Integrand size = 165, antiderivative size = 30 \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\frac {-2+e^4+\frac {x (2-\log (2))}{e^{e^x}-x+\log (3)}}{x^3} \]
Timed out. \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\text {\$Aborted} \]
Integrate[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-1 6*x + 6*E^4*x + 2*x*Log[2])*Log[3] + (6 - 3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4*x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3]))/ (E^(2*E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4 *Log[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.
| \(\displaystyle \int \frac {-3 e^4 x^2+12 x^2-3 x^2 \log (2)+e^{e^x} \left (e^x \left (x^2 \log (2)-2 x^2\right )+6 e^4 x-16 x+2 x \log (2)+\left (12-6 e^4\right ) \log (3)\right )+\left (6-3 e^4\right ) e^{2 e^x}+\log (3) \left (6 e^4 x-16 x+2 x \log (2)\right )+\left (6-3 e^4\right ) \log ^2(3)}{x^6-2 x^5 \log (3)+e^{2 e^x} x^4+x^4 \log ^2(3)+e^{e^x} \left (2 x^4 \log (3)-2 x^5\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (12-3 e^4\right ) x^2-3 x^2 \log (2)+e^{e^x} \left (e^x \left (x^2 \log (2)-2 x^2\right )+6 e^4 x-16 x+2 x \log (2)+\left (12-6 e^4\right ) \log (3)\right )+\left (6-3 e^4\right ) e^{2 e^x}+\log (3) \left (6 e^4 x-16 x+2 x \log (2)\right )+\left (6-3 e^4\right ) \log ^2(3)}{x^6-2 x^5 \log (3)+e^{2 e^x} x^4+x^4 \log ^2(3)+e^{e^x} \left (2 x^4 \log (3)-2 x^5\right )}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x^2 \left (12-3 e^4-3 \log (2)\right )+e^{e^x} \left (e^x \left (x^2 \log (2)-2 x^2\right )+6 e^4 x-16 x+2 x \log (2)+\left (12-6 e^4\right ) \log (3)\right )+\left (6-3 e^4\right ) e^{2 e^x}+\log (3) \left (6 e^4 x-16 x+2 x \log (2)\right )+\left (6-3 e^4\right ) \log ^2(3)}{x^6-2 x^5 \log (3)+e^{2 e^x} x^4+x^4 \log ^2(3)+e^{e^x} \left (2 x^4 \log (3)-2 x^5\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {x^2 \left (12-3 e^4-3 \log (2)\right )+e^{e^x} \left (e^x \left (x^2 \log (2)-2 x^2\right )+6 e^4 x-16 x+2 x \log (2)+\left (12-6 e^4\right ) \log (3)\right )+\left (6-3 e^4\right ) e^{2 e^x}+\log (3) \left (6 e^4 x-16 x+2 x \log (2)\right )+\left (6-3 e^4\right ) \log ^2(3)}{x^4 \left (-x+e^{e^x}+\log (3)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {3 \left (e^4-2\right ) \log ^2(3)}{x^4 \left (x-e^{e^x}-\log (3)\right )^2}-\frac {3 \left (e^4-2\right ) e^{2 e^x}}{x^4 \left (-x+e^{e^x}+\log (3)\right )^2}-\frac {6 \left (e^4-2\right ) e^{e^x} \log (3)}{x^4 \left (-x+e^{e^x}+\log (3)\right )^2}-\frac {16 e^{e^x} \left (1+\frac {1}{8} \left (-3 e^4-\log (2)\right )\right )}{x^3 \left (-x+e^{e^x}+\log (3)\right )^2}+\frac {2 \left (-8+3 e^4+\log (2)\right ) \log (3)}{x^3 \left (x-e^{e^x}-\log (3)\right )^2}+\frac {e^{x+e^x} (\log (2)-2)}{x^2 \left (-x+e^{e^x}+\log (3)\right )^2}-\frac {3 \left (-4+e^4+\log (2)\right )}{x^2 \left (x-e^{e^x}-\log (3)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (2-e^4\right ) \log ^2(3) \int \frac {1}{x^4 \left (x-e^{e^x}-\log (3)\right )^2}dx+6 \left (2-e^4\right ) \log (3) \int \frac {e^{e^x}}{x^4 \left (-x+e^{e^x}+\log (3)\right )^2}dx+3 \left (2-e^4\right ) \int \frac {e^{2 e^x}}{x^4 \left (-x+e^{e^x}+\log (3)\right )^2}dx-2 \left (8-3 e^4-\log (2)\right ) \log (3) \int \frac {1}{x^3 \left (x-e^{e^x}-\log (3)\right )^2}dx-2 \left (8-3 e^4-\log (2)\right ) \int \frac {e^{e^x}}{x^3 \left (-x+e^{e^x}+\log (3)\right )^2}dx+3 \left (4-e^4-\log (2)\right ) \int \frac {1}{x^2 \left (x-e^{e^x}-\log (3)\right )^2}dx-(2-\log (2)) \int \frac {e^{x+e^x}}{x^2 \left (-x+e^{e^x}+\log (3)\right )^2}dx\) |
