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

Optimal. Leaf size=19 \[ \frac{1}{4} \tanh ^{-1}(x)-\frac{x}{4 \left (x^2+1\right )} \]

[Out]

-x/(4*(1 + x^2)) + ArcTanh[x]/4

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Rubi [A]  time = 0.0125273, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {471, 206} \[ \frac{1}{4} \tanh ^{-1}(x)-\frac{x}{4 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((1 - x^2)*(1 + x^2)^2),x]

[Out]

-x/(4*(1 + x^2)) + ArcTanh[x]/4

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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

Mathematica [A]  time = 0.0112309, size = 27, normalized size = 1.42 \[ \frac{1}{8} \left (-\frac{2 x}{x^2+1}-\log (1-x)+\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*x)/(1 + x^2) - Log[1 - x] + Log[1 + x])/8

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Maple [A]  time = 0.008, size = 24, normalized size = 1.3 \begin{align*} -{\frac{x}{4\,{x}^{2}+4}}-{\frac{\ln \left ( x-1 \right ) }{8}}+{\frac{\ln \left ( 1+x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x/(x^2+1)-1/8*ln(x-1)+1/8*ln(1+x)

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Maxima [A]  time = 1.17111, size = 31, normalized size = 1.63 \begin{align*} -\frac{x}{4 \,{\left (x^{2} + 1\right )}} + \frac{1}{8} \, \log \left (x + 1\right ) - \frac{1}{8} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x/(x^2 + 1) + 1/8*log(x + 1) - 1/8*log(x - 1)

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Fricas [B]  time = 1.43784, size = 90, normalized size = 4.74 \begin{align*} \frac{{\left (x^{2} + 1\right )} \log \left (x + 1\right ) -{\left (x^{2} + 1\right )} \log \left (x - 1\right ) - 2 \, x}{8 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((x^2 + 1)*log(x + 1) - (x^2 + 1)*log(x - 1) - 2*x)/(x^2 + 1)

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Sympy [A]  time = 0.109978, size = 20, normalized size = 1.05 \begin{align*} - \frac{x}{4 x^{2} + 4} - \frac{\log{\left (x - 1 \right )}}{8} + \frac{\log{\left (x + 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(4*x**2 + 4) - log(x - 1)/8 + log(x + 1)/8

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Giac [B]  time = 1.12892, size = 41, normalized size = 2.16 \begin{align*} -\frac{1}{4 \,{\left (x + \frac{1}{x}\right )}} + \frac{1}{16} \, \log \left ({\left | x + \frac{1}{x} + 2 \right |}\right ) - \frac{1}{16} \, \log \left ({\left | x + \frac{1}{x} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(x + 1/x) + 1/16*log(abs(x + 1/x + 2)) - 1/16*log(abs(x + 1/x - 2))