Integrand size = 111, antiderivative size = 29 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=e^{3+\frac {4-e^{\left (1+e^{e^5}+2 x\right )^2}}{x^2}+x}+x \]
Time = 0.70 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=e^{3-\frac {-4+e^{\left (1+e^{e^5}+2 x\right )^2}}{x^2}+x}+x \]
Integrate[(x^3 + E^((4 - E^(1 + E^(2*E^5) + 4*x + 4*x^2 + E^E^5*(2 + 4*x)) + 3*x^2 + x^3)/x^2)*(-8 + x^3 + E^(1 + E^(2*E^5) + 4*x + 4*x^2 + E^E^5*(2 + 4*x))*(2 - 4*x - 4*E^E^5*x - 8*x^2)))/x^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.
\(\displaystyle \int \frac {\left (x^3+e^{4 x^2+4 x+e^{e^5} (4 x+2)+e^{2 e^5}+1} \left (-8 x^2-4 e^{e^5} x-4 x+2\right )-8\right ) \exp \left (\frac {x^3+3 x^2-e^{4 x^2+4 x+e^{e^5} (4 x+2)+e^{2 e^5}+1}+4}{x^2}\right )+x^3}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {2 \left (-4 x^2-2 \left (1+e^{e^5}\right ) x+1\right ) \exp \left (4 x^2-\frac {e^{\left (2 x+e^{e^5}+1\right )^2}}{x^2}+\frac {4}{x^2}+5 x+e^{e^5} (4 x+2)+4 \left (1+\frac {e^{2 e^5}}{4}\right )\right )}{x^3}+\frac {\left (-8 e^{\frac {4}{x^2}+x+3}+e^{\frac {e^{\left (2 x+e^{e^5}+1\right )^2}}{x^2}} x^3+e^{\frac {4}{x^2}+x+3} x^3\right ) \exp \left (-\frac {e^{4 x^2+4 \left (1+e^{e^5}\right ) x+\left (1+e^{e^5}\right )^2}}{x^2}\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {2 \left (-4 x^2-2 \left (1+e^{e^5}\right ) x+1\right ) \exp \left (4 x^2-\frac {e^{\left (2 x+e^{e^5}+1\right )^2}}{x^2}+\frac {4}{x^2}+5 x+e^{e^5} (4 x+2)+4 \left (1+\frac {e^{2 e^5}}{4}\right )\right )}{x^3}+\frac {\left (-8 e^{\frac {4}{x^2}+x+3}+e^{\frac {e^{\left (2 x+e^{e^5}+1\right )^2}}{x^2}} x^3+e^{\frac {4}{x^2}+x+3} x^3\right ) \exp \left (-\frac {e^{4 x^2+4 \left (1+e^{e^5}\right ) x+\left (1+e^{e^5}\right )^2}}{x^2}\right )}{x^3}\right )dx\) |
Int[(x^3 + E^((4 - E^(1 + E^(2*E^5) + 4*x + 4*x^2 + E^E^5*(2 + 4*x)) + 3*x ^2 + x^3)/x^2)*(-8 + x^3 + E^(1 + E^(2*E^5) + 4*x + 4*x^2 + E^E^5*(2 + 4*x ))*(2 - 4*x - 4*E^E^5*x - 8*x^2)))/x^3,x]
3.15.77.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.76 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55
method | result | size |
parallelrisch | \(x +{\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5}}+\left (4 x +2\right ) {\mathrm e}^{{\mathrm e}^{5}}+4 x^{2}+4 x +1}+x^{3}+3 x^{2}+4}{x^{2}}}\) | \(45\) |
parts | \(x +{\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5}}+\left (4 x +2\right ) {\mathrm e}^{{\mathrm e}^{5}}+4 x^{2}+4 x +1}+x^{3}+3 x^{2}+4}{x^{2}}}\) | \(45\) |
risch | \(x +{\mathrm e}^{\frac {-{\mathrm e}^{4 x \,{\mathrm e}^{{\mathrm e}^{5}}+4 x^{2}+{\mathrm e}^{2 \,{\mathrm e}^{5}}+2 \,{\mathrm e}^{{\mathrm e}^{5}}+4 x +1}+x^{3}+3 x^{2}+4}{x^{2}}}\) | \(47\) |
norman | \(\frac {x^{3}+x^{2} {\mathrm e}^{\frac {-{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{5}}+\left (4 x +2\right ) {\mathrm e}^{{\mathrm e}^{5}}+4 x^{2}+4 x +1}+x^{3}+3 x^{2}+4}{x^{2}}}}{x^{2}}\) | \(55\) |
int((((-4*x*exp(exp(5))-8*x^2-4*x+2)*exp(exp(exp(5))^2+(4*x+2)*exp(exp(5)) +4*x^2+4*x+1)+x^3-8)*exp((-exp(exp(exp(5))^2+(4*x+2)*exp(exp(5))+4*x^2+4*x +1)+x^3+3*x^2+4)/x^2)+x^3)/x^3,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=x + e^{\left (\frac {x^{3} + 3 \, x^{2} - e^{\left (4 \, x^{2} + 2 \, {\left (2 \, x + 1\right )} e^{\left (e^{5}\right )} + 4 \, x + e^{\left (2 \, e^{5}\right )} + 1\right )} + 4}{x^{2}}\right )} \]
