3.63 \(\int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx\)

Optimal. Leaf size=42 \[ \frac {\sqrt {\frac {b \cos ^2(x)}{a}+1} F\left (x+\frac {\pi }{2}|-\frac {b}{a}\right )}{\sqrt {a+b \cos ^2(x)}} \]

[Out]

-(sin(x)^2)^(1/2)/sin(x)*EllipticF(cos(x),(-b/a)^(1/2))*(1+b*cos(x)^2/a)^(1/2)/(a+b*cos(x)^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3183, 3182} \[ \frac {\sqrt {\frac {b \cos ^2(x)}{a}+1} F\left (x+\frac {\pi }{2}|-\frac {b}{a}\right )}{\sqrt {a+b \cos ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Cos[x]^2],x]

[Out]

(Sqrt[1 + (b*Cos[x]^2)/a]*EllipticF[Pi/2 + x, -(b/a)])/Sqrt[a + b*Cos[x]^2]

Rule 3182

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1*EllipticF[e + f*x, -(b/a)])/(Sqrt[a]*
f), x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3183

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + (b*Sin[e + f*x]^2)/a]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx &=\frac {\sqrt {1+\frac {b \cos ^2(x)}{a}} \int \frac {1}{\sqrt {1+\frac {b \cos ^2(x)}{a}}} \, dx}{\sqrt {a+b \cos ^2(x)}}\\ &=\frac {\sqrt {1+\frac {b \cos ^2(x)}{a}} F\left (\frac {\pi }{2}+x|-\frac {b}{a}\right )}{\sqrt {a+b \cos ^2(x)}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 46, normalized size = 1.10 \[ \frac {\sqrt {\frac {2 a+b \cos (2 x)+b}{a+b}} F\left (x\left |\frac {b}{a+b}\right .\right )}{\sqrt {2 a+b \cos (2 x)+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Cos[x]^2],x]

[Out]

(Sqrt[(2*a + b + b*Cos[2*x])/(a + b)]*EllipticF[x, b/(a + b)])/Sqrt[2*a + b + b*Cos[2*x]]

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {b \cos \relax (x)^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(b*cos(x)^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cos \relax (x)^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cos(x)^2 + a), x)

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maple [A]  time = 0.70, size = 48, normalized size = 1.14 \[ -\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {\frac {a +b \left (\cos ^{2}\relax (x )\right )}{a}}\, \EllipticF \left (\cos \relax (x ), \sqrt {-\frac {b}{a}}\right )}{\sin \relax (x ) \sqrt {a +b \left (\cos ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(x)^2)^(1/2),x)

[Out]

-(sin(x)^2)^(1/2)*((a+b*cos(x)^2)/a)^(1/2)*EllipticF(cos(x),(-1/a*b)^(1/2))/sin(x)/(a+b*cos(x)^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \cos \relax (x)^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cos(x)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {b\,{\cos \relax (x)}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(x)^2)^(1/2),x)

[Out]

int(1/(a + b*cos(x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b \cos ^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*cos(x)**2), x)

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