Integrand size = 111, antiderivative size = 27 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \]
Time = 4.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {1}{4} e^{\frac {(x+\log (4))^2}{\log ^2\left (\frac {2 x^2}{5}\right )}}} \]
Integrate[(E^(E^((x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)/4 + (x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)*(-6*x^2 - 12*x*Log[4] - 6*Log[4 ]^2 + (3*x^2 + 3*x*Log[4])*Log[(2*x^2)/5]))/(2*x*Log[(2*x^2)/5]^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.
| \(\displaystyle \int \frac {\left (-6 x^2+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )-12 x \log (4)-6 \log ^2(4)\right ) \exp \left (\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {3 \exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) \left (2 x^2+4 \log (4) x-\left (x^2+\log (4) x\right ) \log \left (\frac {2 x^2}{5}\right )+2 \log ^2(4)\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3}{2} \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) \left (2 x^2+4 \log (4) x-\left (x^2+\log (4) x\right ) \log \left (\frac {2 x^2}{5}\right )+2 \log ^2(4)\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {3}{2} \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) (x+\log (4)) \left (-\log \left (\frac {2 x^2}{5}\right ) x+2 x+2 \log (4)\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3}{2} \int \left (\frac {2 \exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) (x+\log (4))^2}{x \log ^3\left (\frac {2 x^2}{5}\right )}+\frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) (-x-\log (4))}{\log ^2\left (\frac {2 x^2}{5}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3}{2} \left (-\log (4) \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right )}{\log ^2\left (\frac {2 x^2}{5}\right )}dx-\int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) x}{\log ^2\left (\frac {2 x^2}{5}\right )}dx+4 \log (4) \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right )}{\log ^3\left (\frac {2 x^2}{5}\right )}dx+2 \log ^2(4) \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right )}{x \log ^3\left (\frac {2 x^2}{5}\right )}dx+2 \int \frac {\exp \left (\frac {x^2+2 \log (4) x+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}+4^{\frac {2 x}{\log ^2\left (\frac {2 x^2}{5}\right )}-1} e^{\frac {x^2+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}\right ) x}{\log ^3\left (\frac {2 x^2}{5}\right )}dx\right )\) |
Int[(E^(E^((x^2 + 2*x*Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)/4 + (x^2 + 2*x* Log[4] + Log[4]^2)/Log[(2*x^2)/5]^2)*(-6*x^2 - 12*x*Log[4] - 6*Log[4]^2 + (3*x^2 + 3*x*Log[4])*Log[(2*x^2)/5]))/(2*x*Log[(2*x^2)/5]^3),x]
3.6.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 24.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
| method | result | size |
| risch | \(3 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right )^{2}}{\ln \left (\frac {2 x^{2}}{5}\right )^{2}}}}{4}}\) | \(24\) |
| parallelrisch | \(3 \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {4 \ln \left (2\right )^{2}+4 x \ln \left (2\right )+x^{2}}{\ln \left (\frac {2 x^{2}}{5}\right )^{2}}}}{4}}\) | \(31\) |
int(1/2*((6*x*ln(2)+3*x^2)*ln(2/5*x^2)-24*ln(2)^2-24*x*ln(2)-6*x^2)*exp((4 *ln(2)^2+4*x*ln(2)+x^2)/ln(2/5*x^2)^2)*exp(1/4*exp((4*ln(2)^2+4*x*ln(2)+x^ 2)/ln(2/5*x^2)^2))/x/ln(2/5*x^2)^3,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (23) = 46\).
Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.33 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 \, e^{\left (\frac {e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \log \left (\frac {2}{5} \, x^{2}\right )^{2} + 4 \, x^{2} + 16 \, x \log \left (2\right ) + 16 \, \log \left (2\right )^{2}}{4 \, \log \left (\frac {2}{5} \, x^{2}\right )^{2}} - \frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )} \]
integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x ^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^ 2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm=\
3*e^(1/4*(e^((x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2)*log(2/5*x^2)^ 2 + 4*x^2 + 16*x*log(2) + 16*log(2)^2)/log(2/5*x^2)^2 - (x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2)
Time = 0.91 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 e^{\frac {e^{\frac {x^{2} + 4 x \log {\left (2 \right )} + 4 \log {\left (2 \right )}^{2}}{\log {\left (\frac {2 x^{2}}{5} \right )}^{2}}}}{4}} \]
integrate(1/2*((6*x*ln(2)+3*x**2)*ln(2/5*x**2)-24*ln(2)**2-24*x*ln(2)-6*x* *2)*exp((4*ln(2)**2+4*x*ln(2)+x**2)/ln(2/5*x**2)**2)*exp(1/4*exp((4*ln(2)* *2+4*x*ln(2)+x**2)/ln(2/5*x**2)**2))/x/ln(2/5*x**2)**3,x)
Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (23) = 46\).
