\(\int \frac {1}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \, dx\) [216]

Optimal result

Integrand size = 24, antiderivative size = 87 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \, dx=\frac {\sqrt {c} \sqrt {1+\frac {b x^2}{a}} \sqrt {1-\frac {d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {a+b x^2} \sqrt {c-d x^2}} \]

[Out]

EllipticF(x*d^(1/2)/c^(1/2),(-b*c/a/d)^(1/2))*c^(1/2)*(1+b*x^2/a)^(1/2)*(1-d*x^2/c)^(1/2)/d^(1/2)/(b*x^2+a)^(1
/2)/(-d*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {432, 430} \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \, dx=\frac {\sqrt {c} \sqrt {\frac {b x^2}{a}+1} \sqrt {1-\frac {d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {a+b x^2} \sqrt {c-d x^2}} \]

[In]

Int[1/(Sqrt[a + b*x^2]*Sqrt[c - d*x^2]),x]

[Out]

(Sqrt[c]*Sqrt[1 + (b*x^2)/a]*Sqrt[1 - (d*x^2)/c]*EllipticF[ArcSin[(Sqrt[d]*x)/Sqrt[c]], -((b*c)/(a*d))])/(Sqrt
[d]*Sqrt[a + b*x^2]*Sqrt[c - d*x^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])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-\frac {d x^2}{c}} \int \frac {1}{\sqrt {a+b x^2} \sqrt {1-\frac {d x^2}{c}}} \, dx}{\sqrt {c-d x^2}} \\ & = \frac {\left (\sqrt {1+\frac {b x^2}{a}} \sqrt {1-\frac {d x^2}{c}}\right ) \int \frac {1}{\sqrt {1+\frac {b x^2}{a}} \sqrt {1-\frac {d x^2}{c}}} \, dx}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \\ & = \frac {\sqrt {c} \sqrt {1+\frac {b x^2}{a}} \sqrt {1-\frac {d x^2}{c}} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|-\frac {b c}{a d}\right )}{\sqrt {d} \sqrt {a+b x^2} \sqrt {c-d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \, dx=\frac {\sqrt {\frac {a+b x^2}{a}} \sqrt {\frac {c-d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {b}{a}} x\right ),-\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}} \sqrt {a+b x^2} \sqrt {c-d x^2}} \]

[In]

Integrate[1/(Sqrt[a + b*x^2]*Sqrt[c - d*x^2]),x]

[Out]

(Sqrt[(a + b*x^2)/a]*Sqrt[(c - d*x^2)/c]*EllipticF[ArcSin[Sqrt[-(b/a)]*x], -((a*d)/(b*c))])/(Sqrt[-(b/a)]*Sqrt
[a + b*x^2]*Sqrt[c - d*x^2])

Maple [A] (verified)

Time = 3.60 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {F\left (x \sqrt {\frac {d}{c}}, \sqrt {-\frac {b c}{a d}}\right ) \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {\frac {-d \,x^{2}+c}{c}}\, \sqrt {b \,x^{2}+a}\, \sqrt {-d \,x^{2}+c}}{\sqrt {\frac {d}{c}}\, \left (-b d \,x^{4}-a d \,x^{2}+c b \,x^{2}+a c \right )}\)	\(103\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (-d \,x^{2}+c \right )}\, \sqrt {1-\frac {d \,x^{2}}{c}}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, F\left (x \sqrt {\frac {d}{c}}, \sqrt {-1-\frac {-a d +b c}{a d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {-d \,x^{2}+c}\, \sqrt {\frac {d}{c}}\, \sqrt {-b d \,x^{4}-a d \,x^{2}+c b \,x^{2}+a c}}\)	\(127\)

[In]

int(1/(b*x^2+a)^(1/2)/(-d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

EllipticF(x*(d/c)^(1/2),(-b*c/a/d)^(1/2))*((b*x^2+a)/a)^(1/2)*((-d*x^2+c)/c)^(1/2)*(b*x^2+a)^(1/2)*(-d*x^2+c)^
(1/2)/(d/c)^(1/2)/(-b*d*x^4-a*d*x^2+b*c*x^2+a*c)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c-d x^2}} \, dx=\frac {\sqrt {a c} \sqrt {\frac {d}{c}} F(\arcsin \left (x \sqrt {\frac {d}{c}}\right )\,|\,-\frac {b c}{a d})}{a d} \]

[In]

integrate(1/(b*x^2+a)^(1/2)/(-d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

sqrt(a*c)*sqrt(d/c)*elliptic_f(arcsin(x*sqrt(d/c)), -b*c/(a*d))/(a*d)

Sympy [F]

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

[In]

integrate(1/(b*x**2+a)**(1/2)/(-d*x**2+c)**(1/2),x)

[Out]

Integral(1/(sqrt(a + b*x**2)*sqrt(c - d*x**2)), x)

Maxima [F]

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

[In]

integrate(1/(b*x^2+a)^(1/2)/(-d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^2 + a)*sqrt(-d*x^2 + c)), x)

Giac [F]

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

[In]

integrate(1/(b*x^2+a)^(1/2)/(-d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*x^2 + a)*sqrt(-d*x^2 + c)), x)

Mupad [F(-1)]

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

[In]

int(1/((a + b*x^2)^(1/2)*(c - d*x^2)^(1/2)),x)

[Out]

int(1/((a + b*x^2)^(1/2)*(c - d*x^2)^(1/2)), x)