Integrand size = 160, antiderivative size = 30 \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {5}{6+e^{4 e^{e^{3/x}}} \left (-1+e^{2/x}+x\right )} \]
Time = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {5}{6+e^{4 e^{e^{3/x}}+\frac {2}{x}}+e^{4 e^{e^{3/x}}} (-1+x)} \]
Integrate[(E^(4*E^E^(3/x))*(10*E^(2/x) - 5*x^2 + E^(E^(3/x) + 3/x)*(-60 + 60*E^(2/x) + 60*x)))/(36*x^2 + E^(4*E^E^(3/x))*(-12*x^2 + 12*E^(2/x)*x^2 + 12*x^3) + E^(8*E^E^(3/x))*(x^2 + E^(4/x)*x^2 - 2*x^3 + x^4 + E^(2/x)*(-2* x^2 + 2*x^3))),x]
Time = 1.72 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {7292, 7237}
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.
\(\displaystyle \int \frac {e^{4 e^{e^{3/x}}} \left (-5 x^2+10 e^{2/x}+e^{e^{3/x}+\frac {3}{x}} \left (60 x+60 e^{2/x}-60\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (12 x^3+12 e^{2/x} x^2-12 x^2\right )+e^{8 e^{e^{3/x}}} \left (x^4-2 x^3+e^{4/x} x^2+x^2+e^{2/x} \left (2 x^3-2 x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{4 e^{e^{3/x}}} \left (-5 x^2+10 e^{2/x}+e^{e^{3/x}+\frac {3}{x}} \left (60 x+60 e^{2/x}-60\right )\right )}{x^2 \left (e^{4 e^{e^{3/x}}} x-e^{4 e^{e^{3/x}}}+e^{4 e^{e^{3/x}}+\frac {2}{x}}+6\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {5}{e^{4 e^{e^{3/x}}} x-e^{4 e^{e^{3/x}}}+e^{4 e^{e^{3/x}}+\frac {2}{x}}+6}\) |
Int[(E^(4*E^E^(3/x))*(10*E^(2/x) - 5*x^2 + E^(E^(3/x) + 3/x)*(-60 + 60*E^( 2/x) + 60*x)))/(36*x^2 + E^(4*E^E^(3/x))*(-12*x^2 + 12*E^(2/x)*x^2 + 12*x^ 3) + E^(8*E^E^(3/x))*(x^2 + E^(4/x)*x^2 - 2*x^3 + x^4 + E^(2/x)*(-2*x^2 + 2*x^3))),x]
3.27.79.3.1 Defintions of rubi rules used
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 34.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60
method | result | size |
parallelrisch | \(\frac {5}{{\mathrm e}^{\frac {2}{x}} {\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}}}+{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}}} x -{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}}}+6}\) | \(48\) |
risch | \(\frac {5}{{\mathrm e}^{\frac {4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}} x +2}{x}}+{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}}} x -{\mathrm e}^{4 \,{\mathrm e}^{{\mathrm e}^{\frac {3}{x}}}}+6}\) | \(49\) |
int(((60*exp(2/x)+60*x-60)*exp(3/x)*exp(exp(3/x))+10*exp(2/x)-5*x^2)*exp(4 *exp(exp(3/x)))/((x^2*exp(2/x)^2+(2*x^3-2*x^2)*exp(2/x)+x^4-2*x^3+x^2)*exp (4*exp(exp(3/x)))^2+(12*x^2*exp(2/x)+12*x^3-12*x^2)*exp(4*exp(exp(3/x)))+3 6*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {5}{{\left (x + e^{\frac {2}{x}} - 1\right )} e^{\left (4 \, e^{\left (\frac {x e^{\frac {3}{x}} + 3}{x} - \frac {3}{x}\right )}\right )} + 6} \]