Time = 0.54 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.59 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3 \, e^{\left (\frac {1}{4} \, e^{\left (\frac {x^{2}}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}} + \frac {4 \, x \log \left (2\right )}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}} + \frac {4 \, \log \left (2\right )^{2}}{\log \left (5\right )^{2} - 2 \, \log \left (5\right ) \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (x\right ) + 4 \, \log \left (x\right )^{2}}\right )}\right )} \]
integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x ^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^ 2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm=\
3*e^(1/4*e^(x^2/(log(5)^2 - 2*log(5)*log(2) + log(2)^2 - 4*(log(5) - log(2 ))*log(x) + 4*log(x)^2) + 4*x*log(2)/(log(5)^2 - 2*log(5)*log(2) + log(2)^ 2 - 4*(log(5) - log(2))*log(x) + 4*log(x)^2) + 4*log(2)^2/(log(5)^2 - 2*lo g(5)*log(2) + log(2)^2 - 4*(log(5) - log(2))*log(x) + 4*log(x)^2)))
\[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=\int { -\frac {3 \, {\left (2 \, x^{2} + 8 \, x \log \left (2\right ) + 8 \, \log \left (2\right )^{2} - {\left (x^{2} + 2 \, x \log \left (2\right )\right )} \log \left (\frac {2}{5} \, x^{2}\right )\right )} e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}} + \frac {1}{4} \, e^{\left (\frac {x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left (\frac {2}{5} \, x^{2}\right )^{2}}\right )}\right )}}{2 \, x \log \left (\frac {2}{5} \, x^{2}\right )^{3}} \,d x } \]
integrate(1/2*((6*x*log(2)+3*x^2)*log(2/5*x^2)-24*log(2)^2-24*x*log(2)-6*x ^2)*exp((4*log(2)^2+4*x*log(2)+x^2)/log(2/5*x^2)^2)*exp(1/4*exp((4*log(2)^ 2+4*x*log(2)+x^2)/log(2/5*x^2)^2))/x/log(2/5*x^2)^3,x, algorithm=\
integrate(-3/2*(2*x^2 + 8*x*log(2) + 8*log(2)^2 - (x^2 + 2*x*log(2))*log(2 /5*x^2))*e^((x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2 + 1/4*e^((x^2 + 4*x*log(2) + 4*log(2)^2)/log(2/5*x^2)^2))/(x*log(2/5*x^2)^3), x)
Time = 13.73 (sec) , antiderivative size = 139, normalized size of antiderivative = 5.15 \[ \int \frac {e^{\frac {1}{4} e^{\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}}+\frac {x^2+2 x \log (4)+\log ^2(4)}{\log ^2\left (\frac {2 x^2}{5}\right )}} \left (-6 x^2-12 x \log (4)-6 \log ^2(4)+\left (3 x^2+3 x \log (4)\right ) \log \left (\frac {2 x^2}{5}\right )\right )}{2 x \log ^3\left (\frac {2 x^2}{5}\right )} \, dx=3\,{\mathrm {e}}^{\frac {2^{\frac {4\,x}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {4\,{\ln \left (2\right )}^2}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}\,{\mathrm {e}}^{\frac {x^2}{2\,\ln \left (x^2\right )\,\ln \left (2\right )-2\,\ln \left (x^2\right )\,\ln \left (5\right )-2\,\ln \left (2\right )\,\ln \left (5\right )+{\ln \left (x^2\right )}^2+{\ln \left (2\right )}^2+{\ln \left (5\right )}^2}}}{4}} \]
int(-(exp(exp((4*x*log(2) + 4*log(2)^2 + x^2)/log((2*x^2)/5)^2)/4)*exp((4* x*log(2) + 4*log(2)^2 + x^2)/log((2*x^2)/5)^2)*(24*x*log(2) + 24*log(2)^2 + 6*x^2 - log((2*x^2)/5)*(6*x*log(2) + 3*x^2)))/(2*x*log((2*x^2)/5)^3),x)
3*exp((2^((4*x)/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2)^2 + log(2)^2 + log(5)^2))*exp((4*log(2)^2)/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2)^2 + log(2)^2 + log(5)^2))*e xp(x^2/(2*log(x^2)*log(2) - 2*log(x^2)*log(5) - 2*log(2)*log(5) + log(x^2) ^2 + log(2)^2 + log(5)^2)))/4)