Integrand size = 152, antiderivative size = 30 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 e^{e^5} x}{\log \left (e^5+x^2+3 \log \left (-2+\frac {x^2}{2}\right )\right )} \]
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 e^{e^5} x}{\log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \]
Integrate[(E^E^5*(4*x^2 - 4*x^4) + (E^E^5*(-8*x^2 + 2*x^4 + E^5*(-8 + 2*x^ 2)) + E^E^5*(-24 + 6*x^2)*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x ^2)/2]])/((-4*x^2 + x^4 + E^5*(-4 + x^2) + (-12 + 3*x^2)*Log[(-4 + x^2)/2] )*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2),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.
| \(\displaystyle \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (6 x^2-24\right ) \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^{e^5} \left (2 x^4-8 x^2+e^5 \left (2 x^2-8\right )\right )\right ) \log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}{\left (x^4-4 x^2+e^5 \left (x^2-4\right )+\left (3 x^2-12\right ) \log \left (\frac {1}{2} \left (x^2-4\right )\right )\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {2 e^{e^5} \left (\log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )-\frac {2 x^2 \left (x^2-1\right )}{\left (x^2-4\right ) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}\right )}{\log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 e^{e^5} \int -\frac {\frac {2 x^2 \left (1-x^2\right )}{\left (4-x^2\right ) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}-\log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}{\log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 e^{e^5} \int \frac {\frac {2 x^2 \left (1-x^2\right )}{\left (4-x^2\right ) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}-\log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}{\log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 e^{e^5} \int \left (\frac {2 (x-1) x^2 (x+1)}{(x-2) (x+2) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}-\frac {1}{\log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 e^{e^5} \left (6 \int \frac {1}{\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx+6 \int \frac {1}{(x-2) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx+2 \int \frac {x^2}{\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx-6 \int \frac {1}{(x+2) \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right ) \log ^2\left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx-\int \frac {1}{\log \left (x^2+3 \log \left (\frac {1}{2} \left (x^2-4\right )\right )+e^5\right )}dx\right )\) |
Int[(E^E^5*(4*x^2 - 4*x^4) + (E^E^5*(-8*x^2 + 2*x^4 + E^5*(-8 + 2*x^2)) + E^E^5*(-24 + 6*x^2)*Log[(-4 + x^2)/2])*Log[E^5 + x^2 + 3*Log[(-4 + x^2)/2] ])/((-4*x^2 + x^4 + E^5*(-4 + x^2) + (-12 + 3*x^2)*Log[(-4 + x^2)/2])*Log[ E^5 + x^2 + 3*Log[(-4 + x^2)/2]]^2),x]
3.6.59.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 15.35 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87
| method | result | size |
| risch | \(\frac {2 \,{\mathrm e}^{{\mathrm e}^{5}} x}{\ln \left (3 \ln \left (\frac {x^{2}}{2}-2\right )+x^{2}+{\mathrm e}^{5}\right )}\) | \(26\) |
| parallelrisch | \(\frac {2 \,{\mathrm e}^{{\mathrm e}^{5}} x}{\ln \left (3 \ln \left (\frac {x^{2}}{2}-2\right )+x^{2}+{\mathrm e}^{5}\right )}\) | \(26\) |
int((((6*x^2-24)*exp(exp(5))*ln(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4-8*x^2)* exp(exp(5)))*ln(3*ln(1/2*x^2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(exp(5)))/(( 3*x^2-12)*ln(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/ln(3*ln(1/2*x^2-2)+x^2+e xp(5))^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 \, x e^{\left (e^{5}\right )}}{\log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right )} \]
integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4- 8*x^2)*exp(exp(5)))*log(3*log(1/2*x^2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(ex p(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/2* x^2-2)+x^2+exp(5))^2,x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 x e^{e^{5}}}{\log {\left (x^{2} + 3 \log {\left (\frac {x^{2}}{2} - 2 \right )} + e^{5} \right )}} \]
integrate((((6*x**2-24)*exp(exp(5))*ln(1/2*x**2-2)+((2*x**2-8)*exp(5)+2*x* *4-8*x**2)*exp(exp(5)))*ln(3*ln(1/2*x**2-2)+x**2+exp(5))+(-4*x**4+4*x**2)* exp(exp(5)))/((3*x**2-12)*ln(1/2*x**2-2)+(x**2-4)*exp(5)+x**4-4*x**2)/ln(3 *ln(1/2*x**2-2)+x**2+exp(5))**2,x)
Time = 0.33 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 \, x e^{\left (e^{5}\right )}}{\log \left (x^{2} + e^{5} - 3 \, \log \left (2\right ) + 3 \, \log \left (x + 2\right ) + 3 \, \log \left (x - 2\right )\right )} \]
integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4- 8*x^2)*exp(exp(5)))*log(3*log(1/2*x^2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(ex p(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/2* x^2-2)+x^2+exp(5))^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (25) = 50\).
