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

Optimal. Leaf size=14 \[ \frac {\tanh ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \]

[Out]

1/2*arctanh(x*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {28, 21, 206} \[ \frac {\tanh ^{-1}\left (\sqrt {2} x\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sqrt[2]*x]/Sqrt[2]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

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Mathematica [B]  time = 0.01, size = 32, normalized size = 2.29 \[ \frac {\log \left (2 x+\sqrt {2}\right )-\log \left (\sqrt {2}-2 x\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-Log[Sqrt[2] - 2*x] + Log[Sqrt[2] + 2*x])/(2*Sqrt[2])

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fricas [B]  time = 0.40, size = 29, normalized size = 2.07 \[ \frac {1}{4} \, \sqrt {2} \log \left (\frac {2 \, x^{2} + 2 \, \sqrt {2} x + 1}{2 \, x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 29, normalized size = 2.07 \[ \frac {1}{4} \, \sqrt {2} \log \left ({\left | x + \frac {1}{2} \, \sqrt {2} \right |}\right ) - \frac {1}{4} \, \sqrt {2} \log \left ({\left | x - \frac {1}{2} \, \sqrt {2} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(abs(x + 1/2*sqrt(2))) - 1/4*sqrt(2)*log(abs(x - 1/2*sqrt(2)))

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maple [A]  time = 0.00, size = 12, normalized size = 0.86 \[ \frac {\sqrt {2}\, \arctanh \left (\sqrt {2}\, x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctanh(2^(1/2)*x)*2^(1/2)

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maxima [B]  time = 2.35, size = 25, normalized size = 1.79 \[ -\frac {1}{4} \, \sqrt {2} \log \left (\frac {2 \, x - \sqrt {2}}{2 \, x + \sqrt {2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*log((2*x - sqrt(2))/(2*x + sqrt(2)))

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mupad [B]  time = 4.33, size = 11, normalized size = 0.79 \[ \frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*atanh(2^(1/2)*x))/2

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sympy [B]  time = 0.11, size = 32, normalized size = 2.29 \[ - \frac {\sqrt {2} \log {\left (x - \frac {\sqrt {2}}{2} \right )}}{4} + \frac {\sqrt {2} \log {\left (x + \frac {\sqrt {2}}{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*log(x - sqrt(2)/2)/4 + sqrt(2)*log(x + sqrt(2)/2)/4

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