\(\int \frac {17 e^{-2/x}+x^2}{x^2} \, dx\) [2293]

Optimal result

Integrand size = 17, antiderivative size = 13 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=\frac {17 e^{-2/x}}{2}+x \]

[Out]

x+17/2*exp(-2/x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2240} \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=x+\frac {17 e^{-2/x}}{2} \]

[In]

Int[(17/E^(2/x) + x^2)/x^2,x]

[Out]

17/(2*E^(2/x)) + x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {17 e^{-2/x}}{x^2}\right ) \, dx \\ & = x+17 \int \frac {e^{-2/x}}{x^2} \, dx \\ & = \frac {17 e^{-2/x}}{2}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=\frac {17 e^{-2/x}}{2}+x \]

[In]

Integrate[(17/E^(2/x) + x^2)/x^2,x]

[Out]

17/(2*E^(2/x)) + x

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\)	\(11\)
default	\(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\)	\(11\)
risch	\(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\)	\(11\)
parallelrisch	\(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\)	\(11\)
parts	\(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\)	\(11\)
norman	\(\frac {x^{2}+\frac {17 x \,{\mathrm e}^{-\frac {2}{x}}}{2}}{x}\)	\(18\)

[In]

int((17*exp(-2/x)+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x+17/2*exp(-2/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=x + \frac {17}{2} \, e^{\left (-\frac {2}{x}\right )} \]

[In]

integrate((17*exp(-2/x)+x^2)/x^2,x, algorithm="fricas")

[Out]

x + 17/2*e^(-2/x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=x + \frac {17 e^{- \frac {2}{x}}}{2} \]

[In]

integrate((17*exp(-2/x)+x**2)/x**2,x)

[Out]

x + 17*exp(-2/x)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=x + \frac {17}{2} \, e^{\left (-\frac {2}{x}\right )} \]

[In]

integrate((17*exp(-2/x)+x^2)/x^2,x, algorithm="maxima")

[Out]

x + 17/2*e^(-2/x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=\frac {1}{2} \, x {\left (\frac {17 \, e^{\left (-\frac {2}{x}\right )}}{x} + 2\right )} \]

[In]

integrate((17*exp(-2/x)+x^2)/x^2,x, algorithm="giac")

[Out]

1/2*x*(17*e^(-2/x)/x + 2)

Mupad [B] (verification not implemented)

Time = 7.96 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {17 e^{-2/x}+x^2}{x^2} \, dx=x+\frac {17\,{\mathrm {e}}^{-\frac {2}{x}}}{2} \]

[In]

int((17*exp(-2/x) + x^2)/x^2,x)

[Out]

x + (17*exp(-2/x))/2