Int[(E^(2*E^x)*(6 - 3*E^4) + 12*x^2 - 3*E^4*x^2 - 3*x^2*Log[2] + (-16*x + 6*E^4*x + 2*x*Log[2])*Log[3] + (6 - 3*E^4)*Log[3]^2 + E^E^x*(-16*x + 6*E^4 *x + 2*x*Log[2] + E^x*(-2*x^2 + x^2*Log[2]) + (12 - 6*E^4)*Log[3]))/(E^(2* E^x)*x^4 + x^6 - 2*x^5*Log[3] + x^4*Log[3]^2 + E^E^x*(-2*x^5 + 2*x^4*Log[3 ])),x]
3.5.14.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(\frac {{\mathrm e}^{4}}{x^{3}}-\frac {2}{x^{3}}-\frac {\ln \left (2\right )-2}{x^{2} \left (\ln \left (3\right )+{\mathrm e}^{{\mathrm e}^{x}}-x \right )}\) | \(33\) |
| parallelrisch | \(\frac {{\mathrm e}^{4} \ln \left (3\right )+{\mathrm e}^{4} {\mathrm e}^{{\mathrm e}^{x}}-x \,{\mathrm e}^{4}-x \ln \left (2\right )-2 \ln \left (3\right )-2 \,{\mathrm e}^{{\mathrm e}^{x}}+4 x}{x^{3} \left (\ln \left (3\right )+{\mathrm e}^{{\mathrm e}^{x}}-x \right )}\) | \(50\) |
int(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*ln(2)-2*x^2)*exp(x)+(-6*exp(4)+12)* ln(3)+2*x*ln(2)+6*x*exp(4)-16*x)*exp(exp(x))+(-3*exp(4)+6)*ln(3)^2+(2*x*ln (2)+6*x*exp(4)-16*x)*ln(3)-3*x^2*ln(2)-3*x^2*exp(4)+12*x^2)/(x^4*exp(exp(x ))^2+(2*x^4*ln(3)-2*x^5)*exp(exp(x))+x^4*ln(3)^2-2*x^5*ln(3)+x^6),x,method =_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\frac {x e^{4} - {\left (e^{4} - 2\right )} e^{\left (e^{x}\right )} - {\left (e^{4} - 2\right )} \log \left (3\right ) + x \log \left (2\right ) - 4 \, x}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \left (3\right )} \]
integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp( 4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3) ^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)/( x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log( 3)+x^6),x, algorithm=\
(x*e^4 - (e^4 - 2)*e^(e^x) - (e^4 - 2)*log(3) + x*log(2) - 4*x)/(x^4 - x^3 *e^(e^x) - x^3*log(3))
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\frac {2 - \log {\left (2 \right )}}{- x^{3} + x^{2} e^{e^{x}} + x^{2} \log {\left (3 \right )}} - \frac {6 - 3 e^{4}}{3 x^{3}} \]
integrate(((-3*exp(4)+6)*exp(exp(x))**2+((x**2*ln(2)-2*x**2)*exp(x)+(-6*ex p(4)+12)*ln(3)+2*x*ln(2)+6*x*exp(4)-16*x)*exp(exp(x))+(-3*exp(4)+6)*ln(3)* *2+(2*x*ln(2)+6*x*exp(4)-16*x)*ln(3)-3*x**2*ln(2)-3*x**2*exp(4)+12*x**2)/( x**4*exp(exp(x))**2+(2*x**4*ln(3)-2*x**5)*exp(exp(x))+x**4*ln(3)**2-2*x**5 *ln(3)+x**6),x)
Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\frac {x {\left (e^{4} + \log \left (2\right ) - 4\right )} - {\left (e^{4} - 2\right )} e^{\left (e^{x}\right )} - e^{4} \log \left (3\right ) + 2 \, \log \left (3\right )}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \left (3\right )} \]
integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp( 4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3) ^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)/( x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log( 3)+x^6),x, algorithm=\
(x*(e^4 + log(2) - 4) - (e^4 - 2)*e^(e^x) - e^4*log(3) + 2*log(3))/(x^4 - x^3*e^(e^x) - x^3*log(3))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).
