\(\int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx\) [218]

Optimal result

Integrand size = 23, antiderivative size = 12 \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\frac {\operatorname {EllipticF}\left (\arcsin (x),-\frac {5}{2}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {430} \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\frac {\operatorname {EllipticF}\left (\arcsin (x),-\frac {5}{2}\right )}{\sqrt {2}} \]

[In]

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

[Out]

EllipticF[ArcSin[x], -5/2]/Sqrt[2]

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rubi steps \begin{align*} \text {integral}& = \frac {F\left (\sin ^{-1}(x)|-\frac {5}{2}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\frac {\operatorname {EllipticF}\left (\arcsin (x),-\frac {5}{2}\right )}{\sqrt {2}} \]

[In]

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

[Out]

EllipticF[ArcSin[x], -5/2]/Sqrt[2]

Maple [A] (verified)

Time = 3.60 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {F\left (x , \frac {i \sqrt {10}}{2}\right ) \sqrt {2}}{2}\)	\(14\)
elliptic	\(\frac {\sqrt {-\left (x^{2}-1\right ) \left (5 x^{2}+2\right )}\, \sqrt {10 x^{2}+4}\, F\left (x , \frac {i \sqrt {10}}{2}\right )}{2 \sqrt {5 x^{2}+2}\, \sqrt {-5 x^{4}+3 x^{2}+2}}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\frac {1}{2} \, \sqrt {2} F(\arcsin \left (x\right )\,|\,-\frac {5}{2}) \]

[In]

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

[Out]

1/2*sqrt(2)*elliptic_f(arcsin(x), -5/2)

Sympy [A] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\begin {cases} \frac {\sqrt {2} F\left (\operatorname {asin}{\left (x \right )}\middle | - \frac {5}{2}\right )}{2} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(2)*elliptic_f(asin(x), -5/2)/2, (x > -1) & (x < 1)))

Maxima [F]

\[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\int { \frac {1}{\sqrt {5 \, x^{2} + 2} \sqrt {-x^{2} + 1}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\int { \frac {1}{\sqrt {5 \, x^{2} + 2} \sqrt {-x^{2} + 1}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {1-x^2} \sqrt {2+5 x^2}} \, dx=\int \frac {1}{\sqrt {1-x^2}\,\sqrt {5\,x^2+2}} \,d x \]

[In]

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

[Out]

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