\(\int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} (40-96 x^2)}{x^2} \, dx\) [6619]

Optimal result

Integrand size = 45, antiderivative size = 19 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=e^{5+e^{8 \left (\frac {-5+x}{x}-12 x\right )}} \]

[Out]

exp(5+exp(8*(-5+x)/x-96*x))

Rubi [F]

\[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=\int \frac {\exp \left (5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}\right ) \left (40-96 x^2\right )}{x^2} \, dx \]

[In]

Int[(E^(5 + E^((-40 + 8*x - 96*x^2)/x) + (-40 + 8*x - 96*x^2)/x)*(40 - 96*x^2))/x^2,x]

[Out]

-96*Defer[Int][E^(13 + E^(8 - 40/x - 96*x) - 40/x - 96*x), x] + 40*Defer[Int][E^(13 + E^(8 - 40/x - 96*x) - 40
/x - 96*x)/x^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-96 \exp \left (5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}\right )+\frac {40 \exp \left (5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}\right )}{x^2}\right ) \, dx \\ & = 40 \int \frac {\exp \left (5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}\right )}{x^2} \, dx-96 \int \exp \left (5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}\right ) \, dx \\ & = 40 \int \frac {e^{13+e^{8-\frac {40}{x}-96 x}-\frac {40}{x}-96 x}}{x^2} \, dx-96 \int e^{13+e^{8-\frac {40}{x}-96 x}-\frac {40}{x}-96 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=e^{5+e^{8-\frac {40}{x}-96 x}} \]

[In]

Integrate[(E^(5 + E^((-40 + 8*x - 96*x^2)/x) + (-40 + 8*x - 96*x^2)/x)*(40 - 96*x^2))/x^2,x]

[Out]

E^(5 + E^(8 - 40/x - 96*x))

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
norman	\({\mathrm e}^{{\mathrm e}^{\frac {-96 x^{2}+8 x -40}{x}}+5}\)	\(19\)
risch	\({\mathrm e}^{{\mathrm e}^{-\frac {8 \left (12 x^{2}-x +5\right )}{x}}+5}\)	\(20\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{-\frac {8 \left (12 x^{2}-x +5\right )}{x}}+5}\)	\(20\)

[In]

int((-96*x^2+40)*exp((-96*x^2+8*x-40)/x)*exp(exp((-96*x^2+8*x-40)/x)+5)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(exp((-96*x^2+8*x-40)/x)+5)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (16) = 32\).

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.68 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=e^{\left (-\frac {96 \, x^{2} - x e^{\left (-\frac {8 \, {\left (12 \, x^{2} - x + 5\right )}}{x}\right )} - 13 \, x + 40}{x} + \frac {8 \, {\left (12 \, x^{2} - x + 5\right )}}{x}\right )} \]

[In]

integrate((-96*x^2+40)*exp((-96*x^2+8*x-40)/x)*exp(exp((-96*x^2+8*x-40)/x)+5)/x^2,x, algorithm="fricas")

[Out]

e^(-(96*x^2 - x*e^(-8*(12*x^2 - x + 5)/x) - 13*x + 40)/x + 8*(12*x^2 - x + 5)/x)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=e^{e^{\frac {- 96 x^{2} + 8 x - 40}{x}} + 5} \]

[In]

integrate((-96*x**2+40)*exp((-96*x**2+8*x-40)/x)*exp(exp((-96*x**2+8*x-40)/x)+5)/x**2,x)

[Out]

exp(exp((-96*x**2 + 8*x - 40)/x) + 5)

Maxima [A] (verification not implemented)

none

Time = 1.80 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=e^{\left (e^{\left (-96 \, x - \frac {40}{x} + 8\right )} + 5\right )} \]

[In]

integrate((-96*x^2+40)*exp((-96*x^2+8*x-40)/x)*exp(exp((-96*x^2+8*x-40)/x)+5)/x^2,x, algorithm="maxima")

[Out]

e^(e^(-96*x - 40/x + 8) + 5)

Giac [F]

\[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx=\int { -\frac {8 \, {\left (12 \, x^{2} - 5\right )} e^{\left (-\frac {8 \, {\left (12 \, x^{2} - x + 5\right )}}{x} + e^{\left (-\frac {8 \, {\left (12 \, x^{2} - x + 5\right )}}{x}\right )} + 5\right )}}{x^{2}} \,d x } \]

[In]

integrate((-96*x^2+40)*exp((-96*x^2+8*x-40)/x)*exp(exp((-96*x^2+8*x-40)/x)+5)/x^2,x, algorithm="giac")

[Out]

integrate(-8*(12*x^2 - 5)*e^(-8*(12*x^2 - x + 5)/x + e^(-8*(12*x^2 - x + 5)/x) + 5)/x^2, x)

Mupad [B] (verification not implemented)

Time = 11.97 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{5+e^{\frac {-40+8 x-96 x^2}{x}}+\frac {-40+8 x-96 x^2}{x}} \left (40-96 x^2\right )}{x^2} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-96\,x}\,{\mathrm {e}}^8\,{\mathrm {e}}^{-\frac {40}{x}}}\,{\mathrm {e}}^5 \]

[In]

int(-(exp(-(96*x^2 - 8*x + 40)/x)*exp(exp(-(96*x^2 - 8*x + 40)/x) + 5)*(96*x^2 - 40))/x^2,x)

[Out]

exp(exp(-96*x)*exp(8)*exp(-40/x))*exp(5)