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

Optimal. Leaf size=14 \[ -4+\frac {4 x \log (4)}{x+x^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 9, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 27, 32} \begin {gather*} \frac {4 \log (4)}{x+1} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 9, normalized size = 0.64 \begin {gather*} \frac {4 \log (4)}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Log[4])/(1 + x)

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fricas [A]  time = 0.66, size = 9, normalized size = 0.64 \begin {gather*} \frac {8 \, \log \relax (2)}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*log(2)/(x + 1)

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giac [A]  time = 0.13, size = 9, normalized size = 0.64 \begin {gather*} \frac {8 \, \log \relax (2)}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*log(2)/(x + 1)

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maple [A]  time = 0.46, size = 10, normalized size = 0.71

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 9, normalized size = 0.64 \begin {gather*} \frac {8 \, \log \relax (2)}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*log(2)/(x + 1)

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mupad [B]  time = 0.05, size = 9, normalized size = 0.64 \begin {gather*} \frac {8\,\ln \relax (2)}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*log(2))/(2*x + x^2 + 1),x)

[Out]

(8*log(2))/(x + 1)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.50 \begin {gather*} \frac {8 \log {\relax (2 )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*log(2)/(x + 1)

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