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]

1/4*(1+x)/(1-(1+x)^2)^2+3/8*(1+x)/(1-(1+x)^2)+3/8*arctanh(1+x)

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [B]  time = 0.83, size = 71, normalized size = 1.58 \[ -\frac {6 \, x^{3} + 18 \, x^{2} - 3 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \left (x + 2\right ) + 3 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} \log \relax (x) + 8 \, x - 4}{16 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} \]

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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giac [A]  time = 0.34, size = 39, normalized size = 0.87 \[ -\frac {3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \, {\left (x^{2} + 2 \, x\right )}^{2}} + \frac {3}{16} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {3}{16} \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 36, normalized size = 0.80 \[ -\frac {3 \ln \relax (x )}{16}+\frac {3 \ln \left (x +2\right )}{16}-\frac {3}{16 x}+\frac {1}{16 x^{2}}-\frac {1}{16 \left (x +2\right )^{2}}-\frac {3}{16 \left (x +2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.64, size = 44, normalized size = 0.98 \[ -\frac {3 \, x^{3} + 9 \, x^{2} + 4 \, x - 2}{8 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )}} + \frac {3}{16} \, \log \left (x + 2\right ) - \frac {3}{16} \, \log \relax (x) \]

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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mupad [B]  time = 2.09, size = 36, normalized size = 0.80 \[ \frac {3\,\mathrm {atanh}\left (x+1\right )}{8}+\frac {\frac {5\,x}{8}-\frac {3\,{\left (x+1\right )}^3}{8}+\frac {5}{8}}{{\left (x+1\right )}^4-2\,{\left (x+1\right )}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((x + 1)^2 - 1)^3,x)

[Out]

(3*atanh(x + 1))/8 + ((5*x)/8 - (3*(x + 1)^3)/8 + 5/8)/((x + 1)^4 - 2*(x + 1)^2 + 1)

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sympy [A]  time = 0.14, size = 44, normalized size = 0.98 \[ - \frac {3 \log {\relax (x )}}{16} + \frac {3 \log {\left (x + 2 \right )}}{16} - \frac {3 x^{3} + 9 x^{2} + 4 x - 2}{8 x^{4} + 32 x^{3} + 32 x^{2}} \]

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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