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

Optimal. Leaf size=4 \[ \tanh ^{-1}(x+1) \]

[Out]

arctanh(1+x)

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Rubi [A]  time = 0.00, antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {247, 206} \[ \tanh ^{-1}(x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 - (1 + x)^2)^(-1),x]

[Out]

ArcTanh[1 + x]

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

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Mathematica [B]  time = 0.00, size = 15, normalized size = 3.75 \[ \frac {1}{2} \log (x+2)-\frac {\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - (1 + x)^2)^(-1),x]

[Out]

-1/2*Log[x] + Log[2 + x]/2

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fricas [B]  time = 0.89, size = 11, normalized size = 2.75 \[ \frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x + 2) - 1/2*log(x)

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giac [B]  time = 0.35, size = 13, normalized size = 3.25 \[ \frac {1}{2} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(x + 2)) - 1/2*log(abs(x))

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maple [B]  time = 0.00, size = 12, normalized size = 3.00 \[ -\frac {\ln \relax (x )}{2}+\frac {\ln \left (x +2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.57, size = 11, normalized size = 2.75 \[ \frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x + 2) - 1/2*log(x)

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mupad [B]  time = 0.15, size = 4, normalized size = 1.00 \[ \mathrm {atanh}\left (x+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((x + 1)^2 - 1),x)

[Out]

atanh(x + 1)

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sympy [B]  time = 0.10, size = 10, normalized size = 2.50 \[ - \frac {\log {\relax (x )}}{2} + \frac {\log {\left (x + 2 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)/2 + log(x + 2)/2

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