\(\int \frac {e^{e^{e^x}} (30 x-15 x^2+e^x (-15-5 x+5 x^2)+e^{e^x} (e^{2 x} (15 x-5 x^2)+e^x (-15 x^2+5 x^3)))}{9 x^4-6 x^5+x^6+e^{2 x} (9 x^2-6 x^3+x^4)+e^x (-18 x^3+12 x^4-2 x^5)} \, dx\) [7526]

Optimal result

Integrand size = 124, antiderivative size = 28 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\frac {5 e^{e^{e^x}}}{(3-x) \left (e^x-x\right ) x} \]

[Out]

5*exp(exp(exp(x)))/x/(exp(x)-x)/(-x+3)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(100\) vs. \(2(28)=56\).

Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.57, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2326} \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\frac {5 e^{e^{e^x}-x} \left (e^{2 x} \left (3 x-x^2\right )-e^x \left (3 x^2-x^3\right )\right )}{x^6-6 x^5+9 x^4-2 e^x \left (x^5-6 x^4+9 x^3\right )+e^{2 x} \left (x^4-6 x^3+9 x^2\right )} \]

[In]

Int[(E^E^E^x*(30*x - 15*x^2 + E^x*(-15 - 5*x + 5*x^2) + E^E^x*(E^(2*x)*(15*x - 5*x^2) + E^x*(-15*x^2 + 5*x^3))
))/(9*x^4 - 6*x^5 + x^6 + E^(2*x)*(9*x^2 - 6*x^3 + x^4) + E^x*(-18*x^3 + 12*x^4 - 2*x^5)),x]

[Out]

(5*E^(E^E^x - x)*(E^(2*x)*(3*x - x^2) - E^x*(3*x^2 - x^3)))/(9*x^4 - 6*x^5 + x^6 + E^(2*x)*(9*x^2 - 6*x^3 + x^
4) - 2*E^x*(9*x^3 - 6*x^4 + x^5))

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {5 e^{e^{e^x}-x} \left (e^{2 x} \left (3 x-x^2\right )-e^x \left (3 x^2-x^3\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )-2 e^x \left (9 x^3-6 x^4+x^5\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=-\frac {5 e^{e^{e^x}}}{\left (e^x-x\right ) (-3+x) x} \]

[In]

Integrate[(E^E^E^x*(30*x - 15*x^2 + E^x*(-15 - 5*x + 5*x^2) + E^E^x*(E^(2*x)*(15*x - 5*x^2) + E^x*(-15*x^2 + 5
*x^3))))/(9*x^4 - 6*x^5 + x^6 + E^(2*x)*(9*x^2 - 6*x^3 + x^4) + E^x*(-18*x^3 + 12*x^4 - 2*x^5)),x]

[Out]

(-5*E^E^E^x)/((E^x - x)*(-3 + x)*x)

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {5 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}}{x \left (x^{2}-{\mathrm e}^{x} x -3 x +3 \,{\mathrm e}^{x}\right )}\)	\(28\)
parallelrisch	\(\frac {5 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}}{x \left (x^{2}-{\mathrm e}^{x} x -3 x +3 \,{\mathrm e}^{x}\right )}\)	\(28\)

[In]

int((((-5*x^2+15*x)*exp(x)^2+(5*x^3-15*x^2)*exp(x))*exp(exp(x))+(5*x^2-5*x-15)*exp(x)-15*x^2+30*x)*exp(exp(exp
(x)))/((x^4-6*x^3+9*x^2)*exp(x)^2+(-2*x^5+12*x^4-18*x^3)*exp(x)+x^6-6*x^5+9*x^4),x,method=_RETURNVERBOSE)

[Out]

5/x/(x^2-exp(x)*x-3*x+3*exp(x))*exp(exp(exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\frac {5 \, e^{\left (e^{\left (e^{x}\right )}\right )}}{x^{3} - 3 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x}} \]

[In]

integrate((((-5*x^2+15*x)*exp(x)^2+(5*x^3-15*x^2)*exp(x))*exp(exp(x))+(5*x^2-5*x-15)*exp(x)-15*x^2+30*x)*exp(e
xp(exp(x)))/((x^4-6*x^3+9*x^2)*exp(x)^2+(-2*x^5+12*x^4-18*x^3)*exp(x)+x^6-6*x^5+9*x^4),x, algorithm="fricas")

[Out]

