\(\int \frac {e^{e^{-4+2 x}} (1+8 x+e^{-4+2 x} (10-2 x-8 x^2))}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx\) [6472]

Optimal result

Integrand size = 55, antiderivative size = 22 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=\frac {e^{e^{-4+2 x}}}{5-x-4 x^2} \]

[Out]

1/(-4*x^2-x+5)*exp(exp(-2+x)^2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {2326} \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=\frac {e^{e^{2 x-4}} \left (-4 x^2-x+5\right )}{16 x^4+8 x^3-39 x^2-10 x+25} \]

[In]

Int[(E^E^(-4 + 2*x)*(1 + 8*x + E^(-4 + 2*x)*(10 - 2*x - 8*x^2)))/(25 - 10*x - 39*x^2 + 8*x^3 + 16*x^4),x]

[Out]

(E^E^(-4 + 2*x)*(5 - x - 4*x^2))/(25 - 10*x - 39*x^2 + 8*x^3 + 16*x^4)

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 {e^{e^{-4+2 x}} \left (5-x-4 x^2\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=-\frac {e^{e^{-4+2 x}}}{-5+x+4 x^2} \]

[In]

Integrate[(E^E^(-4 + 2*x)*(1 + 8*x + E^(-4 + 2*x)*(10 - 2*x - 8*x^2)))/(25 - 10*x - 39*x^2 + 8*x^3 + 16*x^4),x
]

[Out]

-(E^E^(-4 + 2*x)/(-5 + x + 4*x^2))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

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

[In]

int(((-8*x^2-2*x+10)*exp(-2+x)^2+8*x+1)*exp(exp(-2+x)^2)/(16*x^4+8*x^3-39*x^2-10*x+25),x,method=_RETURNVERBOSE
)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=-\frac {e^{\left (e^{\left (2 \, x - 4\right )}\right )}}{4 \, x^{2} + x - 5} \]

[In]

integrate(((-8*x^2-2*x+10)*exp(-2+x)^2+8*x+1)*exp(exp(-2+x)^2)/(16*x^4+8*x^3-39*x^2-10*x+25),x, algorithm="fri
cas")

[Out]

-e^(e^(2*x - 4))/(4*x^2 + x - 5)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=- \frac {e^{e^{2 x - 4}}}{4 x^{2} + x - 5} \]

[In]

integrate(((-8*x**2-2*x+10)*exp(-2+x)**2+8*x+1)*exp(exp(-2+x)**2)/(16*x**4+8*x**3-39*x**2-10*x+25),x)

[Out]

-exp(exp(2*x - 4))/(4*x**2 + x - 5)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=-\frac {e^{\left (e^{\left (2 \, x - 4\right )}\right )}}{4 \, x^{2} + x - 5} \]

[In]

integrate(((-8*x^2-2*x+10)*exp(-2+x)^2+8*x+1)*exp(exp(-2+x)^2)/(16*x^4+8*x^3-39*x^2-10*x+25),x, algorithm="max
ima")

[Out]

-e^(e^(2*x - 4))/(4*x^2 + x - 5)

Giac [F]

\[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=\int { -\frac {{\left (2 \, {\left (4 \, x^{2} + x - 5\right )} e^{\left (2 \, x - 4\right )} - 8 \, x - 1\right )} e^{\left (e^{\left (2 \, x - 4\right )}\right )}}{16 \, x^{4} + 8 \, x^{3} - 39 \, x^{2} - 10 \, x + 25} \,d x } \]

[In]

integrate(((-8*x^2-2*x+10)*exp(-2+x)^2+8*x+1)*exp(exp(-2+x)^2)/(16*x^4+8*x^3-39*x^2-10*x+25),x, algorithm="gia
c")

[Out]

integrate(-(2*(4*x^2 + x - 5)*e^(2*x - 4) - 8*x - 1)*e^(e^(2*x - 4))/(16*x^4 + 8*x^3 - 39*x^2 - 10*x + 25), x)

Mupad [B] (verification not implemented)

Time = 12.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{e^{-4+2 x}} \left (1+8 x+e^{-4+2 x} \left (10-2 x-8 x^2\right )\right )}{25-10 x-39 x^2+8 x^3+16 x^4} \, dx=-\frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-4}}}{4\,x^2+x-5} \]

[In]

int((exp(exp(2*x - 4))*(8*x - exp(2*x - 4)*(2*x + 8*x^2 - 10) + 1))/(8*x^3 - 39*x^2 - 10*x + 16*x^4 + 25),x)

[Out]

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