3.47.46 \(\int -\frac {8}{x \log ^2(\frac {4 x^3}{3})} \, dx\)

Optimal. Leaf size=14 \[ \frac {8}{3 \log \left (\frac {4 x^3}{3}\right )} \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2302, 30} \begin {gather*} \frac {8}{3 \log \left (\frac {4 x^3}{3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-8/(x*Log[(4*x^3)/3]^2),x]

[Out]

8/(3*Log[(4*x^3)/3])

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]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (8 \int \frac {1}{x \log ^2\left (\frac {4 x^3}{3}\right )} \, dx\right )\\ &=-\left (\frac {8}{3} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {4 x^3}{3}\right )\right )\right )\\ &=\frac {8}{3 \log \left (\frac {4 x^3}{3}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {8}{3 \log \left (\frac {4 x^3}{3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-8/(x*Log[(4*x^3)/3]^2),x]

[Out]

8/(3*Log[(4*x^3)/3])

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fricas [A]  time = 0.62, size = 10, normalized size = 0.71 \begin {gather*} \frac {8}{3 \, \log \left (\frac {4}{3} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/x/log(4/3*x^3)^2,x, algorithm="fricas")

[Out]

8/3/log(4/3*x^3)

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giac [A]  time = 0.14, size = 10, normalized size = 0.71 \begin {gather*} \frac {8}{3 \, \log \left (\frac {4}{3} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/x/log(4/3*x^3)^2,x, algorithm="giac")

[Out]

8/3/log(4/3*x^3)

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maple [A]  time = 0.02, size = 11, normalized size = 0.79

method	result	size
derivativedivides	\(\frac {8}{3 \ln \left (\frac {4 x^{3}}{3}\right )}\)	\(11\)
default	\(\frac {8}{3 \ln \left (\frac {4 x^{3}}{3}\right )}\)	\(11\)
norman	\(\frac {8}{3 \ln \left (\frac {4 x^{3}}{3}\right )}\)	\(11\)
risch	\(\frac {8}{3 \ln \left (\frac {4 x^{3}}{3}\right )}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8/x/ln(4/3*x^3)^2,x,method=_RETURNVERBOSE)

[Out]

8/3/ln(4/3*x^3)

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maxima [A]  time = 0.36, size = 10, normalized size = 0.71 \begin {gather*} \frac {8}{3 \, \log \left (\frac {4}{3} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/x/log(4/3*x^3)^2,x, algorithm="maxima")

[Out]

8/3/log(4/3*x^3)

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mupad [B]  time = 3.16, size = 10, normalized size = 0.71 \begin {gather*} \frac {8}{3\,\ln \left (\frac {4\,x^3}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8/(x*log((4*x^3)/3)^2),x)

[Out]

8/(3*log((4*x^3)/3))

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sympy [A]  time = 0.09, size = 10, normalized size = 0.71 \begin {gather*} \frac {8}{3 \log {\left (\frac {4 x^{3}}{3} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8/x/ln(4/3*x**3)**2,x)

[Out]

8/(3*log(4*x**3/3))

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