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

Optimal. Leaf size=28 \[ \frac{\sqrt{x^2+1} \tan ^{-1}(x)}{\sqrt{2} \sqrt{-x^2-1}} \]

[Out]

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

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Rubi [A]  time = 0.0035139, antiderivative size = 28, 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 = {23, 203} \[ \frac{\sqrt{x^2+1} \tan ^{-1}(x)}{\sqrt{2} \sqrt{-x^2-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x^2]*ArcTan[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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0096165, size = 26, normalized size = 0.93 \[ \frac{\left (x^2+1\right ) \tan ^{-1}(x)}{\sqrt{2} \sqrt{-\left (x^2+1\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 24, normalized size = 0.9 \begin{align*} -{\frac{\arctan \left ( x \right ) \sqrt{2}}{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/(-x^2-1)^(1/2)/(2*x^2+2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.84894, size = 258, normalized size = 9.21 \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (\frac{2 \,{\left (2 \, \sqrt{2 \, x^{2} + 2} \sqrt{-x^{2} - 1} x + \sqrt{2}{\left (x^{4} - 1\right )}\right )}}{x^{4} + 2 \, x^{2} + 1}\right ) - \frac{1}{8} \, \sqrt{2} \log \left (\frac{2 \,{\left (2 \, \sqrt{2 \, x^{2} + 2} \sqrt{-x^{2} - 1} x - \sqrt{2}{\left (x^{4} - 1\right )}\right )}}{x^{4} + 2 \, x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(2*(2*sqrt(2*x^2 + 2)*sqrt(-x^2 - 1)*x + sqrt(2)*(x^4 - 1))/(x^4 + 2*x^2 + 1)) - 1/8*sqrt(2)*lo
g(2*(2*sqrt(2*x^2 + 2)*sqrt(-x^2 - 1)*x - sqrt(2)*(x^4 - 1))/(x^4 + 2*x^2 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*Integral(1/(sqrt(-x**2 - 1)*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{2 \, x^{2} + 2} \sqrt{-x^{2} - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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