3.24.51 \(\int -\frac {16 e^{\frac {16-20 x}{x}}}{x^2} \, dx\)

Optimal. Leaf size=12 \[ e^{\frac {4 (4-5 x)}{x}} \]

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Rubi [A]  time = 0.04, antiderivative size = 9, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2225, 2209} \begin {gather*} e^{\frac {16}{x}-20} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*E^((16 - 20*x)/x))/x^2,x]

[Out]

E^(-20 + 16/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 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]

Rule 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (16 \int \frac {e^{\frac {16-20 x}{x}}}{x^2} \, dx\right )\\ &=-\left (16 \int \frac {e^{-20+\frac {16}{x}}}{x^2} \, dx\right )\\ &=e^{-20+\frac {16}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 9, normalized size = 0.75 \begin {gather*} e^{-20+\frac {16}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^((16 - 20*x)/x))/x^2,x]

[Out]

E^(-20 + 16/x)

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fricas [A]  time = 0.45, size = 11, normalized size = 0.92 \begin {gather*} e^{\left (-\frac {4 \, {\left (5 \, x - 4\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-16*exp((-20*x+16)/x)/x^2,x, algorithm="fricas")

[Out]

e^(-4*(5*x - 4)/x)

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giac [A]  time = 0.23, size = 8, normalized size = 0.67 \begin {gather*} e^{\left (\frac {16}{x} - 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-16*exp((-20*x+16)/x)/x^2,x, algorithm="giac")

[Out]

e^(16/x - 20)

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maple [A]  time = 0.09, size = 9, normalized size = 0.75

method	result	size
derivativedivides	\({\mathrm e}^{-20+\frac {16}{x}}\)	\(9\)
default	\({\mathrm e}^{-20+\frac {16}{x}}\)	\(9\)
norman	\({\mathrm e}^{\frac {-20 x +16}{x}}\)	\(11\)
gosper	\({\mathrm e}^{-\frac {4 \left (5 x -4\right )}{x}}\)	\(12\)
risch	\({\mathrm e}^{-\frac {4 \left (5 x -4\right )}{x}}\)	\(12\)
meijerg	\(-{\mathrm e}^{-20} \left (1-{\mathrm e}^{\frac {16}{x}}\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-16*exp((-20*x+16)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(-20+16/x)

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maxima [A]  time = 0.40, size = 8, normalized size = 0.67 \begin {gather*} e^{\left (\frac {16}{x} - 20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-16*exp((-20*x+16)/x)/x^2,x, algorithm="maxima")

[Out]

e^(16/x - 20)

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mupad [B]  time = 1.39, size = 9, normalized size = 0.75 \begin {gather*} {\mathrm {e}}^{-20}\,{\mathrm {e}}^{16/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(-(20*x - 16)/x))/x^2,x)

[Out]

exp(-20)*exp(16/x)

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sympy [A]  time = 0.09, size = 7, normalized size = 0.58 \begin {gather*} e^{\frac {16 - 20 x}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-16*exp((-20*x+16)/x)/x**2,x)

[Out]

exp((16 - 20*x)/x)

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