Time = 0.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 4.03 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=\frac {2 \, {\left (x^{3} e^{\left (e^{5}\right )} + 3 \, x e^{\left (e^{5}\right )} \log \left (\frac {1}{2} \, x^{2} - 2\right ) + x e^{\left (e^{5} + 5\right )}\right )}}{x^{2} \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) + e^{5} \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) - 3 \, \log \left (2\right ) \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) + 3 \, \log \left (x^{2} + e^{5} + 3 \, \log \left (\frac {1}{2} \, x^{2} - 2\right )\right ) \log \left (x^{2} - 4\right )} \]
integrate((((6*x^2-24)*exp(exp(5))*log(1/2*x^2-2)+((2*x^2-8)*exp(5)+2*x^4- 8*x^2)*exp(exp(5)))*log(3*log(1/2*x^2-2)+x^2+exp(5))+(-4*x^4+4*x^2)*exp(ex p(5)))/((3*x^2-12)*log(1/2*x^2-2)+(x^2-4)*exp(5)+x^4-4*x^2)/log(3*log(1/2* x^2-2)+x^2+exp(5))^2,x, algorithm=\
2*(x^3*e^(e^5) + 3*x*e^(e^5)*log(1/2*x^2 - 2) + x*e^(e^5 + 5))/(x^2*log(x^ 2 + e^5 + 3*log(1/2*x^2 - 2)) + e^5*log(x^2 + e^5 + 3*log(1/2*x^2 - 2)) - 3*log(2)*log(x^2 + e^5 + 3*log(1/2*x^2 - 2)) + 3*log(x^2 + e^5 + 3*log(1/2 *x^2 - 2))*log(x^2 - 4))
Time = 9.96 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.13 \[ \int \frac {e^{e^5} \left (4 x^2-4 x^4\right )+\left (e^{e^5} \left (-8 x^2+2 x^4+e^5 \left (-8+2 x^2\right )\right )+e^{e^5} \left (-24+6 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log \left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )}{\left (-4 x^2+x^4+e^5 \left (-4+x^2\right )+\left (-12+3 x^2\right ) \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right ) \log ^2\left (e^5+x^2+3 \log \left (\frac {1}{2} \left (-4+x^2\right )\right )\right )} \, dx=x\,{\mathrm {e}}^{{\mathrm {e}}^5}+\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^5}}{\ln \left (3\,\ln \left (\frac {x^2}{2}-2\right )+{\mathrm {e}}^5+x^2\right )}-\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x-x^3}+\frac {x^4\,{\mathrm {e}}^{{\mathrm {e}}^5}}{x-x^3} \]
int((log(3*log(x^2/2 - 2) + exp(5) + x^2)*(exp(exp(5))*(exp(5)*(2*x^2 - 8) - 8*x^2 + 2*x^4) + log(x^2/2 - 2)*exp(exp(5))*(6*x^2 - 24)) + exp(exp(5)) *(4*x^2 - 4*x^4))/(log(3*log(x^2/2 - 2) + exp(5) + x^2)^2*(x^4 - 4*x^2 + l og(x^2/2 - 2)*(3*x^2 - 12) + exp(5)*(x^2 - 4))),x)