3.96.92 \(\int \frac {-3-\log (3)-e^3 \log (16)}{x^2} \, dx\)

Optimal. Leaf size=20 \[ \log (3)+\frac {3+\log (3)}{x}+\frac {e^3 \log (16)}{x} \]

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Rubi [A]  time = 0.00, antiderivative size = 14, normalized size of antiderivative = 0.70, 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, 30} \begin {gather*} \frac {3+\log (3)+e^3 \log (16)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 - Log[3] - E^3*Log[16])/x^2,x]

[Out]

(3 + Log[3] + E^3*Log[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 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 (-3-\log (3)-e^3 \log (16)\right ) \int \frac {1}{x^2} \, dx\\ &=\frac {3+\log (3)+e^3 \log (16)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.70 \begin {gather*} \frac {3+\log (3)+e^3 \log (16)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 - Log[3] - E^3*Log[16])/x^2,x]

[Out]

(3 + Log[3] + E^3*Log[16])/x

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fricas [A]  time = 0.44, size = 14, normalized size = 0.70 \begin {gather*} \frac {4 \, e^{3} \log \relax (2) + \log \relax (3) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

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giac [A]  time = 0.13, size = 14, normalized size = 0.70 \begin {gather*} \frac {4 \, e^{3} \log \relax (2) + \log \relax (3) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

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maple [A]  time = 0.02, size = 15, normalized size = 0.75

method	result	size
gosper	\(\frac {4 \,{\mathrm e}^{3} \ln \relax (2)+\ln \relax (3)+3}{x}\)	\(15\)
norman	\(\frac {4 \,{\mathrm e}^{3} \ln \relax (2)+\ln \relax (3)+3}{x}\)	\(15\)
default	\(-\frac {-4 \,{\mathrm e}^{3} \ln \relax (2)-\ln \relax (3)-3}{x}\)	\(18\)
risch	\(\frac {4 \,{\mathrm e}^{3} \ln \relax (2)}{x}+\frac {\ln \relax (3)}{x}+\frac {3}{x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(3)*ln(2)+ln(3)+3)/x

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maxima [A]  time = 0.36, size = 14, normalized size = 0.70 \begin {gather*} \frac {4 \, e^{3} \log \relax (2) + \log \relax (3) + 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*e^3*log(2) + log(3) + 3)/x

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mupad [B]  time = 8.85, size = 14, normalized size = 0.70 \begin {gather*} \frac {\ln \relax (3)+4\,{\mathrm {e}}^3\,\ln \relax (2)+3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(3) + 4*exp(3)*log(2) + 3)/x

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sympy [A]  time = 0.06, size = 17, normalized size = 0.85 \begin {gather*} - \frac {- 4 e^{3} \log {\relax (2 )} - 3 - \log {\relax (3 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-4*exp(3)*log(2) - 3 - log(3))/x

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