3.25.7 \(\int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx\)

Optimal. Leaf size=14 \[ 10+\frac {4 e^{-58-4 x}}{x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2197} \begin {gather*} \frac {4 e^{-4 x-58}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(-58 - 4*x)*(-2 - 4*x))/x^3,x]

[Out]

(4*E^(-58 - 4*x))/x^2

Rule 12

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

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=4 \int \frac {e^{-58-4 x} (-2-4 x)}{x^3} \, dx\\ &=\frac {4 e^{-58-4 x}}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.86 \begin {gather*} \frac {4 e^{-58-4 x}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(-58 - 4*x)*(-2 - 4*x))/x^3,x]

[Out]

(4*E^(-58 - 4*x))/x^2

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fricas [A]  time = 0.52, size = 14, normalized size = 1.00 \begin {gather*} \frac {e^{\left (-4 \, x + 2 \, \log \relax (2) - 58\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="fricas")

[Out]

e^(-4*x + 2*log(2) - 58)/x^2

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giac [A]  time = 0.36, size = 11, normalized size = 0.79 \begin {gather*} \frac {4 \, e^{\left (-4 \, x - 58\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="giac")

[Out]

4*e^(-4*x - 58)/x^2

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maple [A]  time = 0.15, size = 12, normalized size = 0.86

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(12\)
gosper	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
derivativedivides	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
default	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
norman	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x-2)*exp(ln(2)-2*x-29)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

4*exp(-4*x-58)/x^2

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maxima [C]  time = 0.50, size = 19, normalized size = 1.36 \begin {gather*} 64 \, e^{\left (-58\right )} \Gamma \left (-1, 4 \, x\right ) + 128 \, e^{\left (-58\right )} \Gamma \left (-2, 4 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="maxima")

[Out]

64*e^(-58)*gamma(-1, 4*x) + 128*e^(-58)*gamma(-2, 4*x)

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mupad [B]  time = 0.07, size = 11, normalized size = 0.79 \begin {gather*} \frac {4\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-58}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*log(2) - 4*x - 58)*(4*x + 2))/x^3,x)

[Out]

(4*exp(-4*x)*exp(-58))/x^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.86 \begin {gather*} \frac {4 e^{- 4 x - 58}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x-2)*exp(ln(2)-2*x-29)**2/x**3,x)

[Out]

4*exp(-4*x - 58)/x**2

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