Integrand size = 83, antiderivative size = 30 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=-e^4+x-\frac {2 e^x \left (1-e^{e^x}\right )}{\frac {2}{x}+x} \]
Time = 0.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=\frac {x \left (2-2 e^x+2 e^{e^x+x}+x^2\right )}{2+x^2} \]
Integrate[(4 + 4*x^2 + x^4 + E^x*(-4 - 4*x + 2*x^2 - 2*x^3) + E^E^x*(E^(2* x)*(4*x + 2*x^3) + E^x*(4 + 4*x - 2*x^2 + 2*x^3)))/(4 + 4*x^2 + x^4),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.
\(\displaystyle \int \frac {x^4+4 x^2+e^x \left (-2 x^3+2 x^2-4 x-4\right )+e^{e^x} \left (e^{2 x} \left (2 x^3+4 x\right )+e^x \left (2 x^3-2 x^2+4 x+4\right )\right )+4}{x^4+4 x^2+4} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {x^4+4 x^2-2 e^x \left (x^3-x^2+2 x+2\right )+2 e^{e^x} \left (e^{2 x} \left (x^3+2 x\right )+e^x \left (x^3-x^2+2 x+2\right )\right )+4}{\left (x^2+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{2 x+e^x} x}{x^2+2}+\frac {2 e^x \left (e^{e^x}-1\right ) \left (x^3-x^2+2 x+2\right )}{\left (x^2+2\right )^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{x+e^x}}{\left (i \sqrt {2}-x\right )^2}dx-\frac {1}{2} \left (2+i \sqrt {2}\right ) \int \frac {e^{x+e^x}}{i \sqrt {2}-x}dx+\frac {i \int \frac {e^{x+e^x}}{i \sqrt {2}-x}dx}{\sqrt {2}}-\int \frac {e^{2 x+e^x}}{i \sqrt {2}-x}dx-\int \frac {e^{x+e^x}}{\left (x+i \sqrt {2}\right )^2}dx+\frac {1}{2} \left (2-i \sqrt {2}\right ) \int \frac {e^{x+e^x}}{x+i \sqrt {2}}dx+\frac {i \int \frac {e^{x+e^x}}{x+i \sqrt {2}}dx}{\sqrt {2}}+\int \frac {e^{2 x+e^x}}{x+i \sqrt {2}}dx-\frac {2 e^x \left (x^3+2 x\right )}{\left (x^2+2\right )^2}+x\) |
Int[(4 + 4*x^2 + x^4 + E^x*(-4 - 4*x + 2*x^2 - 2*x^3) + E^E^x*(E^(2*x)*(4* x + 2*x^3) + E^x*(4 + 4*x - 2*x^2 + 2*x^3)))/(4 + 4*x^2 + x^4),x]
3.7.43.3.1 Defintions of rubi rules used
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {x^{3}+2 x -2 \,{\mathrm e}^{x} x +2 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}}{x^{2}+2}\) | \(29\) |
parallelrisch | \(\frac {x^{3}+2 x -2 \,{\mathrm e}^{x} x +2 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}}{x^{2}+2}\) | \(29\) |
risch | \(x -\frac {2 \,{\mathrm e}^{x} x}{x^{2}+2}+\frac {2 x \,{\mathrm e}^{{\mathrm e}^{x}+x}}{x^{2}+2}\) | \(30\) |
int((((2*x^3+4*x)*exp(x)^2+(2*x^3-2*x^2+4*x+4)*exp(x))*exp(exp(x))+(-2*x^3 +2*x^2-4*x-4)*exp(x)+x^4+4*x^2+4)/(x^4+4*x^2+4),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=\frac {x^{3} + 2 \, x e^{\left (x + e^{x}\right )} - 2 \, x e^{x} + 2 \, x}{x^{2} + 2} \]
integrate((((2*x^3+4*x)*exp(x)^2+(2*x^3-2*x^2+4*x+4)*exp(x))*exp(exp(x))+( -2*x^3+2*x^2-4*x-4)*exp(x)+x^4+4*x^2+4)/(x^4+4*x^2+4),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=x + \frac {2 x e^{x} e^{e^{x}}}{x^{2} + 2} - \frac {2 x e^{x}}{x^{2} + 2} \]
integrate((((2*x**3+4*x)*exp(x)**2+(2*x**3-2*x**2+4*x+4)*exp(x))*exp(exp(x ))+(-2*x**3+2*x**2-4*x-4)*exp(x)+x**4+4*x**2+4)/(x**4+4*x**2+4),x)
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=x + \frac {2 \, {\left (x e^{\left (x + e^{x}\right )} - x e^{x}\right )}}{x^{2} + 2} \]
integrate((((2*x^3+4*x)*exp(x)^2+(2*x^3-2*x^2+4*x+4)*exp(x))*exp(exp(x))+( -2*x^3+2*x^2-4*x-4)*exp(x)+x^4+4*x^2+4)/(x^4+4*x^2+4),x, algorithm=\
\[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=\int { \frac {x^{4} + 4 \, x^{2} - 2 \, {\left (x^{3} - x^{2} + 2 \, x + 2\right )} e^{x} + 2 \, {\left ({\left (x^{3} + 2 \, x\right )} e^{\left (2 \, x\right )} + {\left (x^{3} - x^{2} + 2 \, x + 2\right )} e^{x}\right )} e^{\left (e^{x}\right )} + 4}{x^{4} + 4 \, x^{2} + 4} \,d x } \]
integrate((((2*x^3+4*x)*exp(x)^2+(2*x^3-2*x^2+4*x+4)*exp(x))*exp(exp(x))+( -2*x^3+2*x^2-4*x-4)*exp(x)+x^4+4*x^2+4)/(x^4+4*x^2+4),x, algorithm=\
integrate((x^4 + 4*x^2 - 2*(x^3 - x^2 + 2*x + 2)*e^x + 2*((x^3 + 2*x)*e^(2 *x) + (x^3 - x^2 + 2*x + 2)*e^x)*e^(e^x) + 4)/(x^4 + 4*x^2 + 4), x)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {4+4 x^2+x^4+e^x \left (-4-4 x+2 x^2-2 x^3\right )+e^{e^x} \left (e^{2 x} \left (4 x+2 x^3\right )+e^x \left (4+4 x-2 x^2+2 x^3\right )\right )}{4+4 x^2+x^4} \, dx=x-\frac {2\,x\,{\mathrm {e}}^x}{x^2+2}+\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x}{x^2+2} \]