integrate(((60*exp(2/x)+60*x-60)*exp(3/x)*exp(exp(3/x))+10*exp(2/x)-5*x^2) *exp(4*exp(exp(3/x)))/((x^2*exp(2/x)^2+(2*x^3-2*x^2)*exp(2/x)+x^4-2*x^3+x^ 2)*exp(4*exp(exp(3/x)))^2+(12*x^2*exp(2/x)+12*x^3-12*x^2)*exp(4*exp(exp(3/ x)))+36*x^2),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {5}{\left (x + e^{\frac {2}{x}} - 1\right ) e^{4 e^{e^{\frac {3}{x}}}} + 6} \]
integrate(((60*exp(2/x)+60*x-60)*exp(3/x)*exp(exp(3/x))+10*exp(2/x)-5*x**2 )*exp(4*exp(exp(3/x)))/((x**2*exp(2/x)**2+(2*x**3-2*x**2)*exp(2/x)+x**4-2* x**3+x**2)*exp(4*exp(exp(3/x)))**2+(12*x**2*exp(2/x)+12*x**3-12*x**2)*exp( 4*exp(exp(3/x)))+36*x**2),x)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\frac {5}{{\left (x + e^{\frac {2}{x}} - 1\right )} e^{\left (4 \, e^{\left (e^{\frac {3}{x}}\right )}\right )} + 6} \]
integrate(((60*exp(2/x)+60*x-60)*exp(3/x)*exp(exp(3/x))+10*exp(2/x)-5*x^2) *exp(4*exp(exp(3/x)))/((x^2*exp(2/x)^2+(2*x^3-2*x^2)*exp(2/x)+x^4-2*x^3+x^ 2)*exp(4*exp(exp(3/x)))^2+(12*x^2*exp(2/x)+12*x^3-12*x^2)*exp(4*exp(exp(3/ x)))+36*x^2),x, algorithm=\
Timed out. \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\text {Timed out} \]
integrate(((60*exp(2/x)+60*x-60)*exp(3/x)*exp(exp(3/x))+10*exp(2/x)-5*x^2) *exp(4*exp(exp(3/x)))/((x^2*exp(2/x)^2+(2*x^3-2*x^2)*exp(2/x)+x^4-2*x^3+x^ 2)*exp(4*exp(exp(3/x)))^2+(12*x^2*exp(2/x)+12*x^3-12*x^2)*exp(4*exp(exp(3/ x)))+36*x^2),x, algorithm=\
Timed out. \[ \int \frac {e^{4 e^{e^{3/x}}} \left (10 e^{2/x}-5 x^2+e^{e^{3/x}+\frac {3}{x}} \left (-60+60 e^{2/x}+60 x\right )\right )}{36 x^2+e^{4 e^{e^{3/x}}} \left (-12 x^2+12 e^{2/x} x^2+12 x^3\right )+e^{8 e^{e^{3/x}}} \left (x^2+e^{4/x} x^2-2 x^3+x^4+e^{2/x} \left (-2 x^2+2 x^3\right )\right )} \, dx=\int \frac {{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^{3/x}}}\,\left (10\,{\mathrm {e}}^{2/x}-5\,x^2+{\mathrm {e}}^{{\mathrm {e}}^{3/x}}\,{\mathrm {e}}^{3/x}\,\left (60\,x+60\,{\mathrm {e}}^{2/x}-60\right )\right )}{{\mathrm {e}}^{8\,{\mathrm {e}}^{{\mathrm {e}}^{3/x}}}\,\left (x^2\,{\mathrm {e}}^{4/x}-{\mathrm {e}}^{2/x}\,\left (2\,x^2-2\,x^3\right )+x^2-2\,x^3+x^4\right )+{\mathrm {e}}^{4\,{\mathrm {e}}^{{\mathrm {e}}^{3/x}}}\,\left (12\,x^2\,{\mathrm {e}}^{2/x}-12\,x^2+12\,x^3\right )+36\,x^2} \,d x \]
int((exp(4*exp(exp(3/x)))*(10*exp(2/x) - 5*x^2 + exp(exp(3/x))*exp(3/x)*(6 0*x + 60*exp(2/x) - 60)))/(exp(8*exp(exp(3/x)))*(x^2*exp(4/x) - exp(2/x)*( 2*x^2 - 2*x^3) + x^2 - 2*x^3 + x^4) + exp(4*exp(exp(3/x)))*(12*x^2*exp(2/x ) - 12*x^2 + 12*x^3) + 36*x^2),x)