3.3.51 \(\int \frac {1}{\sqrt {-1-x^2} \sqrt {2+2 x^2}} \, dx\) [251]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 28, 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 = {23, 209} \begin {gather*} \frac {\sqrt {x^2+1} \text {ArcTan}(x)}{\sqrt {2} \sqrt {-x^2-1}} \end {gather*}

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 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*}

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Mathematica [A]
time = 0.01, size = 26, normalized size = 0.93 \begin {gather*} \frac {\left (1+x^2\right ) \tan ^{-1}(x)}{\sqrt {2} \sqrt {-\left (1+x^2\right )^2}} \end {gather*}

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 = 0.08, size = 24, normalized size = 0.86

method	result	size
meijerg	\(-\frac {i \sqrt {2}\, \arctan \left (x \right )}{2}\)	\(9\)
default	\(-\frac {\sqrt {-x^{2}-1}\, \sqrt {2}\, \arctan \left (x \right )}{2 \sqrt {x^{2}+1}}\)	\(24\)
risch	\(\frac {\sqrt {\frac {\left (-x^{2}-1\right ) \left (2 x^{2}+2\right )}{\left (x^{2}+1\right )^{2}}}\, \left (x^{2}+1\right ) \left (-\frac {i \sqrt {-2}\, \ln \left (x +i\right )}{4}+\frac {i \sqrt {-2}\, \ln \left (x -i\right )}{4}\right )}{\sqrt {-x^{2}-1}\, \sqrt {2 x^{2}+2}}\)	\(72\)

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,method=_RETURNVERBOSE)

[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 = 0.00, 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/(-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] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (23) = 46\).
time = 1.07, size = 104, normalized size = 3.71 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\sqrt {2} \int \frac {1}{\sqrt {- x^{2} - 1} \sqrt {x^{2} + 1}}\, dx}{2} \end {gather*}

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 = 0.00, 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/(-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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\sqrt {-x^2-1}\,\sqrt {2\,x^2+2}} \,d x \end {gather*}

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]

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

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