integrate((((-4*x*exp(exp(5))-8*x^2-4*x+2)*exp(exp(exp(5))^2+(4*x+2)*exp(e xp(5))+4*x^2+4*x+1)+x^3-8)*exp((-exp(exp(exp(5))^2+(4*x+2)*exp(exp(5))+4*x ^2+4*x+1)+x^3+3*x^2+4)/x^2)+x^3)/x^3,x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=x + e^{\frac {x^{3} + 3 x^{2} - e^{4 x^{2} + 4 x + \left (4 x + 2\right ) e^{e^{5}} + 1 + e^{2 e^{5}}} + 4}{x^{2}}} \]
integrate((((-4*x*exp(exp(5))-8*x**2-4*x+2)*exp(exp(exp(5))**2+(4*x+2)*exp (exp(5))+4*x**2+4*x+1)+x**3-8)*exp((-exp(exp(exp(5))**2+(4*x+2)*exp(exp(5) )+4*x**2+4*x+1)+x**3+3*x**2+4)/x**2)+x**3)/x**3,x)
x + exp((x**3 + 3*x**2 - exp(4*x**2 + 4*x + (4*x + 2)*exp(exp(5)) + 1 + ex p(2*exp(5))) + 4)/x**2)
Time = 0.43 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=x + e^{\left (x - \frac {e^{\left (4 \, x^{2} + 4 \, x e^{\left (e^{5}\right )} + 4 \, x + e^{\left (2 \, e^{5}\right )} + 2 \, e^{\left (e^{5}\right )} + 1\right )}}{x^{2}} + \frac {4}{x^{2}} + 3\right )} \]
integrate((((-4*x*exp(exp(5))-8*x^2-4*x+2)*exp(exp(exp(5))^2+(4*x+2)*exp(e xp(5))+4*x^2+4*x+1)+x^3-8)*exp((-exp(exp(exp(5))^2+(4*x+2)*exp(exp(5))+4*x ^2+4*x+1)+x^3+3*x^2+4)/x^2)+x^3)/x^3,x, algorithm=\
\[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=\int { \frac {x^{3} + {\left (x^{3} - 2 \, {\left (4 \, x^{2} + 2 \, x e^{\left (e^{5}\right )} + 2 \, x - 1\right )} e^{\left (4 \, x^{2} + 2 \, {\left (2 \, x + 1\right )} e^{\left (e^{5}\right )} + 4 \, x + e^{\left (2 \, e^{5}\right )} + 1\right )} - 8\right )} e^{\left (\frac {x^{3} + 3 \, x^{2} - e^{\left (4 \, x^{2} + 2 \, {\left (2 \, x + 1\right )} e^{\left (e^{5}\right )} + 4 \, x + e^{\left (2 \, e^{5}\right )} + 1\right )} + 4}{x^{2}}\right )}}{x^{3}} \,d x } \]
integrate((((-4*x*exp(exp(5))-8*x^2-4*x+2)*exp(exp(exp(5))^2+(4*x+2)*exp(e xp(5))+4*x^2+4*x+1)+x^3-8)*exp((-exp(exp(exp(5))^2+(4*x+2)*exp(exp(5))+4*x ^2+4*x+1)+x^3+3*x^2+4)/x^2)+x^3)/x^3,x, algorithm=\
integrate((x^3 + (x^3 - 2*(4*x^2 + 2*x*e^(e^5) + 2*x - 1)*e^(4*x^2 + 2*(2* x + 1)*e^(e^5) + 4*x + e^(2*e^5) + 1) - 8)*e^((x^3 + 3*x^2 - e^(4*x^2 + 2* (2*x + 1)*e^(e^5) + 4*x + e^(2*e^5) + 1) + 4)/x^2))/x^3, x)
Time = 8.55 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {x^3+e^{\frac {4-e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)}+3 x^2+x^3}{x^2}} \left (-8+x^3+e^{1+e^{2 e^5}+4 x+4 x^2+e^{e^5} (2+4 x)} \left (2-4 x-4 e^{e^5} x-8 x^2\right )\right )}{x^3} \, dx=x+{\mathrm {e}}^{-\frac {{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{{\mathrm {e}}^5}}\,{\mathrm {e}}^{4\,x}\,\mathrm {e}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^5}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^5}}\,{\mathrm {e}}^{4\,x^2}}{x^2}}\,{\mathrm {e}}^3\,{\mathrm {e}}^{\frac {4}{x^2}}\,{\mathrm {e}}^x \]
int(-(exp((3*x^2 - exp(4*x + exp(2*exp(5)) + exp(exp(5))*(4*x + 2) + 4*x^2 + 1) + x^3 + 4)/x^2)*(exp(4*x + exp(2*exp(5)) + exp(exp(5))*(4*x + 2) + 4 *x^2 + 1)*(4*x + 4*x*exp(exp(5)) + 8*x^2 - 2) - x^3 + 8) - x^3)/x^3,x)