Integrand size = 114, antiderivative size = 32 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=e^{e^{5 e^{e^{\frac {3}{x}+\frac {5+3 e^5+\frac {9}{x^2}}{x}}}}} \]
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=e^{e^{5 e^{e^{\frac {9}{x^3}+\frac {8+3 e^5}{x}}}}} \]
Integrate[(E^(E^(5*E^E^((9 + 8*x^2 + 3*E^5*x^2)/x^3)) + 5*E^E^((9 + 8*x^2 + 3*E^5*x^2)/x^3) + E^((9 + 8*x^2 + 3*E^5*x^2)/x^3) + (9 + 8*x^2 + 3*E^5*x ^2)/x^3)*(-135 - 40*x^2 - 15*E^5*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.
\(\displaystyle \int \frac {\left (-15 e^5 x^2-40 x^2-135\right ) \exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (\left (-40-15 e^5\right ) x^2-135\right ) \exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^4}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 \left (8+3 e^5\right ) \exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^2}-\frac {135 \exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (8+3 e^5\right ) \int \frac {\exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^2}dx-135 \int \frac {\exp \left (\frac {3 e^5 x^2+8 x^2+9}{x^3}+5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}+e^{5 e^{e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}}}+e^{\frac {3 e^5 x^2+8 x^2+9}{x^3}}\right )}{x^4}dx\) |
Int[(E^(E^(5*E^E^((9 + 8*x^2 + 3*E^5*x^2)/x^3)) + 5*E^E^((9 + 8*x^2 + 3*E^ 5*x^2)/x^3) + E^((9 + 8*x^2 + 3*E^5*x^2)/x^3) + (9 + 8*x^2 + 3*E^5*x^2)/x^ 3)*(-135 - 40*x^2 - 15*E^5*x^2))/x^4,x]
3.9.53.3.1 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 = 11.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
\[{\mathrm e}^{{\mathrm e}^{5 \,{\mathrm e}^{{\mathrm e}^{\frac {3 x^{2} {\mathrm e}^{5}+8 x^{2}+9}{x^{3}}}}}}\]
int((-15*x^2*exp(5)-40*x^2-135)*exp((3*x^2*exp(5)+8*x^2+9)/x^3)*exp(exp((3 *x^2*exp(5)+8*x^2+9)/x^3))*exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3)))*exp (exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3))))/x^4,x)
Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (27) = 54\).
Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 4.91 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=e^{\left (\frac {x^{3} e^{\left (5 \, e^{\left (e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )}\right )}\right )} + x^{3} e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )} + 5 \, x^{3} e^{\left (e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )}\right )} + 3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}} - \frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}} - e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )} - 5 \, e^{\left (e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )}\right )}\right )} \]
integrate((-15*x^2*exp(5)-40*x^2-135)*exp((3*x^2*exp(5)+8*x^2+9)/x^3)*exp( exp((3*x^2*exp(5)+8*x^2+9)/x^3))*exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3) ))*exp(exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3))))/x^4,x, algorithm=\
e^((x^3*e^(5*e^(e^((3*x^2*e^5 + 8*x^2 + 9)/x^3))) + x^3*e^((3*x^2*e^5 + 8* x^2 + 9)/x^3) + 5*x^3*e^(e^((3*x^2*e^5 + 8*x^2 + 9)/x^3)) + 3*x^2*e^5 + 8* x^2 + 9)/x^3 - (3*x^2*e^5 + 8*x^2 + 9)/x^3 - e^((3*x^2*e^5 + 8*x^2 + 9)/x^ 3) - 5*e^(e^((3*x^2*e^5 + 8*x^2 + 9)/x^3)))
Time = 0.66 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=e^{e^{5 e^{e^{\frac {8 x^{2} + 3 x^{2} e^{5} + 9}{x^{3}}}}}} \]
integrate((-15*x**2*exp(5)-40*x**2-135)*exp((3*x**2*exp(5)+8*x**2+9)/x**3) *exp(exp((3*x**2*exp(5)+8*x**2+9)/x**3))*exp(5*exp(exp((3*x**2*exp(5)+8*x* *2+9)/x**3)))*exp(exp(5*exp(exp((3*x**2*exp(5)+8*x**2+9)/x**3))))/x**4,x)
Time = 0.68 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=e^{\left (e^{\left (5 \, e^{\left (e^{\left (\frac {3 \, e^{5}}{x} + \frac {8}{x} + \frac {9}{x^{3}}\right )}\right )}\right )}\right )} \]
integrate((-15*x^2*exp(5)-40*x^2-135)*exp((3*x^2*exp(5)+8*x^2+9)/x^3)*exp( exp((3*x^2*exp(5)+8*x^2+9)/x^3))*exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3) ))*exp(exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3))))/x^4,x, algorithm=\
\[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx=\int { -\frac {5 \, {\left (3 \, x^{2} e^{5} + 8 \, x^{2} + 27\right )} e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}} + e^{\left (5 \, e^{\left (e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )}\right )}\right )} + e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )} + 5 \, e^{\left (e^{\left (\frac {3 \, x^{2} e^{5} + 8 \, x^{2} + 9}{x^{3}}\right )}\right )}\right )}}{x^{4}} \,d x } \]
integrate((-15*x^2*exp(5)-40*x^2-135)*exp((3*x^2*exp(5)+8*x^2+9)/x^3)*exp( exp((3*x^2*exp(5)+8*x^2+9)/x^3))*exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3) ))*exp(exp(5*exp(exp((3*x^2*exp(5)+8*x^2+9)/x^3))))/x^4,x, algorithm=\
integrate(-5*(3*x^2*e^5 + 8*x^2 + 27)*e^((3*x^2*e^5 + 8*x^2 + 9)/x^3 + e^( 5*e^(e^((3*x^2*e^5 + 8*x^2 + 9)/x^3))) + e^((3*x^2*e^5 + 8*x^2 + 9)/x^3) + 5*e^(e^((3*x^2*e^5 + 8*x^2 + 9)/x^3)))/x^4, x)
Time = 10.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{e^{5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}}+5 e^{e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}}+e^{\frac {9+8 x^2+3 e^5 x^2}{x^3}}+\frac {9+8 x^2+3 e^5 x^2}{x^3}} \left (-135-40 x^2-15 e^5 x^2\right )}{x^4} \, dx={\mathrm {e}}^{{\mathrm {e}}^{5\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^{8/x}\,{\mathrm {e}}^{\frac {9}{x^3}}}}} \]