3.6.89 \(\int \frac {-4-8 x-6 x^2-2 \log (\frac {16 x^2}{9})}{4 x^4+4 x^5+x^6+(4 x^3+2 x^4) \log (\frac {16 x^2}{9})+x^2 \log ^2(\frac {16 x^2}{9})} \, dx\)

Optimal. Leaf size=24 \[ 5+\frac {2}{x^2 \left (2+x+\frac {\log \left (\frac {16 x^2}{9}\right )}{x}\right )} \]

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Rubi [A]  time = 0.18, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6688, 12, 6687} \begin {gather*} \frac {2}{x \left (\log \left (\frac {16 x^2}{9}\right )+x (x+2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 8*x - 6*x^2 - 2*Log[(16*x^2)/9])/(4*x^4 + 4*x^5 + x^6 + (4*x^3 + 2*x^4)*Log[(16*x^2)/9] + x^2*Log[(1
6*x^2)/9]^2),x]

[Out]

2/(x*(x*(2 + x) + Log[(16*x^2)/9]))

Rule 12

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

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-2-4 x-3 x^2-\log \left (\frac {16 x^2}{9}\right )\right )}{x^2 \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )^2} \, dx\\ &=2 \int \frac {-2-4 x-3 x^2-\log \left (\frac {16 x^2}{9}\right )}{x^2 \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )^2} \, dx\\ &=\frac {2}{x \left (x (2+x)+\log \left (\frac {16 x^2}{9}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.34, size = 22, normalized size = 0.92 \begin {gather*} \frac {2}{x \left (2 x+x^2+\log \left (\frac {16 x^2}{9}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 8*x - 6*x^2 - 2*Log[(16*x^2)/9])/(4*x^4 + 4*x^5 + x^6 + (4*x^3 + 2*x^4)*Log[(16*x^2)/9] + x^2*
Log[(16*x^2)/9]^2),x]

[Out]

2/(x*(2*x + x^2 + Log[(16*x^2)/9]))

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fricas [A]  time = 0.71, size = 21, normalized size = 0.88 \begin {gather*} \frac {2}{x^{3} + 2 \, x^{2} + x \log \left (\frac {16}{9} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16/9*x^2)-6*x^2-8*x-4)/(x^2*log(16/9*x^2)^2+(2*x^4+4*x^3)*log(16/9*x^2)+x^6+4*x^5+4*x^4),x,
algorithm="fricas")

[Out]

2/(x^3 + 2*x^2 + x*log(16/9*x^2))

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giac [A]  time = 0.62, size = 21, normalized size = 0.88 \begin {gather*} \frac {2}{x^{3} + 2 \, x^{2} + x \log \left (\frac {16}{9} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16/9*x^2)-6*x^2-8*x-4)/(x^2*log(16/9*x^2)^2+(2*x^4+4*x^3)*log(16/9*x^2)+x^6+4*x^5+4*x^4),x,
algorithm="giac")

[Out]

2/(x^3 + 2*x^2 + x*log(16/9*x^2))

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maple [A]  time = 0.06, size = 21, normalized size = 0.88

method	result	size
norman	\(\frac {2}{x \left (x^{2}+\ln \left (\frac {16 x^{2}}{9}\right )+2 x \right )}\)	\(21\)
risch	\(\frac {2}{x \left (x^{2}+\ln \left (\frac {16 x^{2}}{9}\right )+2 x \right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(16/9*x^2)-6*x^2-8*x-4)/(x^2*ln(16/9*x^2)^2+(2*x^4+4*x^3)*ln(16/9*x^2)+x^6+4*x^5+4*x^4),x,method=_RE
TURNVERBOSE)

[Out]

2/x/(x^2+ln(16/9*x^2)+2*x)

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maxima [A]  time = 0.73, size = 28, normalized size = 1.17 \begin {gather*} \frac {2}{x^{3} + 2 \, x^{2} - 2 \, x {\left (\log \relax (3) - 2 \, \log \relax (2)\right )} + 2 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(16/9*x^2)-6*x^2-8*x-4)/(x^2*log(16/9*x^2)^2+(2*x^4+4*x^3)*log(16/9*x^2)+x^6+4*x^5+4*x^4),x,
algorithm="maxima")

[Out]

2/(x^3 + 2*x^2 - 2*x*(log(3) - 2*log(2)) + 2*x*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {8\,x+2\,\ln \left (\frac {16\,x^2}{9}\right )+6\,x^2+4}{x^2\,{\ln \left (\frac {16\,x^2}{9}\right )}^2+\ln \left (\frac {16\,x^2}{9}\right )\,\left (2\,x^4+4\,x^3\right )+4\,x^4+4\,x^5+x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + 2*log((16*x^2)/9) + 6*x^2 + 4)/(x^2*log((16*x^2)/9)^2 + log((16*x^2)/9)*(4*x^3 + 2*x^4) + 4*x^4 +
4*x^5 + x^6),x)

[Out]

int(-(8*x + 2*log((16*x^2)/9) + 6*x^2 + 4)/(x^2*log((16*x^2)/9)^2 + log((16*x^2)/9)*(4*x^3 + 2*x^4) + 4*x^4 +
4*x^5 + x^6), x)

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sympy [A]  time = 0.15, size = 19, normalized size = 0.79 \begin {gather*} \frac {2}{x^{3} + 2 x^{2} + x \log {\left (\frac {16 x^{2}}{9} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(16/9*x**2)-6*x**2-8*x-4)/(x**2*ln(16/9*x**2)**2+(2*x**4+4*x**3)*ln(16/9*x**2)+x**6+4*x**5+4*x
**4),x)

[Out]

2/(x**3 + 2*x**2 + x*log(16*x**2/9))

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