3.34.63 \(\int \frac {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{x^2-2 x \log (4)+\log ^2(4)} \, dx\)

Optimal. Leaf size=18 \[ 2 x+\frac {-3-x}{x-\log (4)} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {27, 1850} \begin {gather*} 2 x-\frac {3+\log (4)}{x-\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x^2 + (1 - 4*x)*Log[4] + 2*Log[4]^2)/(x^2 - 2*x*Log[4] + Log[4]^2),x]

[Out]

2*x - (3 + Log[4])/(x - Log[4])

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 {3+2 x^2+(1-4 x) \log (4)+2 \log ^2(4)}{(x-\log (4))^2} \, dx\\ &=\int \left (2+\frac {3+\log (4)}{(x-\log (4))^2}\right ) \, dx\\ &=2 x-\frac {3+\log (4)}{x-\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.33 \begin {gather*} \frac {-3-\log (4)}{x-\log (4)}+2 (x-\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*x^2 + (1 - 4*x)*Log[4] + 2*Log[4]^2)/(x^2 - 2*x*Log[4] + Log[4]^2),x]

[Out]

(-3 - Log[4])/(x - Log[4]) + 2*(x - Log[4])

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fricas [A]  time = 0.66, size = 25, normalized size = 1.39 \begin {gather*} \frac {2 \, x^{2} - 2 \, {\left (2 \, x + 1\right )} \log \relax (2) - 3}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="fricas")

[Out]

(2*x^2 - 2*(2*x + 1)*log(2) - 3)/(x - 2*log(2))

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giac [A]  time = 0.27, size = 20, normalized size = 1.11 \begin {gather*} 2 \, x - \frac {2 \, \log \relax (2) + 3}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="giac")

[Out]

2*x - (2*log(2) + 3)/(x - 2*log(2))

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maple [A]  time = 0.53, size = 21, normalized size = 1.17

method	result	size
default	\(2 x -\frac {2 \ln \relax (2)+3}{x -2 \ln \relax (2)}\)	\(21\)
risch	\(2 x +\frac {\ln \relax (2)}{\ln \relax (2)-\frac {x}{2}}+\frac {3}{2 \left (\ln \relax (2)-\frac {x}{2}\right )}\)	\(26\)
gosper	\(\frac {-2 x^{2}+3+8 \ln \relax (2)^{2}+2 \ln \relax (2)}{2 \ln \relax (2)-x}\)	\(29\)
norman	\(\frac {-2 x^{2}+3+8 \ln \relax (2)^{2}+2 \ln \relax (2)}{2 \ln \relax (2)-x}\)	\(29\)
meijerg	\(\frac {2 x}{1-\frac {x}{2 \ln \relax (2)}}+\frac {3 x}{4 \ln \relax (2)^{2} \left (1-\frac {x}{2 \ln \relax (2)}\right )}+\frac {x}{2 \left (1-\frac {x}{2 \ln \relax (2)}\right ) \ln \relax (2)}-8 \ln \relax (2) \left (\frac {x}{2 \left (1-\frac {x}{2 \ln \relax (2)}\right ) \ln \relax (2)}+\ln \left (1-\frac {x}{2 \ln \relax (2)}\right )\right )-4 \ln \relax (2) \left (-\frac {x \left (-\frac {3 x}{2 \ln \relax (2)}+6\right )}{6 \ln \relax (2) \left (1-\frac {x}{2 \ln \relax (2)}\right )}-2 \ln \left (1-\frac {x}{2 \ln \relax (2)}\right )\right )\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*ln(2)^2+2*(-4*x+1)*ln(2)+2*x^2+3)/(4*ln(2)^2-4*x*ln(2)+x^2),x,method=_RETURNVERBOSE)

[Out]

2*x-(2*ln(2)+3)/(x-2*ln(2))

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maxima [A]  time = 0.48, size = 20, normalized size = 1.11 \begin {gather*} 2 \, x - \frac {2 \, \log \relax (2) + 3}{x - 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(2)^2+2*(-4*x+1)*log(2)+2*x^2+3)/(4*log(2)^2-4*x*log(2)+x^2),x, algorithm="maxima")

[Out]

2*x - (2*log(2) + 3)/(x - 2*log(2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {8\,{\ln \relax (2)}^2-2\,\ln \relax (2)\,\left (4\,x-1\right )+2\,x^2+3}{x^2-4\,\ln \relax (2)\,x+4\,{\ln \relax (2)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*log(2)^2 - 2*log(2)*(4*x - 1) + 2*x^2 + 3)/(4*log(2)^2 - 4*x*log(2) + x^2),x)

[Out]

int((8*log(2)^2 - 2*log(2)*(4*x - 1) + 2*x^2 + 3)/(4*log(2)^2 - 4*x*log(2) + x^2), x)

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sympy [A]  time = 0.14, size = 17, normalized size = 0.94 \begin {gather*} 2 x + \frac {-3 - 2 \log {\relax (2 )}}{x - 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*ln(2)**2+2*(-4*x+1)*ln(2)+2*x**2+3)/(4*ln(2)**2-4*x*ln(2)+x**2),x)

[Out]

2*x + (-3 - 2*log(2))/(x - 2*log(2))

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