\(\int -\frac {4 \log (4)}{1+2 x+x^2} \, dx\) [3376]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=-4+\frac {4 x \log (4)}{x+x^2} \]

[Out]

8*x*ln(2)/(x^2+x)-4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 27, 32} \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {4 \log (4)}{x+1} \]

[In]

Int[(-4*Log[4])/(1 + 2*x + x^2),x]

[Out]

(4*Log[4])/(1 + x)

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}& = -\left ((4 \log (4)) \int \frac {1}{1+2 x+x^2} \, dx\right ) \\ & = -\left ((4 \log (4)) \int \frac {1}{(1+x)^2} \, dx\right ) \\ & = \frac {4 \log (4)}{1+x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {4 \log (4)}{1+x} \]

[In]

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

[Out]

(4*Log[4])/(1 + x)

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {8 \ln \left (2\right )}{1+x}\)	\(10\)
default	\(\frac {8 \ln \left (2\right )}{1+x}\)	\(10\)
norman	\(\frac {8 \ln \left (2\right )}{1+x}\)	\(10\)
risch	\(\frac {8 \ln \left (2\right )}{1+x}\)	\(10\)
parallelrisch	\(\frac {8 \ln \left (2\right )}{1+x}\)	\(10\)
meijerg	\(-\frac {8 \ln \left (2\right ) x}{1+x}\)	\(11\)

[In]

int(-8*ln(2)/(x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

8*ln(2)/(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {8 \, \log \left (2\right )}{x + 1} \]

[In]

integrate(-8*log(2)/(x^2+2*x+1),x, algorithm="fricas")

[Out]

8*log(2)/(x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.50 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {8 \log {\left (2 \right )}}{x + 1} \]

[In]

integrate(-8*ln(2)/(x**2+2*x+1),x)

[Out]

8*log(2)/(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {8 \, \log \left (2\right )}{x + 1} \]

[In]

integrate(-8*log(2)/(x^2+2*x+1),x, algorithm="maxima")

[Out]

8*log(2)/(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {8 \, \log \left (2\right )}{x + 1} \]

[In]

integrate(-8*log(2)/(x^2+2*x+1),x, algorithm="giac")

[Out]

8*log(2)/(x + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int -\frac {4 \log (4)}{1+2 x+x^2} \, dx=\frac {8\,\ln \left (2\right )}{x+1} \]

[In]

int(-(8*log(2))/(2*x + x^2 + 1),x)

[Out]

(8*log(2))/(x + 1)