3.75.9 \(\int -\frac {8 x}{5 x^4+10 x^2 \log (3)+5 \log ^2(3)} \, dx\)

Optimal. Leaf size=24 \[ \frac {4}{5 x \left (x+\frac {\log (3)}{x}\right )}+\frac {\log (5)}{4} \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.50, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 28, 261} \begin {gather*} \frac {4}{5 x^2+\log (243)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x)/(5*x^4 + 10*x^2*Log[3] + 5*Log[3]^2),x]

[Out]

4/(5*x^2 + Log[243])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (8 \int \frac {x}{5 x^4+10 x^2 \log (3)+5 \log ^2(3)} \, dx\right )\\ &=-\left (40 \int \frac {x}{\left (5 x^2+5 \log (3)\right )^2} \, dx\right )\\ &=\frac {4}{5 x^2+\log (243)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.50 \begin {gather*} \frac {4}{5 \left (x^2+\log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x)/(5*x^4 + 10*x^2*Log[3] + 5*Log[3]^2),x]

[Out]

4/(5*(x^2 + Log[3]))

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fricas [A]  time = 0.99, size = 10, normalized size = 0.42 \begin {gather*} \frac {4}{5 \, {\left (x^{2} + \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*x/(5*log(3)^2+10*x^2*log(3)+5*x^4),x, algorithm="fricas")

[Out]

4/5/(x^2 + log(3))

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giac [A]  time = 0.72, size = 10, normalized size = 0.42 \begin {gather*} \frac {4}{5 \, {\left (x^{2} + \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*x/(5*log(3)^2+10*x^2*log(3)+5*x^4),x, algorithm="giac")

[Out]

4/5/(x^2 + log(3))

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maple [A]  time = 0.03, size = 11, normalized size = 0.46

method	result	size
gosper	\(\frac {4}{5 \left (\ln \relax (3)+x^{2}\right )}\)	\(11\)
default	\(\frac {4}{5 \left (\ln \relax (3)+x^{2}\right )}\)	\(11\)
norman	\(\frac {4}{5 \left (\ln \relax (3)+x^{2}\right )}\)	\(11\)
risch	\(\frac {4}{5 \left (\ln \relax (3)+x^{2}\right )}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-8*x/(5*ln(3)^2+10*x^2*ln(3)+5*x^4),x,method=_RETURNVERBOSE)

[Out]

4/5/(ln(3)+x^2)

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maxima [A]  time = 0.38, size = 10, normalized size = 0.42 \begin {gather*} \frac {4}{5 \, {\left (x^{2} + \log \relax (3)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*x/(5*log(3)^2+10*x^2*log(3)+5*x^4),x, algorithm="maxima")

[Out]

4/5/(x^2 + log(3))

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mupad [B]  time = 0.05, size = 14, normalized size = 0.58 \begin {gather*} \frac {4}{5\,\left (x^2+\ln \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x)/(10*x^2*log(3) + 5*log(3)^2 + 5*x^4),x)

[Out]

4/(5*(log(3) + x^2))

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sympy [A]  time = 0.14, size = 10, normalized size = 0.42 \begin {gather*} \frac {8}{10 x^{2} + 10 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-8*x/(5*ln(3)**2+10*x**2*ln(3)+5*x**4),x)

[Out]

8/(10*x**2 + 10*log(3))

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