3.3.24 \(\int \frac {1}{\sqrt {2-2 x^2} \sqrt {1-x^2}} \, dx\) [224]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 8, normalized size of antiderivative = 1.00, 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, 212} \begin {gather*} \frac {\tanh ^{-1}(x)}{\sqrt {2}} \end {gather*}

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 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])

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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(26\) vs. \(2(8)=16\).
time = 0.01, size = 26, normalized size = 3.25 \begin {gather*} -\frac {\frac {1}{2} \log (1-x)-\frac {1}{2} \log (1+x)}{\sqrt {2}} \end {gather*}

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 = 0.08, size = 8, normalized size = 1.00

method	result	size
meijerg	\(\frac {\arctanh \left (x \right ) \sqrt {2}}{2}\)	\(8\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (7) = 14\).
time = 1.25, size = 68, normalized size = 8.50 \begin {gather*} \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 {gather*}

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 = 0.92, size = 22, normalized size = 2.75 \begin {gather*} - \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 {gather*}

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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.12 \begin {gather*} \int \frac {1}{\sqrt {1-x^2}\,\sqrt {2-2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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