\(\int \frac {e^{\frac {1-6 x+x^2}{x}} (-16+16 x^2)}{x^2} \, dx\) [512]

Optimal result

Integrand size = 25, antiderivative size = 10 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 e^{-6+\frac {1}{x}+x} \]

[Out]

16*exp(1/x+x-6)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6820, 12, 6838} \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 e^{x+\frac {1}{x}-6} \]

[In]

Int[(E^((1 - 6*x + x^2)/x)*(-16 + 16*x^2))/x^2,x]

[Out]

16*E^(-6 + x^(-1) + x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}& = \int \frac {16 e^{-6+\frac {1}{x}+x} \left (-1+x^2\right )}{x^2} \, dx \\ & = 16 \int \frac {e^{-6+\frac {1}{x}+x} \left (-1+x^2\right )}{x^2} \, dx \\ & = 16 e^{-6+\frac {1}{x}+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 e^{-6+\frac {1}{x}+x} \]

[In]

Integrate[(E^((1 - 6*x + x^2)/x)*(-16 + 16*x^2))/x^2,x]

[Out]

16*E^(-6 + x^(-1) + x)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.60

method	result	size
gosper	\(16 \,{\mathrm e}^{\frac {x^{2}-6 x +1}{x}}\)	\(16\)
default	\(16 \,{\mathrm e}^{\frac {x^{2}-6 x +1}{x}}\)	\(16\)
norman	\(16 \,{\mathrm e}^{\frac {x^{2}-6 x +1}{x}}\)	\(16\)
risch	\(16 \,{\mathrm e}^{\frac {x^{2}-6 x +1}{x}}\)	\(16\)
parallelrisch	\(16 \,{\mathrm e}^{\frac {x^{2}-6 x +1}{x}}\)	\(16\)

[In]

int((16*x^2-16)*exp((x^2-6*x+1)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

16*exp((x^2-6*x+1)/x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 \, e^{\left (\frac {x^{2} - 6 \, x + 1}{x}\right )} \]

[In]

integrate((16*x^2-16)*exp((x^2-6*x+1)/x)/x^2,x, algorithm="fricas")

[Out]

16*e^((x^2 - 6*x + 1)/x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 e^{\frac {x^{2} - 6 x + 1}{x}} \]

[In]

integrate((16*x**2-16)*exp((x**2-6*x+1)/x)/x**2,x)

[Out]

16*exp((x**2 - 6*x + 1)/x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 \, e^{\left (x + \frac {1}{x} - 6\right )} \]

[In]

integrate((16*x^2-16)*exp((x^2-6*x+1)/x)/x^2,x, algorithm="maxima")

[Out]

16*e^(x + 1/x - 6)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16 \, e^{\left (x + \frac {1}{x} - 6\right )} \]

[In]

integrate((16*x^2-16)*exp((x^2-6*x+1)/x)/x^2,x, algorithm="giac")

[Out]

16*e^(x + 1/x - 6)

Mupad [B] (verification not implemented)

Time = 7.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1-6 x+x^2}{x}} \left (-16+16 x^2\right )}{x^2} \, dx=16\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{-6}\,{\mathrm {e}}^x \]

[In]

int((exp((x^2 - 6*x + 1)/x)*(16*x^2 - 16))/x^2,x)

[Out]

16*exp(1/x)*exp(-6)*exp(x)