\(\int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} (16-4 x+304 x^3+128 x^4)}{x^3} \, dx\) [2150]

Optimal result

Integrand size = 46, antiderivative size = 30 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=3+4 e^{-2-\frac {2-x}{x^2}-4 x+4 (5+2 x)^2} \]

[Out]

4/exp((2-x)/x^2+4*x-2*(5+2*x)*(4*x+10)+2)+3

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6838} \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 e^{-\frac {-16 x^4-76 x^3-98 x^2-x+2}{x^2}} \]

[In]

Int[(16 - 4*x + 304*x^3 + 128*x^4)/(E^((2 - x - 98*x^2 - 76*x^3 - 16*x^4)/x^2)*x^3),x]

[Out]

4/E^((2 - x - 98*x^2 - 76*x^3 - 16*x^4)/x^2)

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}& = 4 e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 e^{98-\frac {2}{x^2}+\frac {1}{x}+76 x+16 x^2} \]

[In]

Integrate[(16 - 4*x + 304*x^3 + 128*x^4)/(E^((2 - x - 98*x^2 - 76*x^3 - 16*x^4)/x^2)*x^3),x]

[Out]

4*E^(98 - 2/x^2 + x^(-1) + 76*x + 16*x^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
risch	\(4 \,{\mathrm e}^{\frac {16 x^{4}+76 x^{3}+98 x^{2}+x -2}{x^{2}}}\)	\(26\)
gosper	\(4 \,{\mathrm e}^{\frac {16 x^{4}+76 x^{3}+98 x^{2}+x -2}{x^{2}}}\)	\(29\)
parallelrisch	\(4 \,{\mathrm e}^{\frac {16 x^{4}+76 x^{3}+98 x^{2}+x -2}{x^{2}}}\)	\(29\)
default	\(4 \,{\mathrm e}^{-\frac {-16 x^{4}-76 x^{3}-98 x^{2}-x +2}{x^{2}}}\)	\(30\)
norman	\(4 \,{\mathrm e}^{-\frac {-16 x^{4}-76 x^{3}-98 x^{2}-x +2}{x^{2}}}\)	\(30\)

[In]

int((128*x^4+304*x^3-4*x+16)/x^3/exp((-16*x^4-76*x^3-98*x^2-x+2)/x^2),x,method=_RETURNVERBOSE)

[Out]

4*exp((16*x^4+76*x^3+98*x^2+x-2)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 \, e^{\left (\frac {16 \, x^{4} + 76 \, x^{3} + 98 \, x^{2} + x - 2}{x^{2}}\right )} \]

[In]

integrate((128*x^4+304*x^3-4*x+16)/x^3/exp((-16*x^4-76*x^3-98*x^2-x+2)/x^2),x, algorithm="fricas")

[Out]

4*e^((16*x^4 + 76*x^3 + 98*x^2 + x - 2)/x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 e^{- \frac {- 16 x^{4} - 76 x^{3} - 98 x^{2} - x + 2}{x^{2}}} \]

[In]

integrate((128*x**4+304*x**3-4*x+16)/x**3/exp((-16*x**4-76*x**3-98*x**2-x+2)/x**2),x)

[Out]

4*exp(-(-16*x**4 - 76*x**3 - 98*x**2 - x + 2)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 \, e^{\left (16 \, x^{2} + 76 \, x + \frac {1}{x} - \frac {2}{x^{2}} + 98\right )} \]

[In]

integrate((128*x^4+304*x^3-4*x+16)/x^3/exp((-16*x^4-76*x^3-98*x^2-x+2)/x^2),x, algorithm="maxima")

[Out]

4*e^(16*x^2 + 76*x + 1/x - 2/x^2 + 98)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4 \, e^{\left (16 \, x^{2} + 76 \, x + \frac {1}{x} - \frac {2}{x^{2}} + 98\right )} \]

[In]

integrate((128*x^4+304*x^3-4*x+16)/x^3/exp((-16*x^4-76*x^3-98*x^2-x+2)/x^2),x, algorithm="giac")

[Out]

4*e^(16*x^2 + 76*x + 1/x - 2/x^2 + 98)

Mupad [B] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-\frac {2-x-98 x^2-76 x^3-16 x^4}{x^2}} \left (16-4 x+304 x^3+128 x^4\right )}{x^3} \, dx=4\,{\mathrm {e}}^{76\,x}\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{98}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\mathrm {e}}^{16\,x^2} \]

[In]

int((exp((x + 98*x^2 + 76*x^3 + 16*x^4 - 2)/x^2)*(304*x^3 - 4*x + 128*x^4 + 16))/x^3,x)

[Out]

4*exp(76*x)*exp(1/x)*exp(98)*exp(-2/x^2)*exp(16*x^2)