3.248 \(\int \frac {1}{\sqrt {-1-x^2} \sqrt {2+5 x^2}} \, dx\)

Optimal. Leaf size=53 \[ \frac {\sqrt {5 x^2+2} \operatorname {EllipticF}\left (\tan ^{-1}(x),-\frac {3}{2}\right )}{\sqrt {2} \sqrt {-x^2-1} \sqrt {\frac {5 x^2+2}{x^2+1}}} \]

[Out]

1/2*(1/(x^2+1))^(1/2)*(x^2+1)^(1/2)*EllipticF(x/(x^2+1)^(1/2),1/2*I*6^(1/2))*(5*x^2+2)^(1/2)*2^(1/2)/(-x^2-1)^
(1/2)/((5*x^2+2)/(x^2+1))^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {418} \[ \frac {\sqrt {5 x^2+2} F\left (\tan ^{-1}(x)|-\frac {3}{2}\right )}{\sqrt {2} \sqrt {-x^2-1} \sqrt {\frac {5 x^2+2}{x^2+1}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-1 - x^2]*Sqrt[2 + 5*x^2]),x]

[Out]

(Sqrt[2 + 5*x^2]*EllipticF[ArcTan[x], -3/2])/(Sqrt[2]*Sqrt[-1 - x^2]*Sqrt[(2 + 5*x^2)/(1 + x^2)])

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), 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+5 x^2}} \, dx &=\frac {\sqrt {2+5 x^2} F\left (\tan ^{-1}(x)|-\frac {3}{2}\right )}{\sqrt {2} \sqrt {-1-x^2} \sqrt {\frac {2+5 x^2}{1+x^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 39, normalized size = 0.74 \[ -\frac {i \sqrt {x^2+1} \operatorname {EllipticF}\left (i \sinh ^{-1}(x),\frac {5}{2}\right )}{\sqrt {2} \sqrt {-x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[-1 - x^2]*Sqrt[2 + 5*x^2]),x]

[Out]

((-I)*Sqrt[1 + x^2]*EllipticF[I*ArcSinh[x], 5/2])/(Sqrt[2]*Sqrt[-1 - x^2])

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {5 \, x^{2} + 2} \sqrt {-x^{2} - 1}}{5 \, x^{4} + 7 \, x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-1)^(1/2)/(5*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(5*x^2 + 2)*sqrt(-x^2 - 1)/(5*x^4 + 7*x^2 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {5 \, x^{2} + 2} \sqrt {-x^{2} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-1)^(1/2)/(5*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(5*x^2 + 2)*sqrt(-x^2 - 1)), x)

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maple [A]  time = 0.03, size = 36, normalized size = 0.68 \[ \frac {i \sqrt {5}\, \sqrt {-x^{2}-1}\, \EllipticF \left (\frac {i \sqrt {10}\, x}{2}, \frac {\sqrt {10}}{5}\right )}{5 \sqrt {x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-x^2-1)^(1/2)/(5*x^2+2)^(1/2),x)

[Out]

1/5*I*EllipticF(1/2*I*10^(1/2)*x,1/5*10^(1/2))/(x^2+1)^(1/2)*5^(1/2)*(-x^2-1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {5 \, x^{2} + 2} \sqrt {-x^{2} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x^2-1)^(1/2)/(5*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(5*x^2 + 2)*sqrt(-x^2 - 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {-x^2-1}\,\sqrt {5\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((- x^2 - 1)^(1/2)*(5*x^2 + 2)^(1/2)),x)

[Out]

int(1/((- x^2 - 1)^(1/2)*(5*x^2 + 2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- x^{2} - 1} \sqrt {5 x^{2} + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-x**2-1)**(1/2)/(5*x**2+2)**(1/2),x)

[Out]

Integral(1/(sqrt(-x**2 - 1)*sqrt(5*x**2 + 2)), x)

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