3.3.29 \(\int \frac {1}{\sqrt {1+x^2} \sqrt {2+4 x^2}} \, dx\) [229]

Optimal. Leaf size=49 \[ \frac {\sqrt {1+2 x^2} F\left (\left .\tan ^{-1}(x)\right |-1\right )}{\sqrt {2} \sqrt {1+x^2} \sqrt {\frac {1+2 x^2}{1+x^2}}} \]

[Out]

1/2*(1/(x^2+1))^(1/2)*EllipticF(x/(x^2+1)^(1/2),I)*(2*x^2+1)^(1/2)*2^(1/2)/((2*x^2+1)/(x^2+1))^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {429} \begin {gather*} \frac {\sqrt {2 x^2+1} F(\text {ArcTan}(x)|-1)}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {2 x^2+1}{x^2+1}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 + x^2]*Sqrt[2 + 4*x^2]),x]

[Out]

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

Rule 429

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*
Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1+x^2} \sqrt {2+4 x^2}} \, dx &=\frac {\sqrt {1+2 x^2} F\left (\left .\tan ^{-1}(x)\right |-1\right )}{\sqrt {2} \sqrt {1+x^2} \sqrt {\frac {1+2 x^2}{1+x^2}}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.07, size = 17, normalized size = 0.35 \begin {gather*} -\frac {i F\left (\left .i \sinh ^{-1}(x)\right |2\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 + x^2]*Sqrt[2 + 4*x^2]),x]

[Out]

((-I)*EllipticF[I*ArcSinh[x], 2])/Sqrt[2]

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Maple [A]
time = 0.08, size = 15, normalized size = 0.31

method	result	size
default	\(-\frac {i \EllipticF \left (i x , \sqrt {2}\right ) \sqrt {2}}{2}\)	\(15\)
elliptic	\(-\frac {i \sqrt {\left (x^{2}+1\right ) \left (2 x^{2}+1\right )}\, \EllipticF \left (i x , \sqrt {2}\right )}{\sqrt {4 x^{4}+6 x^{2}+2}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2+1)^(1/2)/(4*x^2+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*I*EllipticF(I*x,2^(1/2))*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/(x^2+1)^(1/2)/(4*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(4*x^2 + 2)*sqrt(x^2 + 1)), x)

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Fricas [A]
time = 0.34, size = 10, normalized size = 0.20 \begin {gather*} -\frac {1}{2} i \, \sqrt {2} {\rm ellipticF}\left (i \, x, 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2+1)^(1/2)/(4*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*I*sqrt(2)*ellipticF(I*x, 2)

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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 {2 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)/(4*x**2+2)**(1/2),x)

[Out]

sqrt(2)*Integral(1/(sqrt(x**2 + 1)*sqrt(2*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)/(4*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(4*x^2 + 2)*sqrt(x^2 + 1)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {x^2+1}\,\sqrt {4\,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)*(4*x^2 + 2)^(1/2)),x)

[Out]

int(1/((x^2 + 1)^(1/2)*(4*x^2 + 2)^(1/2)), x)

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