3.1.95 \(\int \frac {1}{1-(1+x)^2} \, dx\) [95]

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

[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 = {253, 212} \begin {gather*} \tanh ^{-1}(x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - (1 + x)^2)^(-1),x]

[Out]

ArcTanh[1 + x]

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*} \int \frac {1}{1-(1+x)^2} \, dx &=\text {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] Leaf count is larger than twice the leaf count of optimal. \(15\) vs. \(2(4)=8\).
time = 0.00, size = 15, normalized size = 3.75 \begin {gather*} -\frac {\log (x)}{2}+\frac {1}{2} \log (2+x) \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(11\) vs. \(2(4)=8\).
time = 0.23, size = 12, normalized size = 3.00

method	result	size
default	\(\frac {\ln \left (x +2\right )}{2}-\frac {\ln \left (x \right )}{2}\)	\(12\)
norman	\(\frac {\ln \left (x +2\right )}{2}-\frac {\ln \left (x \right )}{2}\)	\(12\)
risch	\(\frac {\ln \left (x +2\right )}{2}-\frac {\ln \left (x \right )}{2}\)	\(12\)
meijerg	\(\frac {\ln \left (1+\frac {x}{2}\right )}{2}-\frac {\ln \left (x \right )}{2}+\frac {\ln \left (2\right )}{2}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (4) = 8\).
time = 0.28, size = 11, normalized size = 2.75 \begin {gather*} \frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x\right ) \end {gather*}

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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 11 vs. \(2 (4) = 8\).
time = 0.36, size = 11, normalized size = 2.75 \begin {gather*} \frac {1}{2} \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x\right ) \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 10 vs. \(2 (3) = 6\).
time = 0.03, size = 10, normalized size = 2.50 \begin {gather*} - \frac {\log {\left (x \right )}}{2} + \frac {\log {\left (x + 2 \right )}}{2} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 13 vs. \(2 (4) = 8\).
time = 2.64, size = 13, normalized size = 3.25 \begin {gather*} \frac {1}{2} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \end {gather*}

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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Mupad [B]
time = 0.15, size = 4, normalized size = 1.00 \begin {gather*} \mathrm {atanh}\left (x+1\right ) \end {gather*}

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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