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

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

[Out]

1/2*arctan(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 = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {22, 203} \[ \frac {\tan ^{-1}(x)}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.00, size = 8, normalized size = 1.00 \[ \frac {\tan ^{-1}(x)}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[x]/Sqrt[2]

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fricas [B]  time = 1.04, size = 34, normalized size = 4.25 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {2 \, x^{2} + 2} \sqrt {x^{2} + 1} x}{x^{4} - 1}\right ) \]

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]

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

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giac [B]  time = 0.57, size = 26, normalized size = 3.25 \[ \frac {1}{4} \, \sqrt {2} i \log \left (i x - 1\right ) - \frac {1}{4} \, \sqrt {2} i \log \left (-i x - 1\right ) \]

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]

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

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maple [A]  time = 0.31, size = 8, normalized size = 1.00 \[ \frac {\sqrt {2}\, \arctan \relax (x )}{2} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{2} + 2} \sqrt {x^{2} + 1}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.12 \[ \int \frac {1}{\sqrt {x^2+1}\,\sqrt {2\,x^2+2}} \,d x \]

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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sympy [A]  time = 2.49, size = 8, normalized size = 1.00 \[ \frac {\sqrt {2} \operatorname {atan}{\relax (x )}}{2} \]

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)*atan(x)/2

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