\(\int \frac {1}{(1-(1+x)^2)^2} \, dx\) [96]

Optimal result

Integrand size = 11, antiderivative size = 27 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1+x}{2 \left (1-(1+x)^2\right )}+\frac {1}{2} \text {arctanh}(1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {253, 205, 212} \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1}{2} \text {arctanh}(x+1)+\frac {x+1}{2 \left (1-(x+1)^2\right )} \]

[In]

Int[(1 - (1 + x)^2)^(-2),x]

[Out]

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

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 253

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*} \text {integral}& = \text {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} \text {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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {1}{4} \left (-\frac {2 (1+x)}{x (2+x)}-\log (x)+\log (2+x)\right ) \]

[In]

Integrate[(1 - (1 + x)^2)^(-2),x]

[Out]

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

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {1}{4 x}-\frac {\ln \left (x \right )}{4}-\frac {1}{4 \left (x +2\right )}+\frac {\ln \left (x +2\right )}{4}\)	\(24\)
norman	\(\frac {-\frac {1}{2}-\frac {x}{2}}{x \left (x +2\right )}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x +2\right )}{4}\)	\(26\)
risch	\(\frac {-\frac {1}{2}-\frac {x}{2}}{x \left (x +2\right )}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x +2\right )}{4}\)	\(26\)
meijerg	\(\frac {3 x}{16 \left (\frac {3 x}{2}+3\right )}+\frac {\ln \left (1+\frac {x}{2}\right )}{4}-\frac {1}{8}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (2\right )}{4}-\frac {1}{4 x}\)	\(34\)
parallelrisch	\(-\frac {\ln \left (x \right ) x^{2}-\ln \left (x +2\right ) x^{2}+2+2 \ln \left (x \right ) x -2 \ln \left (x +2\right ) x +2 x}{4 x \left (x +2\right )}\)	\(43\)

[In]

int(1/(1-(x+1)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {{\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) - {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) - 2 \, x - 2}{4 \, {\left (x^{2} + 2 \, x\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {- x - 1}{2 x^{2} + 4 x} - \frac {\log {\left (x \right )}}{4} + \frac {\log {\left (x + 2 \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=-\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left (x + 2\right ) - \frac {1}{4} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=-\frac {x + 1}{2 \, {\left (x^{2} + 2 \, x\right )}} + \frac {1}{4} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (1-(1+x)^2\right )^2} \, dx=\frac {\mathrm {atanh}\left (x+1\right )}{2}-\frac {x+1}{2\,\left ({\left (x+1\right )}^2-1\right )} \]

[In]

int(1/((x + 1)^2 - 1)^2,x)

[Out]

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