3.926 \(\int \frac{1}{\sqrt{1-x} (1+x)} \, dx\)

Optimal. Leaf size=23 \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

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Rubi [A]  time = 0.0061987, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {63, 206} \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - x]*(1 + x)),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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])

Rubi steps

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

Mathematica [A]  time = 0.0044313, size = 23, normalized size = 1. \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - x]*(1 + x)),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[1 - x]/Sqrt[2]])

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Maple [A]  time = 0.041, size = 19, normalized size = 0.8 \begin{align*} -{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1-x}} \right ) \sqrt{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.74477, size = 46, normalized size = 2. \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - \sqrt{-x + 1}}{\sqrt{2} + \sqrt{-x + 1}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8582, size = 80, normalized size = 3.48 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (\frac{x + 2 \, \sqrt{2} \sqrt{-x + 1} - 3}{x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log((x + 2*sqrt(2)*sqrt(-x + 1) - 3)/(x + 1))

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Sympy [A]  time = 1.51749, size = 44, normalized size = 1.91 \begin{align*} \begin{cases} - \sqrt{2} \operatorname{acosh}{\left (\frac{\sqrt{2}}{\sqrt{x + 1}} \right )} & \text{for}\: \frac{2}{\left |{x + 1}\right |} > 1 \\\sqrt{2} i \operatorname{asin}{\left (\frac{\sqrt{2}}{\sqrt{x + 1}} \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-sqrt(2)*acosh(sqrt(2)/sqrt(x + 1)), 2/Abs(x + 1) > 1), (sqrt(2)*I*asin(sqrt(2)/sqrt(x + 1)), True)
)

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Giac [B]  time = 1.20572, size = 51, normalized size = 2.22 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (\sqrt{2} + \sqrt{-x + 1}\right ) + \frac{1}{2} \, \sqrt{2} \log \left ({\left | -\sqrt{2} + \sqrt{-x + 1} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*log(sqrt(2) + sqrt(-x + 1)) + 1/2*sqrt(2)*log(abs(-sqrt(2) + sqrt(-x + 1)))