3.1.63 \(\int \frac {1}{\sqrt {a+b \cos ^2(x)}} \, dx\) [63]

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

[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 = 0.02, 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 = {3262, 3261} \begin {gather*} \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)}} \end {gather*}

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 3261

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

Rule 3262

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 \begin {gather*} \frac {\sqrt {\frac {2 a+b+b \cos (2 x)}{a+b}} F\left (x\left |\frac {b}{a+b}\right .\right )}{\sqrt {2 a+b+b \cos (2 x)}} \end {gather*}

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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Maple [A]
time = 0.27, size = 48, normalized size = 1.14

method	result	size
default	\(-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {\frac {a +b \left (\cos ^{2}\left (x \right )\right )}{a}}\, \EllipticF \left (\cos \left (x \right ), \sqrt {-\frac {b}{a}}\right )}{\sin \left (x \right ) \sqrt {a +b \left (\cos ^{2}\left (x \right )\right )}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [C] Result contains complex when optimal does not.
time = 0.10, size = 276, normalized size = 6.57 \begin {gather*} -\frac {{\left (-2 i \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 i \, a - i \, b\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 i \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 i \, a + i \, b\right )} \sqrt {b} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} - 2 \, a - b}{b}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((-2*I*b*sqrt((a^2 + a*b)/b^2) - 2*I*a - I*b)*sqrt(b)*sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*elliptic_
f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) + I*sin(x))), (8*a^2 + 8*a*b + b^2 + 4*(2*a*b +
b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*I*b*sqrt((a^2 + a*b)/b^2) + 2*I*a + I*b)*sqrt(b)*sqrt((2*b*sqrt((a^2 +
a*b)/b^2) - 2*a - b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) - 2*a - b)/b)*(cos(x) - I*sin(x))),
(8*a^2 + 8*a*b + b^2 + 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + b \cos ^{2}{\left (x \right )}}}\, dx \end {gather*}

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

[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 \begin {gather*} \int \frac {1}{\sqrt {b\,{\cos \left (x\right )}^2+a}} \,d x \end {gather*}

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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