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

Optimal. Leaf size=31 \[ \frac {1}{4} \log \left (2 x^2+2 x+1\right )-\frac {1}{4} \log \left (2 x^2-2 x+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1165, 628} \[ \frac {1}{4} \log \left (2 x^2+2 x+1\right )-\frac {1}{4} \log \left (2 x^2-2 x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x^2)/(1 + 4*x^4),x]

[Out]

-Log[1 - 2*x + 2*x^2]/4 + Log[1 + 2*x + 2*x^2]/4

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 31, normalized size = 1.00 \[ \frac {1}{4} \log \left (2 x^2+2 x+1\right )-\frac {1}{4} \log \left (2 x^2-2 x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x^2)/(1 + 4*x^4),x]

[Out]

-1/4*Log[1 - 2*x + 2*x^2] + Log[1 + 2*x + 2*x^2]/4

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fricas [A]  time = 0.40, size = 27, normalized size = 0.87 \[ \frac {1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 34, normalized size = 1.10 \[ \frac {1}{4} \, \log \left (x^{2} + \sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} x + \frac {1}{2}\right ) - \frac {1}{4} \, \log \left (x^{2} - \sqrt {2} \left (\frac {1}{4}\right )^{\frac {1}{4}} x + \frac {1}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 28, normalized size = 0.90 \[ -\frac {\ln \left (2 x^{2}-2 x +1\right )}{4}+\frac {\ln \left (2 x^{2}+2 x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.06, size = 27, normalized size = 0.87 \[ \frac {1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac {1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 15, normalized size = 0.48 \[ \frac {\mathrm {atanh}\left (\frac {2\,x}{2\,x^2+1}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^2 - 1)/(4*x^4 + 1),x)

[Out]

atanh((2*x)/(2*x^2 + 1))/2

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sympy [A]  time = 0.11, size = 22, normalized size = 0.71 \[ - \frac {\log {\left (x^{2} - x + \frac {1}{2} \right )}}{4} + \frac {\log {\left (x^{2} + x + \frac {1}{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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