\(\int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} (15+e^{8-2 x} (60 x^4-40 x^5))}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx\) [3464]

Optimal result

Integrand size = 71, antiderivative size = 21 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \]

[Out]

exp(5*x/(3-4*x^4*exp(-x+4)^2))

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \]

[In]

Int[(15 + E^(8 - 2*x)*(60*x^4 - 40*x^5))/(E^((5*x)/(-3 + 4*E^(8 - 2*x)*x^4))*(9 - 24*E^(8 - 2*x)*x^4 + 16*E^(1
6 - 4*x)*x^8)),x]

[Out]

E^((5*x)/(3 - 4*E^(8 - 2*x)*x^4))

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{\frac {5 x}{3-4 e^{8-2 x} x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \]

[In]

Integrate[(15 + E^(8 - 2*x)*(60*x^4 - 40*x^5))/(E^((5*x)/(-3 + 4*E^(8 - 2*x)*x^4))*(9 - 24*E^(8 - 2*x)*x^4 + 1
6*E^(16 - 4*x)*x^8)),x]

[Out]

E^((-5*x)/(-3 + 4*E^(8 - 2*x)*x^4))

Maple [A] (verified)

Time = 38.55 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
risch	\({\mathrm e}^{-\frac {5 x}{4 x^{4} {\mathrm e}^{-2 x +8}-3}}\)	\(20\)
parallelrisch	\({\mathrm e}^{-\frac {5 x}{4 x^{4} {\mathrm e}^{-2 x +8}-3}}\)	\(22\)

[In]

int(((-40*x^5+60*x^4)*exp(-x+4)^2+15)*exp(-5*x/(4*x^4*exp(-x+4)^2-3))/(16*x^8*exp(-x+4)^4-24*x^4*exp(-x+4)^2+9
),x,method=_RETURNVERBOSE)

[Out]

exp(-5*x/(4*x^4*exp(-2*x+8)-3))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\left (-\frac {5 \, x}{4 \, x^{4} e^{\left (-2 \, x + 8\right )} - 3}\right )} \]

[In]

integrate(((-40*x^5+60*x^4)*exp(-x+4)^2+15)*exp(-5*x/(4*x^4*exp(-x+4)^2-3))/(16*x^8*exp(-x+4)^4-24*x^4*exp(-x+
4)^2+9),x, algorithm="fricas")

[Out]

e^(-5*x/(4*x^4*e^(-2*x + 8) - 3))

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{- \frac {5 x}{4 x^{4} e^{8 - 2 x} - 3}} \]

[In]

integrate(((-40*x**5+60*x**4)*exp(-x+4)**2+15)*exp(-5*x/(4*x**4*exp(-x+4)**2-3))/(16*x**8*exp(-x+4)**4-24*x**4
*exp(-x+4)**2+9),x)

[Out]

exp(-5*x/(4*x**4*exp(8 - 2*x) - 3))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=e^{\left (-\frac {5 \, x e^{\left (2 \, x\right )}}{4 \, x^{4} e^{8} - 3 \, e^{\left (2 \, x\right )}}\right )} \]

[In]

integrate(((-40*x^5+60*x^4)*exp(-x+4)^2+15)*exp(-5*x/(4*x^4*exp(-x+4)^2-3))/(16*x^8*exp(-x+4)^4-24*x^4*exp(-x+
4)^2+9),x, algorithm="maxima")

[Out]

e^(-5*x*e^(2*x)/(4*x^4*e^8 - 3*e^(2*x)))

Giac [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx=\text {Timed out} \]

[In]

integrate(((-40*x^5+60*x^4)*exp(-x+4)^2+15)*exp(-5*x/(4*x^4*exp(-x+4)^2-3))/(16*x^8*exp(-x+4)^4-24*x^4*exp(-x+
4)^2+9),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-\frac {5 x}{-3+4 e^{8-2 x} x^4}} \left (15+e^{8-2 x} \left (60 x^4-40 x^5\right )\right )}{9-24 e^{8-2 x} x^4+16 e^{16-4 x} x^8} \, dx={\mathrm {e}}^{-\frac {5\,x}{4\,x^4\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^8-3}} \]

[In]

int((exp(-(5*x)/(4*x^4*exp(8 - 2*x) - 3))*(exp(8 - 2*x)*(60*x^4 - 40*x^5) + 15))/(16*x^8*exp(16 - 4*x) - 24*x^
4*exp(8 - 2*x) + 9),x)

[Out]

exp(-(5*x)/(4*x^4*exp(-2*x)*exp(8) - 3))