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

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

[Out]

x/(2*(1 - x^2)) - ArcTanh[x]/2

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Rubi [A]  time = 0.004329, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {73, 288, 207} \[ \frac{x}{2 \left (1-x^2\right )}-\frac{1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^2/((-1 + x)^2*(1 + x)^2),x]

[Out]

x/(2*(1 - x^2)) - ArcTanh[x]/2

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 288

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

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

Mathematica [A]  time = 0.0108394, size = 27, normalized size = 1.29 \[ \frac{1}{4} \left (-\frac{2 x}{x^2-1}+\log (1-x)-\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 28, normalized size = 1.3 \begin{align*} -{\frac{1}{4+4\,x}}-{\frac{\ln \left ( 1+x \right ) }{4}}-{\frac{1}{-4+4\,x}}+{\frac{\ln \left ( -1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03331, size = 31, normalized size = 1.48 \begin{align*} -\frac{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(x^2/(-1+x)^2/(1+x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.05455, 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 ) + 2 \, x}{4 \,{\left (x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.146852, size = 20, normalized size = 0.95 \begin{align*} - \frac{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(x**2/(-1+x)**2/(1+x)**2,x)

[Out]

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

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Giac [B]  time = 1.158, size = 46, normalized size = 2.19 \begin{align*} -\frac{1}{4 \,{\left (x + 1\right )}} + \frac{1}{8 \,{\left (\frac{2}{x + 1} - 1\right )}} + \frac{1}{4} \, \log \left ({\left | -\frac{2}{x + 1} + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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