3.43.6 \(\int \frac {3 e^2}{8 x^3 \log ^2(5)} \, dx\)

Optimal. Leaf size=14 \[ -\frac {3 e^2}{16 x^2 \log ^2(5)} \]

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Rubi [A]  time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 30} \begin {gather*} -\frac {3 e^2}{16 x^2 \log ^2(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*E^2)/(8*x^3*Log[5]^2),x]

[Out]

(-3*E^2)/(16*x^2*Log[5]^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 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 {\left (3 e^2\right ) \int \frac {1}{x^3} \, dx}{8 \log ^2(5)}\\ &=-\frac {3 e^2}{16 x^2 \log ^2(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {3 e^2}{16 x^2 \log ^2(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^2)/(8*x^3*Log[5]^2),x]

[Out]

(-3*E^2)/(16*x^2*Log[5]^2)

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fricas [A]  time = 0.75, size = 11, normalized size = 0.79 \begin {gather*} -\frac {3 \, e^{2}}{16 \, x^{2} \log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="fricas")

[Out]

-3/16*e^2/(x^2*log(5)^2)

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giac [A]  time = 0.16, size = 11, normalized size = 0.79 \begin {gather*} -\frac {3 \, e^{2}}{16 \, x^{2} \log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="giac")

[Out]

-3/16*e^2/(x^2*log(5)^2)

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maple [A]  time = 0.02, size = 12, normalized size = 0.86

method	result	size
gosper	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \relax (5)^{2} x^{2}}\)	\(12\)
default	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \relax (5)^{2} x^{2}}\)	\(12\)
norman	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \relax (5)^{2} x^{2}}\)	\(12\)
risch	\(-\frac {3 \,{\mathrm e}^{2}}{16 \ln \relax (5)^{2} x^{2}}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3/8*exp(2)/x^3/ln(5)^2,x,method=_RETURNVERBOSE)

[Out]

-3/16*exp(2)/ln(5)^2/x^2

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maxima [A]  time = 0.34, size = 11, normalized size = 0.79 \begin {gather*} -\frac {3 \, e^{2}}{16 \, x^{2} \log \relax (5)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/8*exp(2)/x^3/log(5)^2,x, algorithm="maxima")

[Out]

-3/16*e^2/(x^2*log(5)^2)

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mupad [B]  time = 2.98, size = 11, normalized size = 0.79 \begin {gather*} -\frac {3\,{\mathrm {e}}^2}{16\,x^2\,{\ln \relax (5)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(2))/(8*x^3*log(5)^2),x)

[Out]

-(3*exp(2))/(16*x^2*log(5)^2)

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sympy [A]  time = 0.06, size = 15, normalized size = 1.07 \begin {gather*} - \frac {3 e^{2}}{16 x^{2} \log {\relax (5 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3/8*exp(2)/x**3/ln(5)**2,x)

[Out]

-3*exp(2)/(16*x**2*log(5)**2)

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