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

Optimal. Leaf size=16 \[ \frac{1}{2 (x+1)}+\frac{1}{2} \tanh ^{-1}(x) \]

[Out]

1/(2*(1 + x)) + ArcTanh[x]/2

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Rubi [A]  time = 0.0082498, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {77, 207} \[ \frac{1}{2 (x+1)}+\frac{1}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 - x)*(1 + x)^2),x]

[Out]

1/(2*(1 + x)) + ArcTanh[x]/2

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.0087884, size = 24, normalized size = 1.5 \[ \frac{1}{4} \left (\frac{2}{x+1}-\log (1-x)+\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 - x)*(1 + x)^2),x]

[Out]

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

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Maple [A]  time = 0.006, size = 21, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( x-1 \right ) }{4}}+{\frac{1}{2+2\,x}}+{\frac{\ln \left ( 1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5397, size = 27, normalized size = 1.69 \begin{align*} \frac{1}{2 \,{\left (x + 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/(1-x)/(1+x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.40792, size = 80, normalized size = 5. \begin{align*} \frac{{\left (x + 1\right )} \log \left (x + 1\right ) -{\left (x + 1\right )} \log \left (x - 1\right ) + 2}{4 \,{\left (x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.10024, size = 19, normalized size = 1.19 \begin{align*} - \frac{\log{\left (x - 1 \right )}}{4} + \frac{\log{\left (x + 1 \right )}}{4} + \frac{1}{2 x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15128, size = 28, normalized size = 1.75 \begin{align*} \frac{1}{2 \,{\left (x + 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/(1-x)/(1+x)^2,x, algorithm="giac")

[Out]

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