3.68.32 \(\int -\frac {4}{(1+2 x+x^2) \log (\log (5 \log (6)))} \, dx\)

Optimal. Leaf size=18 \[ \frac {4 x}{\left (x+x^2\right ) \log (\log (5 \log (6)))} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 27, 32} \begin {gather*} \frac {4}{(x+1) \log (\log (5 \log (6)))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4/((1 + x)*Log[Log[5*Log[6]]])

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {4 \int \frac {1}{1+2 x+x^2} \, dx}{\log (\log (5 \log (6)))}\\ &=-\frac {4 \int \frac {1}{(1+x)^2} \, dx}{\log (\log (5 \log (6)))}\\ &=\frac {4}{(1+x) \log (\log (5 \log (6)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.83 \begin {gather*} \frac {4}{(1+x) \log (\log (5 \log (6)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-4/((1 + 2*x + x^2)*Log[Log[5*Log[6]]]),x]

[Out]

4/((1 + x)*Log[Log[5*Log[6]]])

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fricas [A]  time = 0.46, size = 15, normalized size = 0.83 \begin {gather*} \frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \relax (6)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/(x^2+2*x+1)/log(log(5*log(6))),x, algorithm="fricas")

[Out]

4/((x + 1)*log(log(5*log(6))))

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giac [A]  time = 1.17, size = 15, normalized size = 0.83 \begin {gather*} \frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \relax (6)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/(x^2+2*x+1)/log(log(5*log(6))),x, algorithm="giac")

[Out]

4/((x + 1)*log(log(5*log(6))))

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maple [A]  time = 0.09, size = 16, normalized size = 0.89

method	result	size
gosper	\(\frac {4}{\ln \left (\ln \left (5 \ln \relax (6)\right )\right ) \left (x +1\right )}\)	\(16\)
default	\(\frac {4}{\ln \left (\ln \left (5 \ln \relax (6)\right )\right ) \left (x +1\right )}\)	\(16\)
norman	\(\frac {4}{\ln \left (\ln \relax (5)+\ln \left (\ln \relax (6)\right )\right ) \left (x +1\right )}\)	\(17\)
meijerg	\(-\frac {4 x}{\ln \left (\ln \left (5 \ln \relax (6)\right )\right ) \left (x +1\right )}\)	\(17\)
risch	\(\frac {4}{\ln \left (\ln \left (5 \ln \relax (2)+5 \ln \relax (3)\right )\right ) \left (x +1\right )}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4/(x^2+2*x+1)/ln(ln(5*ln(6))),x,method=_RETURNVERBOSE)

[Out]

4/ln(ln(5*ln(6)))/(x+1)

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maxima [A]  time = 0.38, size = 15, normalized size = 0.83 \begin {gather*} \frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \relax (6)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/(x^2+2*x+1)/log(log(5*log(6))),x, algorithm="maxima")

[Out]

4/((x + 1)*log(log(5*log(6))))

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mupad [B]  time = 0.06, size = 15, normalized size = 0.83 \begin {gather*} \frac {4}{\ln \left (\ln \left (5\,\ln \relax (6)\right )\right )\,\left (x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-4/(log(log(5*log(6)))*(2*x + x^2 + 1)),x)

[Out]

4/(log(log(5*log(6)))*(x + 1))

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sympy [A]  time = 0.14, size = 22, normalized size = 1.22 \begin {gather*} \frac {4}{x \log {\left (\log {\left (\log {\relax (6 )} \right )} + \log {\relax (5 )} \right )} + \log {\left (\log {\left (\log {\relax (6 )} \right )} + \log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-4/(x**2+2*x+1)/ln(ln(5*ln(6))),x)

[Out]

4/(x*log(log(log(6)) + log(5)) + log(log(log(6)) + log(5)))

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