\(\int -\frac {4}{(1+2 x+x^2) \log (\log (5 \log (6)))} \, dx\) [6732]

Optimal result

Integrand size = 20, antiderivative size = 18 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4 x}{\left (x+x^2\right ) \log (\log (5 \log (6)))} \]

[Out]

4*x/(x^2+x)/ln(ln(5*ln(6)))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 27, 32} \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{(x+1) \log (\log (5 \log (6)))} \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{(1+x) \log (\log (5 \log (6)))} \]

[In]

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

[Out]

4/((1 + x)*Log[Log[5*Log[6]]])

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \left (6\right )\right )\right )} \]

[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))))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{x \log {\left (\log {\left (\log {\left (6 \right )} \right )} + \log {\left (5 \right )} \right )} + \log {\left (\log {\left (\log {\left (6 \right )} \right )} + \log {\left (5 \right )} \right )}} \]

[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)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \left (6\right )\right )\right )} \]

[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))))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{{\left (x + 1\right )} \log \left (\log \left (5 \, \log \left (6\right )\right )\right )} \]

[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))))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int -\frac {4}{\left (1+2 x+x^2\right ) \log (\log (5 \log (6)))} \, dx=\frac {4}{\ln \left (\ln \left (5\,\ln \left (6\right )\right )\right )\,\left (x+1\right )} \]

[In]

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

[Out]

4/(log(log(5*log(6)))*(x + 1))