e^{\frac {x^2-3}{x}} \left (x^2+3\right )-2 x^2\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+24 x}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x-\frac {3}{x}} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+4 x\right )}{x^2 \log \left (\frac {x^4}{16}\right )}-\frac {2 \left (x \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+4 x-12\right )}{x \log \left (\frac {x^4}{16}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+6 \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{x-\frac {3}{x}} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (\frac {3}{x^2}+1\right ) x^2 \log \left (\frac {x^4}{16}\right )}\) |
Int[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x)*(3 + x^2))*Log[x^4/16]*Log[Log[x^4/16]])/(x^2*Log[x^4/16]),x]
6*Log[Log[x^4/16]] - 2*x*Log[Log[x^4/16]] + (E^(-3/x + x)*(3*Log[x^4/16]*L og[Log[x^4/16]] + x^2*Log[x^4/16]*Log[Log[x^4/16]]))/((1 + 3/x^2)*x^2*Log[ x^4/16])
3.2.71.3.1 Defintions of rubi rules used
Time = 3.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62
method | result | size |
parallelrisch | \(-2 \ln \left (\ln \left (\frac {x^{4}}{16}\right )\right ) x +\ln \left (\ln \left (\frac {x^{4}}{16}\right )\right ) {\mathrm e}^{\frac {x^{2}-3}{x}}+6 \ln \left (\ln \left (\frac {x^{4}}{16}\right )\right )\) | \(39\) |
risch | \(\left (-2 x +{\mathrm e}^{\frac {x^{2}-3}{x}}\right ) \ln \left (-4 \ln \left (2\right )+4 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{3}\right )\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}\right )+6 \ln \left (\ln \left (x \right )-\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}+\pi \operatorname {csgn}\left (i x^{3}\right )^{3}-\pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}+\pi \operatorname {csgn}\left (i x^{4}\right )^{3}-8 i \ln \left (2\right )\right )}{8}\right )\) | \(336\) |
int((((x^2+3)*exp((x^2-3)/x)-2*x^2)*ln(1/16*x^4)*ln(ln(1/16*x^4))+4*x*exp( (x^2-3)/x)-8*x^2+24*x)/x^2/ln(1/16*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=-{\left (2 \, x - e^{\left (\frac {x^{2} - 3}{x}\right )} - 6\right )} \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) \]
integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) +4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm=\
Time = 5.77 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=- 2 x \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + e^{\frac {x^{2} - 3}{x}} \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + 6 \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} \]
integrate((((x**2+3)*exp((x**2-3)/x)-2*x**2)*ln(1/16*x**4)*ln(ln(1/16*x**4 ))+4*x*exp((x**2-3)/x)-8*x**2+24*x)/x**2/ln(1/16*x**4),x)
Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=-4 \, x \log \left (2\right ) + 2 \, e^{\left (x - \frac {3}{x}\right )} \log \left (2\right ) - {\left (2 \, x - e^{\left (x - \frac {3}{x}\right )} - 6\right )} \log \left (-\log \left (2\right ) + \log \left (x\right )\right ) \]
integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) +4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm=\
\[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\int { -\frac {{\left (2 \, x^{2} - {\left (x^{2} + 3\right )} e^{\left (\frac {x^{2} - 3}{x}\right )}\right )} \log \left (\frac {1}{16} \, x^{4}\right ) \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) + 8 \, x^{2} - 4 \, x e^{\left (\frac {x^{2} - 3}{x}\right )} - 24 \, x}{x^{2} \log \left (\frac {1}{16} \, x^{4}\right )} \,d x } \]
integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) +4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm=\
integrate(-((2*x^2 - (x^2 + 3)*e^((x^2 - 3)/x))*log(1/16*x^4)*log(log(1/16 *x^4)) + 8*x^2 - 4*x*e^((x^2 - 3)/x) - 24*x)/(x^2*log(1/16*x^4)), x)
Timed out. \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\int \frac {24\,x+4\,x\,{\mathrm {e}}^{\frac {x^2-3}{x}}-8\,x^2+\ln \left (\ln \left (\frac {x^4}{16}\right )\right )\,\ln \left (\frac {x^4}{16}\right )\,\left ({\mathrm {e}}^{\frac {x^2-3}{x}}\,\left (x^2+3\right )-2\,x^2\right )}{x^2\,\ln \left (\frac {x^4}{16}\right )} \,d x \]
int((24*x + 4*x*exp((x^2 - 3)/x) - 8*x^2 + log(log(x^4/16))*log(x^4/16)*(e xp((x^2 - 3)/x)*(x^2 + 3) - 2*x^2))/(x^2*log(x^4/16)),x)