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]

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

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Rubi [A]  time = 0.0149195, antiderivative size = 31, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0046045, size = 31, normalized size = 1. \[ \frac{1}{4} \log \left (2 x^2+2 x+1\right )-\frac{1}{4} \log \left (2 x^2-2 x+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x^2)/(1 + 4*x^4),x]

[Out]

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

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Maple [A]  time = 0.046, size = 28, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( 2\,{x}^{2}-2\,x+1 \right ) }{4}}+{\frac{\ln \left ( 2\,{x}^{2}+2\,x+1 \right ) }{4}} \end{align*}

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 = 0.978012, size = 36, normalized size = 1.16 \begin{align*} \frac{1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac{1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \end{align*}

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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Fricas [A]  time = 1.39473, size = 72, normalized size = 2.32 \begin{align*} \frac{1}{4} \, \log \left (2 \, x^{2} + 2 \, x + 1\right ) - \frac{1}{4} \, \log \left (2 \, x^{2} - 2 \, x + 1\right ) \end{align*}

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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Sympy [A]  time = 0.099135, size = 22, normalized size = 0.71 \begin{align*} - \frac{\log{\left (x^{2} - x + \frac{1}{2} \right )}}{4} + \frac{\log{\left (x^{2} + x + \frac{1}{2} \right )}}{4} \end{align*}

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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Giac [A]  time = 1.1064, size = 46, normalized size = 1.48 \begin{align*} \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 ) \end{align*}

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)