3.38.27 \(\int \frac {-5+(-4-8 x-4 x^2) \log (2)}{1+2 x+x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {4-x}{1+x}-4 x \log (2) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 1850} \begin {gather*} \frac {5}{x+1}-4 x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5/(1 + x) - 4*x*Log[2]

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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5+\left (-4-8 x-4 x^2\right ) \log (2)}{(1+x)^2} \, dx\\ &=\int \left (-\frac {5}{(1+x)^2}-4 \log (2)\right ) \, dx\\ &=\frac {5}{1+x}-4 x \log (2)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 1.24 \begin {gather*} -\frac {-5+\log (16)+x^2 \log (16)+x \log (256)}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + (-4 - 8*x - 4*x^2)*Log[2])/(1 + 2*x + x^2),x]

[Out]

-((-5 + Log[16] + x^2*Log[16] + x*Log[256])/(1 + x))

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fricas [A]  time = 0.46, size = 18, normalized size = 1.06 \begin {gather*} -\frac {4 \, {\left (x^{2} + x\right )} \log \relax (2) - 5}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-8*x-4)*log(2)-5)/(x^2+2*x+1),x, algorithm="fricas")

[Out]

-(4*(x^2 + x)*log(2) - 5)/(x + 1)

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giac [A]  time = 0.18, size = 13, normalized size = 0.76 \begin {gather*} -4 \, x \log \relax (2) + \frac {5}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-8*x-4)*log(2)-5)/(x^2+2*x+1),x, algorithm="giac")

[Out]

-4*x*log(2) + 5/(x + 1)

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maple [A]  time = 0.10, size = 14, normalized size = 0.82

method	result	size
default	\(-4 x \ln \relax (2)+\frac {5}{x +1}\)	\(14\)
risch	\(-4 x \ln \relax (2)+\frac {5}{x +1}\)	\(14\)
norman	\(\frac {-4 x^{2} \ln \relax (2)+5+4 \ln \relax (2)}{x +1}\)	\(20\)
gosper	\(-\frac {4 x^{2} \ln \relax (2)-5-4 \ln \relax (2)}{x +1}\)	\(21\)
meijerg	\(-\frac {5 x}{x +1}-4 \ln \relax (2) \left (\frac {x \left (6+3 x \right )}{3 x +3}-2 \ln \left (x +1\right )\right )-8 \ln \relax (2) \left (-\frac {x}{x +1}+\ln \left (x +1\right )\right )-\frac {4 \ln \relax (2) x}{x +1}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2-8*x-4)*ln(2)-5)/(x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 13, normalized size = 0.76 \begin {gather*} -4 \, x \log \relax (2) + \frac {5}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2-8*x-4)*log(2)-5)/(x^2+2*x+1),x, algorithm="maxima")

[Out]

-4*x*log(2) + 5/(x + 1)

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mupad [B]  time = 0.06, size = 13, normalized size = 0.76 \begin {gather*} \frac {5}{x+1}-4\,x\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(8*x + 4*x^2 + 4) + 5)/(2*x + x^2 + 1),x)

[Out]

5/(x + 1) - 4*x*log(2)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.59 \begin {gather*} - 4 x \log {\relax (2 )} + \frac {5}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2-8*x-4)*ln(2)-5)/(x**2+2*x+1),x)

[Out]

-4*x*log(2) + 5/(x + 1)

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