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

Optimal. Leaf size=29 \[ -\frac{\sqrt{x^2-1} \tanh ^{-1}(x)}{\sqrt{2} \sqrt{1-x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0033172, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {23, 207} \[ -\frac{\sqrt{x^2-1} \tanh ^{-1}(x)}{\sqrt{2} \sqrt{1-x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 23

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

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.0105619, size = 40, normalized size = 1.38 \[ \frac{\left (x^2-1\right ) (\log (1-x)-\log (x+1))}{2 \sqrt{2} \sqrt{-\left (x^2-1\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 24, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}{\it Artanh} \left ( x \right ) }{2}\sqrt{-{x}^{2}+1}{\frac{1}{\sqrt{{x}^{2}-1}}}} \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*(-x^2+1)^(1/2)*2^(1/2)/(x^2-1)^(1/2)*arctanh(x)

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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 [A]  time = 1.83012, size = 97, normalized size = 3.34 \begin{align*} \frac{1}{4} \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{x^{2} - 1} \sqrt{-2 \, x^{2} + 2} x}{x^{4} - 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]

1/4*sqrt(2)*arctan(sqrt(2)*sqrt(x^2 - 1)*sqrt(-2*x^2 + 2)*x/(x^4 - 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2} \int \frac{1}{\sqrt{1 - x^{2}} \sqrt{x^{2} - 1}}\, dx}{2} \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)*Integral(1/(sqrt(1 - x**2)*sqrt(x**2 - 1)), x)/2

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Giac [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="giac")

[Out]

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