3.76.31 \(\int -\frac {4 e^{3-\frac {-3+14 x}{x}}}{3 x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{3} \left (5+\frac {4}{3} e^{-11+\frac {3}{x}}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2225, 2209} \begin {gather*} \frac {4}{9} e^{\frac {3}{x}-11} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4*E^(3 - (-3 + 14*x)/x))/(3*x^2),x]

[Out]

(4*E^(-11 + 3/x))/9

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 (\frac {4}{3} \int \frac {e^{3-\frac {-3+14 x}{x}}}{x^2} \, dx\right )\\ &=-\left (\frac {4}{3} \int \frac {e^{-11+\frac {3}{x}}}{x^2} \, dx\right )\\ &=\frac {4}{9} e^{-11+\frac {3}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.68 \begin {gather*} \frac {4}{9} e^{-11+\frac {3}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(3 - (-3 + 14*x)/x))/(3*x^2),x]

[Out]

(4*E^(-11 + 3/x))/9

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fricas [A]  time = 0.59, size = 13, normalized size = 0.68 \begin {gather*} \frac {4}{9} \, e^{\left (-\frac {11 \, x - 3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/3*exp(3)/x^2/exp((14*x-3)/x),x, algorithm="fricas")

[Out]

4/9*e^(-(11*x - 3)/x)

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giac [A]  time = 0.14, size = 10, normalized size = 0.53 \begin {gather*} \frac {4}{9} \, e^{\left (\frac {3}{x} - 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/3*exp(3)/x^2/exp((14*x-3)/x),x, algorithm="giac")

[Out]

4/9*e^(3/x - 11)

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maple [A]  time = 0.04, size = 14, normalized size = 0.74




method result size



risch \(\frac {4 \,{\mathrm e}^{-\frac {11 x -3}{x}}}{9}\) \(14\)
derivativedivides \(\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{-14+\frac {3}{x}}}{9}\) \(15\)
default \(\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{-14+\frac {3}{x}}}{9}\) \(15\)
gosper \(\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{-\frac {14 x -3}{x}}}{9}\) \(17\)
norman \(\frac {4 \,{\mathrm e}^{3} {\mathrm e}^{-\frac {14 x -3}{x}}}{9}\) \(17\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4/3*exp(3)/x^2/exp((14*x-3)/x),x,method=_RETURNVERBOSE)

[Out]

4/9*exp(-(11*x-3)/x)

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maxima [A]  time = 0.37, size = 10, normalized size = 0.53 \begin {gather*} \frac {4}{9} \, e^{\left (\frac {3}{x} - 11\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/3*exp(3)/x^2/exp((14*x-3)/x),x, algorithm="maxima")

[Out]

4/9*e^(3/x - 11)

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mupad [B]  time = 5.81, size = 10, normalized size = 0.53 \begin {gather*} \frac {4\,{\mathrm {e}}^{-11}\,{\mathrm {e}}^{3/x}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(3)*exp(-(14*x - 3)/x))/(3*x^2),x)

[Out]

(4*exp(-11)*exp(3/x))/9

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sympy [A]  time = 0.11, size = 14, normalized size = 0.74 \begin {gather*} \frac {4 e^{3} e^{- \frac {14 x - 3}{x}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/3*exp(3)/x**2/exp((14*x-3)/x),x)

[Out]

4*exp(3)*exp(-(14*x - 3)/x)/9

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