3.18.59 \(\int -\frac {32 e^{-3+e}}{x^2} \, dx\)

Optimal. Leaf size=10 \[ \frac {32 e^{-3+e}}{x} \]

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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 {32 e^{e-3}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32*E^(-3 + E))/x^2,x]

[Out]

(32*E^(-3 + E))/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} &=-\left (\left (32 e^{-3+e}\right ) \int \frac {1}{x^2} \, dx\right )\\ &=\frac {32 e^{-3+e}}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \frac {32 e^{-3+e}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32*E^(-3 + E))/x^2,x]

[Out]

(32*E^(-3 + E))/x

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fricas [A]  time = 0.90, size = 13, normalized size = 1.30 \begin {gather*} \frac {e^{\left (e + 5 \, \log \relax (2) - 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(x)+5*log(2)+exp(1)-3)/x,x, algorithm="fricas")

[Out]

e^(e + 5*log(2) - 3)/x

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giac [A]  time = 0.26, size = 10, normalized size = 1.00 \begin {gather*} \frac {32 \, e^{\left (e - 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(x)+5*log(2)+exp(1)-3)/x,x, algorithm="giac")

[Out]

32*e^(e - 3)/x

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maple [A]  time = 0.04, size = 11, normalized size = 1.10

method	result	size
norman	\(\frac {32 \,{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-3}}{x}\)	\(11\)
risch	\(\frac {32 \,{\mathrm e}^{{\mathrm e}-3}}{x}\)	\(11\)
gosper	\({\mathrm e}^{-\ln \relax (x )+5 \ln \relax (2)+{\mathrm e}-3}\)	\(14\)
derivativedivides	\({\mathrm e}^{-\ln \relax (x )+5 \ln \relax (2)+{\mathrm e}-3}\)	\(14\)
default	\({\mathrm e}^{-\ln \relax (x )+5 \ln \relax (2)+{\mathrm e}-3}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-ln(x)+5*ln(2)+exp(1)-3)/x,x,method=_RETURNVERBOSE)

[Out]

32*exp(exp(1))*exp(-3)/x

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maxima [A]  time = 0.41, size = 10, normalized size = 1.00 \begin {gather*} \frac {32 \, e^{\left (e - 3\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(x)+5*log(2)+exp(1)-3)/x,x, algorithm="maxima")

[Out]

32*e^(e - 3)/x

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mupad [B]  time = 0.05, size = 10, normalized size = 1.00 \begin {gather*} \frac {32\,{\mathrm {e}}^{\mathrm {e}-3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(exp(1) + 5*log(2) - log(x) - 3)/x,x)

[Out]

(32*exp(exp(1) - 3))/x

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sympy [A]  time = 0.06, size = 10, normalized size = 1.00 \begin {gather*} \frac {32 e^{e}}{x e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-ln(x)+5*ln(2)+exp(1)-3)/x,x)

[Out]

32*exp(-3)*exp(E)/x

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