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

Optimal. Leaf size=25 \[ \frac{1}{2} \log \left (x^2+x+1\right )-\frac{1}{2} \log \left (x^2-x+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.0134529, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1164, 628} \[ \frac{1}{2} \log \left (x^2+x+1\right )-\frac{1}{2} \log \left (x^2-x+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x^2)/(1 + x^2 + x^4),x]

[Out]

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

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

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

Mathematica [A]  time = 0.0057009, size = 25, normalized size = 1. \[ \frac{1}{2} \log \left (x^2+x+1\right )-\frac{1}{2} \log \left (x^2-x+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 22, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ({x}^{2}-x+1 \right ) }{2}}+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.986035, size = 28, normalized size = 1.12 \begin{align*} \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.34951, size = 61, normalized size = 2.44 \begin{align*} \frac{1}{2} \, \log \left (x^{2} + x + 1\right ) - \frac{1}{2} \, \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.097416, size = 19, normalized size = 0.76 \begin{align*} - \frac{\log{\left (x^{2} - x + 1 \right )}}{2} + \frac{\log{\left (x^{2} + x + 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12756, size = 47, normalized size = 1.88 \begin{align*} \frac{1}{4} \, \log \left ({\left | x + \frac{1}{x + \frac{1}{x}} + \frac{1}{x} + 2 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | x + \frac{1}{x + \frac{1}{x}} + \frac{1}{x} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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