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

Optimal. Leaf size=13 \[ \frac {17 e^{-2/x}}{2}+x \]

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2209} \begin {gather*} x+\frac {17 e^{-2/x}}{2} \end {gather*}

Antiderivative was successfully verified.

[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 2209

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} \frac {17 e^{-2/x}}{2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(17/E^(2/x) + x^2)/x^2,x]

[Out]

17/(2*E^(2/x)) + x

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fricas [A]  time = 0.72, size = 10, normalized size = 0.77 \begin {gather*} x + \frac {17}{2} \, e^{\left (-\frac {2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 17/2*e^(-2/x)

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giac [A]  time = 0.15, size = 16, normalized size = 1.23 \begin {gather*} \frac {1}{2} \, x {\left (\frac {17 \, e^{\left (-\frac {2}{x}\right )}}{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*(17*e^(-2/x)/x + 2)

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maple [A]  time = 0.10, size = 11, normalized size = 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 \(x +\frac {17 \,{\mathrm e}^{-\frac {2}{x}}}{2}\) \(11\)
norman \(\frac {x^{2}+\frac {17 x \,{\mathrm e}^{-\frac {2}{x}}}{2}}{x}\) \(18\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 10, normalized size = 0.77 \begin {gather*} x + \frac {17}{2} \, e^{\left (-\frac {2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 17/2*e^(-2/x)

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mupad [B]  time = 1.30, size = 10, normalized size = 0.77 \begin {gather*} x+\frac {17\,{\mathrm {e}}^{-\frac {2}{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((17*exp(-2/x) + x^2)/x^2,x)

[Out]

x + (17*exp(-2/x))/2

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sympy [A]  time = 0.09, size = 8, normalized size = 0.62 \begin {gather*} x + \frac {17 e^{- \frac {2}{x}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 17*exp(-2/x)/2

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