3.33.44 \(\int -\frac {2}{5 x \log (2) \log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=16 \[ \frac {1}{5 \log (2) \log \left (4 x^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2302, 30} \begin {gather*} \frac {1}{5 \log (2) \log \left (4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-2/(5*x*Log[2]*Log[4*x^2]^2),x]

[Out]

1/(5*Log[2]*Log[4*x^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]

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} &=-\frac {2 \int \frac {1}{x \log ^2\left (4 x^2\right )} \, dx}{5 \log (2)}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (4 x^2\right )\right )}{5 \log (2)}\\ &=\frac {1}{5 \log (2) \log \left (4 x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{5 \log (2) \log \left (4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-2/(5*x*Log[2]*Log[4*x^2]^2),x]

[Out]

1/(5*Log[2]*Log[4*x^2])

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fricas [A]  time = 0.55, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{5 \, \log \relax (2) \log \left (4 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(log(2)*log(4*x^2))

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giac [A]  time = 0.23, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{5 \, \log \relax (2) \log \left (4 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(log(2)*log(4*x^2))

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maple [A]  time = 0.03, size = 15, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {1}{5 \ln \left (4 x^{2}\right ) \ln \relax (2)}\)	\(15\)
default	\(\frac {1}{5 \ln \left (4 x^{2}\right ) \ln \relax (2)}\)	\(15\)
norman	\(\frac {1}{5 \ln \left (4 x^{2}\right ) \ln \relax (2)}\)	\(15\)
risch	\(\frac {1}{5 \ln \left (4 x^{2}\right ) \ln \relax (2)}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/ln(4*x^2)/ln(2)

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maxima [A]  time = 0.49, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{5 \, \log \relax (2) \log \left (4 \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(log(2)*log(4*x^2))

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mupad [B]  time = 1.96, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{5\,\ln \relax (2)\,\ln \left (4\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(5*log(2)*log(4*x^2))

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sympy [A]  time = 0.10, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{5 \log {\relax (2 )} \log {\left (4 x^{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(5*log(2)*log(4*x**2))

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