\(\int \frac {e^{\frac {9+4 x-5 x^2}{x}} (-9-5 x^2)}{x^2} \, dx\) [6579]

Optimal result

Integrand size = 27, antiderivative size = 17 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{-5-3 (-3+x)+\frac {9}{x}-2 x} \]

[Out]

exp(9/x-5*x+4)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6838} \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{\frac {-5 x^2+4 x+9}{x}} \]

[In]

Int[(E^((9 + 4*x - 5*x^2)/x)*(-9 - 5*x^2))/x^2,x]

[Out]

E^((9 + 4*x - 5*x^2)/x)

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 {9+4 x-5 x^2}{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{4+\frac {9}{x}-5 x} \]

[In]

Integrate[(E^((9 + 4*x - 5*x^2)/x)*(-9 - 5*x^2))/x^2,x]

[Out]

E^(4 + 9/x - 5*x)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\({\mathrm e}^{-\frac {\left (1+x \right ) \left (5 x -9\right )}{x}}\)	\(15\)
norman	\({\mathrm e}^{\frac {-5 x^{2}+4 x +9}{x}}\)	\(16\)
gosper	\({\mathrm e}^{-\frac {5 x^{2}-4 x -9}{x}}\)	\(17\)
parallelrisch	\({\mathrm e}^{-\frac {5 x^{2}-4 x -9}{x}}\)	\(17\)

[In]

int((-5*x^2-9)*exp((-5*x^2+4*x+9)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(-(1+x)*(5*x-9)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{\left (-\frac {5 \, x^{2} - 4 \, x - 9}{x}\right )} \]

[In]

integrate((-5*x^2-9)*exp((-5*x^2+4*x+9)/x)/x^2,x, algorithm="fricas")

[Out]

e^(-(5*x^2 - 4*x - 9)/x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{\frac {- 5 x^{2} + 4 x + 9}{x}} \]

[In]

integrate((-5*x**2-9)*exp((-5*x**2+4*x+9)/x)/x**2,x)

[Out]

exp((-5*x**2 + 4*x + 9)/x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{\left (-5 \, x + \frac {9}{x} + 4\right )} \]

[In]

integrate((-5*x^2-9)*exp((-5*x^2+4*x+9)/x)/x^2,x, algorithm="maxima")

[Out]

e^(-5*x + 9/x + 4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx=e^{\left (-5 \, x + \frac {9}{x} + 4\right )} \]

[In]

integrate((-5*x^2-9)*exp((-5*x^2+4*x+9)/x)/x^2,x, algorithm="giac")

[Out]

e^(-5*x + 9/x + 4)

Mupad [B] (verification not implemented)

Time = 12.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {9+4 x-5 x^2}{x}} \left (-9-5 x^2\right )}{x^2} \, dx={\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{9/x} \]

[In]

int(-(exp((4*x - 5*x^2 + 9)/x)*(5*x^2 + 9))/x^2,x)

[Out]

exp(-5*x)*exp(4)*exp(9/x)