Integrand size = 78, antiderivative size = 23 \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=e^{e^{\left (-1+\frac {e^4}{4 x^2}+x\right ) \left (x+x^2\right )}} \]
Time = 1.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=e^{e^{\frac {e^4}{4}+\frac {e^4}{4 x}-x+x^3}} \]
Integrate[(E^(E^((-4*x^2 + 4*x^4 + E^4*(1 + x))/(4*x)) + (-4*x^2 + 4*x^4 + E^4*(1 + x))/(4*x))*(-E^4 - 4*x^2 + 12*x^4))/(4*x^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.
\(\displaystyle \int \frac {\left (12 x^4-4 x^2-e^4\right ) \exp \left (\frac {4 x^4-4 x^2+e^4 (x+1)}{4 x}+e^{\frac {4 x^4-4 x^2+e^4 (x+1)}{4 x}}\right )}{4 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {\exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right ) \left (-12 x^4+4 x^2+e^4\right )}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {\exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right ) \left (-12 x^4+4 x^2+e^4\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-12 \exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right ) x^2+4 \exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right )+\frac {\exp \left (-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}+e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}+4\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-4 \int \exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right )dx-\int \frac {\exp \left (-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}+e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}+4\right )}{x^2}dx+12 \int \exp \left (e^{-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}}-\frac {-4 x^4+4 x^2-e^4 (x+1)}{4 x}\right ) x^2dx\right )\) |
Int[(E^(E^((-4*x^2 + 4*x^4 + E^4*(1 + x))/(4*x)) + (-4*x^2 + 4*x^4 + E^4*( 1 + x))/(4*x))*(-E^4 - 4*x^2 + 12*x^4))/(4*x^2),x]
3.6.93.3.1 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 = 0.82 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {\left (1+x \right ) \left (4 x^{3}-4 x^{2}+{\mathrm e}^{4}\right )}{4 x}}}\) | \(24\) |
norman | \({\mathrm e}^{{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{4}+4 x^{4}-4 x^{2}}{4 x}}}\) | \(25\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{4}+4 x^{4}-4 x^{2}}{4 x}}}\) | \(25\) |
int(1/4*(-exp(4)+12*x^4-4*x^2)*exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x)*exp(e xp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x))/x^2,x,method=_RETURNVERBOSE)
Timed out. \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=\text {Timed out} \]
integrate(1/4*(-exp(4)+12*x^4-4*x^2)*exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x) *exp(exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x))/x^2,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=e^{e^{\frac {x^{4} - x^{2} + \frac {\left (x + 1\right ) e^{4}}{4}}{x}}} \]
integrate(1/4*(-exp(4)+12*x**4-4*x**2)*exp(1/4*((1+x)*exp(4)+4*x**4-4*x**2 )/x)*exp(exp(1/4*((1+x)*exp(4)+4*x**4-4*x**2)/x))/x**2,x)
Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=e^{\left (e^{\left (x^{3} - x + \frac {e^{4}}{4 \, x} + \frac {1}{4} \, e^{4}\right )}\right )} \]
integrate(1/4*(-exp(4)+12*x^4-4*x^2)*exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x) *exp(exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x))/x^2,x, algorithm=\
\[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx=\int { \frac {{\left (12 \, x^{4} - 4 \, x^{2} - e^{4}\right )} e^{\left (\frac {4 \, x^{4} - 4 \, x^{2} + {\left (x + 1\right )} e^{4}}{4 \, x} + e^{\left (\frac {4 \, x^{4} - 4 \, x^{2} + {\left (x + 1\right )} e^{4}}{4 \, x}\right )}\right )}}{4 \, x^{2}} \,d x } \]
integrate(1/4*(-exp(4)+12*x^4-4*x^2)*exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x) *exp(exp(1/4*((1+x)*exp(4)+4*x^4-4*x^2)/x))/x^2,x, algorithm=\
integrate(1/4*(12*x^4 - 4*x^2 - e^4)*e^(1/4*(4*x^4 - 4*x^2 + (x + 1)*e^4)/ x + e^(1/4*(4*x^4 - 4*x^2 + (x + 1)*e^4)/x))/x^2, x)
Time = 8.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}}+\frac {-4 x^2+4 x^4+e^4 (1+x)}{4 x}} \left (-e^4-4 x^2+12 x^4\right )}{4 x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{4\,x}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{4}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^3}} \]