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

Optimal. Leaf size=27 \[ \frac {x+1}{2 \left (1-(x+1)^2\right )}+\frac {1}{2} \tanh ^{-1}(x+1) \]

[Out]

1/2*(1+x)/(1-(1+x)^2)+1/2*arctanh(1+x)

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {247, 199, 206} \[ \frac {x+1}{2 \left (1-(x+1)^2\right )}+\frac {1}{2} \tanh ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 - (1 + x)^2)^(-2),x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.96 \[ \frac {1}{4} \left (-\frac {2 (x+1)}{x (x+2)}-\log (x)+\log (x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - (1 + x)^2)^(-2),x]

[Out]

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

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fricas [A]  time = 0.79, size = 39, normalized size = 1.44 \[ \frac {{\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) - {\left (x^{2} + 2 \, x\right )} \log \relax (x) - 2 \, x - 2}{4 \, {\left (x^{2} + 2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 27, normalized size = 1.00 \[ -\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 24, normalized size = 0.89 \[ -\frac {\ln \relax (x )}{4}+\frac {\ln \left (x +2\right )}{4}-\frac {1}{4 x}-\frac {1}{4 \left (x +2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 25, normalized size = 0.93 \[ -\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left (x + 2\right ) - \frac {1}{4} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 23, normalized size = 0.85 \[ \frac {\mathrm {atanh}\left (x+1\right )}{2}-\frac {x+1}{2\,\left ({\left (x+1\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x + 1)^2 - 1)^2,x)

[Out]

atanh(x + 1)/2 - (x + 1)/(2*((x + 1)^2 - 1))

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sympy [A]  time = 0.12, size = 24, normalized size = 0.89 \[ \frac {- x - 1}{2 x^{2} + 4 x} - \frac {\log {\relax (x )}}{4} + \frac {\log {\left (x + 2 \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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