5*e^(e^(e^x))/(x^3 - 3*x^2 - (x^2 - 3*x)*e^x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\frac {5 e^{e^{e^{x}}}}{x^{3} - x^{2} e^{x} - 3 x^{2} + 3 x e^{x}} \]

[In]

integrate((((-5*x**2+15*x)*exp(x)**2+(5*x**3-15*x**2)*exp(x))*exp(exp(x))+(5*x**2-5*x-15)*exp(x)-15*x**2+30*x)
*exp(exp(exp(x)))/((x**4-6*x**3+9*x**2)*exp(x)**2+(-2*x**5+12*x**4-18*x**3)*exp(x)+x**6-6*x**5+9*x**4),x)

[Out]

5*exp(exp(exp(x)))/(x**3 - x**2*exp(x) - 3*x**2 + 3*x*exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\frac {5 \, e^{\left (e^{\left (e^{x}\right )}\right )}}{x^{3} - 3 \, x^{2} - {\left (x^{2} - 3 \, x\right )} e^{x}} \]

[In]

integrate((((-5*x^2+15*x)*exp(x)^2+(5*x^3-15*x^2)*exp(x))*exp(exp(x))+(5*x^2-5*x-15)*exp(x)-15*x^2+30*x)*exp(e
xp(exp(x)))/((x^4-6*x^3+9*x^2)*exp(x)^2+(-2*x^5+12*x^4-18*x^3)*exp(x)+x^6-6*x^5+9*x^4),x, algorithm="maxima")

[Out]

5*e^(e^(e^x))/(x^3 - 3*x^2 - (x^2 - 3*x)*e^x)

Giac [F]

\[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=\int { -\frac {5 \, {\left (3 \, x^{2} - {\left (x^{2} - x - 3\right )} e^{x} + {\left ({\left (x^{2} - 3 \, x\right )} e^{\left (2 \, x\right )} - {\left (x^{3} - 3 \, x^{2}\right )} e^{x}\right )} e^{\left (e^{x}\right )} - 6 \, x\right )} e^{\left (e^{\left (e^{x}\right )}\right )}}{x^{6} - 6 \, x^{5} + 9 \, x^{4} + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{5} - 6 \, x^{4} + 9 \, x^{3}\right )} e^{x}} \,d x } \]

[In]

integrate((((-5*x^2+15*x)*exp(x)^2+(5*x^3-15*x^2)*exp(x))*exp(exp(x))+(5*x^2-5*x-15)*exp(x)-15*x^2+30*x)*exp(e
xp(exp(x)))/((x^4-6*x^3+9*x^2)*exp(x)^2+(-2*x^5+12*x^4-18*x^3)*exp(x)+x^6-6*x^5+9*x^4),x, algorithm="giac")

[Out]

integrate(-5*(3*x^2 - (x^2 - x - 3)*e^x + ((x^2 - 3*x)*e^(2*x) - (x^3 - 3*x^2)*e^x)*e^(e^x) - 6*x)*e^(e^(e^x))
/(x^6 - 6*x^5 + 9*x^4 + (x^4 - 6*x^3 + 9*x^2)*e^(2*x) - 2*(x^5 - 6*x^4 + 9*x^3)*e^x), x)

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e^{e^x}} \left (30 x-15 x^2+e^x \left (-15-5 x+5 x^2\right )+e^{e^x} \left (e^{2 x} \left (15 x-5 x^2\right )+e^x \left (-15 x^2+5 x^3\right )\right )\right )}{9 x^4-6 x^5+x^6+e^{2 x} \left (9 x^2-6 x^3+x^4\right )+e^x \left (-18 x^3+12 x^4-2 x^5\right )} \, dx=-\frac {5\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}{x^2\,{\mathrm {e}}^x-3\,x\,{\mathrm {e}}^x+3\,x^2-x^3} \]

[In]

int((exp(exp(exp(x)))*(30*x + exp(exp(x))*(exp(2*x)*(15*x - 5*x^2) - exp(x)*(15*x^2 - 5*x^3)) - exp(x)*(5*x -
5*x^2 + 15) - 15*x^2))/(exp(2*x)*(9*x^2 - 6*x^3 + x^4) - exp(x)*(18*x^3 - 12*x^4 + 2*x^5) + 9*x^4 - 6*x^5 + x^
6),x)

[Out]

-(5*exp(exp(exp(x))))/(x^2*exp(x) - 3*x*exp(x) + 3*x^2 - x^3)