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

Optimal. Leaf size=21 \[ \frac{5 x}{2 \left (1-x^2\right )}+\frac{1}{2} \tanh ^{-1}(x) \]

[Out]

(5*x)/(2*(1 - x^2)) + ArcTanh[x]/2

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Rubi [A]  time = 0.0050243, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {28, 385, 207} \[ \frac{5 x}{2 \left (1-x^2\right )}+\frac{1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x^2)/(1 - 2*x^2 + x^4),x]

[Out]

(5*x)/(2*(1 - x^2)) + ArcTanh[x]/2

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 207

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

Rubi steps

\begin{align*} \int \frac{3+2 x^2}{1-2 x^2+x^4} \, dx &=\int \frac{3+2 x^2}{\left (-1+x^2\right )^2} \, dx\\ &=\frac{5 x}{2 \left (1-x^2\right )}-\frac{1}{2} \int \frac{1}{-1+x^2} \, dx\\ &=\frac{5 x}{2 \left (1-x^2\right )}+\frac{1}{2} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.009797, size = 27, normalized size = 1.29 \[ \frac{1}{4} \left (-\frac{10 x}{x^2-1}-\log (1-x)+\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-10*x)/(-1 + x^2) - Log[1 - x] + Log[1 + x])/4

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Maple [A]  time = 0.05, size = 28, normalized size = 1.3 \begin{align*} -{\frac{5}{4+4\,x}}+{\frac{\ln \left ( 1+x \right ) }{4}}-{\frac{5}{-4+4\,x}}-{\frac{\ln \left ( -1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4/(1+x)+1/4*ln(1+x)-5/4/(-1+x)-1/4*ln(-1+x)

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Maxima [A]  time = 0.988093, size = 31, normalized size = 1.48 \begin{align*} -\frac{5 \, x}{2 \,{\left (x^{2} - 1\right )}} + \frac{1}{4} \, \log \left (x + 1\right ) - \frac{1}{4} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*x/(x^2 - 1) + 1/4*log(x + 1) - 1/4*log(x - 1)

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Fricas [B]  time = 1.29165, size = 92, normalized size = 4.38 \begin{align*} \frac{{\left (x^{2} - 1\right )} \log \left (x + 1\right ) -{\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 10 \, x}{4 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((x^2 - 1)*log(x + 1) - (x^2 - 1)*log(x - 1) - 10*x)/(x^2 - 1)

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Sympy [A]  time = 0.095955, size = 22, normalized size = 1.05 \begin{align*} - \frac{5 x}{2 x^{2} - 2} - \frac{\log{\left (x - 1 \right )}}{4} + \frac{\log{\left (x + 1 \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*x/(2*x**2 - 2) - log(x - 1)/4 + log(x + 1)/4

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Giac [A]  time = 1.16189, size = 34, normalized size = 1.62 \begin{align*} -\frac{5 \, x}{2 \,{\left (x^{2} - 1\right )}} + \frac{1}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*x/(x^2 - 1) + 1/4*log(abs(x + 1)) - 1/4*log(abs(x - 1))