3.90.3 \(\int \frac {-4+(-160-160 x-40 x^2) \log (2)}{4+4 x+x^2} \, dx\)

Optimal. Leaf size=21 \[ 4 \left (-5-\frac {x}{4+2 x}-10 x \log (2)+\log (5)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.62, 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 {4}{x+2}-40 x \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + (-160 - 160*x - 40*x^2)*Log[2])/(4 + 4*x + x^2),x]

[Out]

4/(2 + x) - 40*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 {-4+\left (-160-160 x-40 x^2\right ) \log (2)}{(2+x)^2} \, dx\\ &=\int \left (-\frac {4}{(2+x)^2}-40 \log (2)\right ) \, dx\\ &=\frac {4}{2+x}-40 x \log (2)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.81 \begin {gather*} -4 \left (-\frac {1}{2+x}+10 (2+x) \log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + (-160 - 160*x - 40*x^2)*Log[2])/(4 + 4*x + x^2),x]

[Out]

-4*(-(2 + x)^(-1) + 10*(2 + x)*Log[2])

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fricas [A]  time = 0.44, size = 20, normalized size = 0.95 \begin {gather*} -\frac {4 \, {\left (10 \, {\left (x^{2} + 2 \, x\right )} \log \relax (2) - 1\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2-160*x-160)*log(2)-4)/(x^2+4*x+4),x, algorithm="fricas")

[Out]

-4*(10*(x^2 + 2*x)*log(2) - 1)/(x + 2)

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giac [A]  time = 0.15, size = 13, normalized size = 0.62 \begin {gather*} -40 \, x \log \relax (2) + \frac {4}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2-160*x-160)*log(2)-4)/(x^2+4*x+4),x, algorithm="giac")

[Out]

-40*x*log(2) + 4/(x + 2)

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maple [A]  time = 0.50, size = 14, normalized size = 0.67

method	result	size
default	\(-40 x \ln \relax (2)+\frac {4}{2+x}\)	\(14\)
risch	\(-40 x \ln \relax (2)+\frac {4}{2+x}\)	\(14\)
norman	\(\frac {-40 x^{2} \ln \relax (2)+4+160 \ln \relax (2)}{2+x}\)	\(20\)
gosper	\(-\frac {4 \left (10 x^{2} \ln \relax (2)-1-40 \ln \relax (2)\right )}{2+x}\)	\(21\)
meijerg	\(-\frac {x}{1+\frac {x}{2}}-80 \ln \relax (2) \left (\frac {x \left (\frac {3 x}{2}+6\right )}{6+3 x}-2 \ln \left (1+\frac {x}{2}\right )\right )-160 \ln \relax (2) \left (-\frac {x}{2 \left (1+\frac {x}{2}\right )}+\ln \left (1+\frac {x}{2}\right )\right )-\frac {40 \ln \relax (2) x}{1+\frac {x}{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x^2-160*x-160)*ln(2)-4)/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

-40*x*ln(2)+4/(2+x)

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maxima [A]  time = 0.38, size = 13, normalized size = 0.62 \begin {gather*} -40 \, x \log \relax (2) + \frac {4}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x^2-160*x-160)*log(2)-4)/(x^2+4*x+4),x, algorithm="maxima")

[Out]

-40*x*log(2) + 4/(x + 2)

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mupad [B]  time = 5.46, size = 13, normalized size = 0.62 \begin {gather*} \frac {4}{x+2}-40\,x\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)*(160*x + 40*x^2 + 160) + 4)/(4*x + x^2 + 4),x)

[Out]

4/(x + 2) - 40*x*log(2)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.48 \begin {gather*} - 40 x \log {\relax (2 )} + \frac {4}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x**2-160*x-160)*ln(2)-4)/(x**2+4*x+4),x)

[Out]

-40*x*log(2) + 4/(x + 2)

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