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

Optimal. Leaf size=45 \[ \frac{3 (x+1)}{8 \left (1-(x+1)^2\right )}+\frac{x+1}{4 \left (1-(x+1)^2\right )^2}+\frac{3}{8} \tanh ^{-1}(x+1) \]

[Out]

(1 + x)/(4*(1 - (1 + x)^2)^2) + (3*(1 + x))/(8*(1 - (1 + x)^2)) + (3*ArcTanh[1 + x])/8

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Rubi [A]  time = 0.0122762, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {247, 199, 206} \[ \frac{3 (x+1)}{8 \left (1-(x+1)^2\right )}+\frac{x+1}{4 \left (1-(x+1)^2\right )^2}+\frac{3}{8} \tanh ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 - (1 + x)^2)^(-3),x]

[Out]

(1 + x)/(4*(1 - (1 + x)^2)^2) + (3*(1 + x))/(8*(1 - (1 + x)^2)) + (3*ArcTanh[1 + x])/8

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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{1}{\left (1-(1+x)^2\right )^3} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3} \, dx,x,1+x\right )\\ &=\frac{1+x}{4 \left (1-(1+x)^2\right )^2}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,1+x\right )\\ &=\frac{1+x}{4 \left (1-(1+x)^2\right )^2}+\frac{3 (1+x)}{8 \left (1-(1+x)^2\right )}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,1+x\right )\\ &=\frac{1+x}{4 \left (1-(1+x)^2\right )^2}+\frac{3 (1+x)}{8 \left (1-(1+x)^2\right )}+\frac{3}{8} \tanh ^{-1}(1+x)\\ \end{align*}

Mathematica [A]  time = 0.0174647, size = 37, normalized size = 0.82 \[ \frac{1}{16} \left (\frac{1}{x^2}-\frac{3}{x}-\frac{3}{x+2}-\frac{1}{(x+2)^2}-3 \log (x)+3 \log (x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - (1 + x)^2)^(-3),x]

[Out]

(x^(-2) - 3/x - (2 + x)^(-2) - 3/(2 + x) - 3*Log[x] + 3*Log[2 + x])/16

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Maple [A]  time = 0.009, size = 36, normalized size = 0.8 \begin{align*}{\frac{1}{16\,{x}^{2}}}-{\frac{3}{16\,x}}-{\frac{3\,\ln \left ( x \right ) }{16}}-{\frac{1}{16\, \left ( 2+x \right ) ^{2}}}-{\frac{3}{32+16\,x}}+{\frac{3\,\ln \left ( 2+x \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/x^2-3/16/x-3/16*ln(x)-1/16/(2+x)^2-3/16/(2+x)+3/16*ln(2+x)

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Maxima [A]  time = 1.18182, size = 59, normalized size = 1.31 \begin{align*} -\frac{3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} + \frac{3}{16} \, \log \left (x + 2\right ) - \frac{3}{16} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*x^3 + 9*x^2 + 4*x - 2)/(x^4 + 4*x^3 + 4*x^2) + 3/16*log(x + 2) - 3/16*log(x)

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Fricas [B]  time = 1.7459, size = 170, normalized size = 3.78 \begin{align*} -\frac{6 \, x^{3} + 18 \, x^{2} - 3 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x + 2\right ) + 3 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x\right ) + 8 \, x - 4}{16 \,{\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*x^3 + 18*x^2 - 3*(x^4 + 4*x^3 + 4*x^2)*log(x + 2) + 3*(x^4 + 4*x^3 + 4*x^2)*log(x) + 8*x - 4)/(x^4 +
4*x^3 + 4*x^2)

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Sympy [A]  time = 0.138179, size = 44, normalized size = 0.98 \begin{align*} - \frac{3 \log{\left (x \right )}}{16} + \frac{3 \log{\left (x + 2 \right )}}{16} - \frac{3 x^{3} + 9 x^{2} + 4 x - 2}{8 x^{4} + 32 x^{3} + 32 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x)/16 + 3*log(x + 2)/16 - (3*x**3 + 9*x**2 + 4*x - 2)/(8*x**4 + 32*x**3 + 32*x**2)

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Giac [A]  time = 1.14151, size = 53, normalized size = 1.18 \begin{align*} -\frac{3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \,{\left (x^{2} + 2 \, x\right )}^{2}} + \frac{3}{16} \, \log \left ({\left | x + 2 \right |}\right ) - \frac{3}{16} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*x^3 + 9*x^2 + 4*x - 2)/(x^2 + 2*x)^2 + 3/16*log(abs(x + 2)) - 3/16*log(abs(x))