Time = 0.33 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=\frac {x e^{4} - e^{4} \log \left (3\right ) + x \log \left (2\right ) - 4 \, x - e^{\left (e^{x} + 4\right )} + 2 \, e^{\left (e^{x}\right )} + 2 \, \log \left (3\right )}{x^{4} - x^{3} e^{\left (e^{x}\right )} - x^{3} \log \left (3\right )} \]
integrate(((-3*exp(4)+6)*exp(exp(x))^2+((x^2*log(2)-2*x^2)*exp(x)+(-6*exp( 4)+12)*log(3)+2*x*log(2)+6*x*exp(4)-16*x)*exp(exp(x))+(-3*exp(4)+6)*log(3) ^2+(2*x*log(2)+6*x*exp(4)-16*x)*log(3)-3*x^2*log(2)-3*x^2*exp(4)+12*x^2)/( x^4*exp(exp(x))^2+(2*x^4*log(3)-2*x^5)*exp(exp(x))+x^4*log(3)^2-2*x^5*log( 3)+x^6),x, algorithm=\
(x*e^4 - e^4*log(3) + x*log(2) - 4*x - e^(e^x + 4) + 2*e^(e^x) + 2*log(3)) /(x^4 - x^3*e^(e^x) - x^3*log(3))
Timed out. \[ \int \frac {e^{2 e^x} \left (6-3 e^4\right )+12 x^2-3 e^4 x^2-3 x^2 \log (2)+\left (-16 x+6 e^4 x+2 x \log (2)\right ) \log (3)+\left (6-3 e^4\right ) \log ^2(3)+e^{e^x} \left (-16 x+6 e^4 x+2 x \log (2)+e^x \left (-2 x^2+x^2 \log (2)\right )+\left (12-6 e^4\right ) \log (3)\right )}{e^{2 e^x} x^4+x^6-2 x^5 \log (3)+x^4 \log ^2(3)+e^{e^x} \left (-2 x^5+2 x^4 \log (3)\right )} \, dx=-\int \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left (3\,{\mathrm {e}}^4-6\right )+{\ln \left (3\right )}^2\,\left (3\,{\mathrm {e}}^4-6\right )-\ln \left (3\right )\,\left (6\,x\,{\mathrm {e}}^4-16\,x+2\,x\,\ln \left (2\right )\right )+3\,x^2\,{\mathrm {e}}^4+3\,x^2\,\ln \left (2\right )-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (6\,x\,{\mathrm {e}}^4-16\,x+2\,x\,\ln \left (2\right )+{\mathrm {e}}^x\,\left (x^2\,\ln \left (2\right )-2\,x^2\right )-\ln \left (3\right )\,\left (6\,{\mathrm {e}}^4-12\right )\right )-12\,x^2}{x^4\,{\ln \left (3\right )}^2-2\,x^5\,\ln \left (3\right )+x^6+x^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (2\,x^4\,\ln \left (3\right )-2\,x^5\right )} \,d x \]
int(-(exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x *exp(4) - 16*x + 2*x*log(2)) + 3*x^2*exp(4) + 3*x^2*log(2) - exp(exp(x))*( 6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6*e xp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp( x)) + exp(exp(x))*(2*x^4*log(3) - 2*x^5)),x)
-int((exp(2*exp(x))*(3*exp(4) - 6) + log(3)^2*(3*exp(4) - 6) - log(3)*(6*x *exp(4) - 16*x + 2*x*log(2)) + 3*x^2*exp(4) + 3*x^2*log(2) - exp(exp(x))*( 6*x*exp(4) - 16*x + 2*x*log(2) + exp(x)*(x^2*log(2) - 2*x^2) - log(3)*(6*e xp(4) - 12)) - 12*x^2)/(x^4*log(3)^2 - 2*x^5*log(3) + x^6 + x^4*exp(2*exp( x)) + exp(exp(x))*(2*x^4*log(3) - 2*x^5)), x)