3.5.61 \(\int -\frac {16}{25 e^2 x^2} \, dx\)

Optimal. Leaf size=10 \[ \frac {16}{25 e^2 x} \]

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Rubi [A]  time = 0.00, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 30} \begin {gather*} \frac {16}{25 e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-16/(25*E^2*x^2),x]

[Out]

16/(25*E^2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {16 \int \frac {1}{x^2} \, dx}{25 e^2}\\ &=\frac {16}{25 e^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \frac {16}{25 e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-16/(25*E^2*x^2),x]

[Out]

16/(25*E^2*x)

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fricas [A]  time = 0.62, size = 16, normalized size = 1.60 \begin {gather*} \frac {4 \, e^{\left (-2 \, \log \relax (5) + 2 \, \log \relax (2) - 2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(-2*log(5)+2*log(2)+x-2)/exp(x)/x^2,x, algorithm="fricas")

[Out]

4*e^(-2*log(5) + 2*log(2) - 2)/x

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giac [A]  time = 0.36, size = 16, normalized size = 1.60 \begin {gather*} \frac {4 \, e^{\left (-2 \, \log \relax (5) + 2 \, \log \relax (2) - 2\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(-2*log(5)+2*log(2)+x-2)/exp(x)/x^2,x, algorithm="giac")

[Out]

4*e^(-2*log(5) + 2*log(2) - 2)/x

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maple [A]  time = 0.04, size = 8, normalized size = 0.80

method	result	size
default	\(\frac {16 \,{\mathrm e}^{-2}}{25 x}\)	\(8\)
risch	\(\frac {16 \,{\mathrm e}^{-2}}{25 x}\)	\(8\)
norman	\(\frac {16 \,{\mathrm e}^{-2}}{25 x}\)	\(10\)
gosper	\(\frac {4 \,{\mathrm e}^{-x} {\mathrm e}^{-2 \ln \relax (5)+2 \ln \relax (2)+x -2}}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4*exp(-2*ln(5)+2*ln(2)+x-2)/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

16/25/x*exp(-2)

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maxima [A]  time = 0.74, size = 7, normalized size = 0.70 \begin {gather*} \frac {16 \, e^{\left (-2\right )}}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(-2*log(5)+2*log(2)+x-2)/exp(x)/x^2,x, algorithm="maxima")

[Out]

16/25*e^(-2)/x

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mupad [B]  time = 0.41, size = 7, normalized size = 0.70 \begin {gather*} \frac {16\,{\mathrm {e}}^{-2}}{25\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(-x)*exp(x + 2*log(2) - 2*log(5) - 2))/x^2,x)

[Out]

(16*exp(-2))/(25*x)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.70 \begin {gather*} \frac {16}{25 x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4*exp(-2*ln(5)+2*ln(2)+x-2)/exp(x)/x**2,x)

[Out]

16*exp(-2)/(25*x)

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