3.224 \(\int \frac{1}{\sqrt{2-2 x^2} \sqrt{1-x^2}} \, dx\)

Optimal. Leaf size=8 \[ \frac{\tanh ^{-1}(x)}{\sqrt{2}} \]

[Out]

ArcTanh[x]/Sqrt[2]

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Rubi [A]  time = 0.0022928, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {22, 206} \[ \frac{\tanh ^{-1}(x)}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 - 2*x^2]*Sqrt[1 - x^2]),x]

[Out]

ArcTanh[x]/Sqrt[2]

Rule 22

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m + n
), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] && GtQ[b/d, 0] &&  !(IntegerQ[m] || IntegerQ[n]
)

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{2-2 x^2} \sqrt{1-x^2}} \, dx &=\frac{\int \frac{1}{1-x^2} \, dx}{\sqrt{2}}\\ &=\frac{\tanh ^{-1}(x)}{\sqrt{2}}\\ \end{align*}

Mathematica [B]  time = 0.0048565, size = 26, normalized size = 3.25 \[ -\frac{\frac{1}{2} \log (1-x)-\frac{1}{2} \log (x+1)}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 - 2*x^2]*Sqrt[1 - x^2]),x]

[Out]

-((Log[1 - x]/2 - Log[1 + x]/2)/Sqrt[2])

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Maple [A]  time = 0.04, size = 8, normalized size = 1. \begin{align*}{\frac{{\it Artanh} \left ( x \right ) \sqrt{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-x^{2} + 1} \sqrt{-2 \, x^{2} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.98489, size = 169, normalized size = 21.12 \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (-\frac{x^{6} + 5 \, x^{4} - 2 \, \sqrt{2}{\left (x^{3} + x\right )} \sqrt{-x^{2} + 1} \sqrt{-2 \, x^{2} + 2} - 5 \, x^{2} - 1}{x^{6} - 3 \, x^{4} + 3 \, x^{2} - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(-(x^6 + 5*x^4 - 2*sqrt(2)*(x^3 + x)*sqrt(-x^2 + 1)*sqrt(-2*x^2 + 2) - 5*x^2 - 1)/(x^6 - 3*x^4
+ 3*x^2 - 1))

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Sympy [A]  time = 2.66071, size = 22, normalized size = 2.75 \begin{align*} - \sqrt{2} \left (\begin{cases} - \frac{\operatorname{acoth}{\left (x \right )}}{2} & \text{for}\: x^{2} > 1 \\- \frac{\operatorname{atanh}{\left (x \right )}}{2} & \text{for}\: x^{2} < 1 \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*Piecewise((-acoth(x)/2, x**2 > 1), (-atanh(x)/2, x**2 < 1))

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Giac [B]  time = 1.0861, size = 26, normalized size = 3.25 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (x + 1\right ) - \frac{1}{4} \, \sqrt{2} \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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