3.21 \(\int \frac {1}{\sqrt {c \cos (a+b x)}} \, dx\)

Optimal. Leaf size=38 \[ \frac {2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b \sqrt {c \cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, 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 = {2642, 2641} \[ \frac {2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b \sqrt {c \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c*Cos[a + b*x]],x]

[Out]

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 38, normalized size = 1.00 \[ \frac {2 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{b \sqrt {c \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c*Cos[a + b*x]],x]

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \cos \left (b x + a\right )}}{c \cos \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*cos(b*x + a))/(c*cos(b*x + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(c*cos(b*x + a)), x)

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maple [C]  time = 0.05, size = 54, normalized size = 1.42 \[ \frac {2 \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \mathrm {am}^{-1}\left (\frac {b x}{2}+\frac {a}{2}| \sqrt {2}\right )}{b \sqrt {c \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b/(c*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)*(2*cos(1/2*b*x+1/2*a)^2-1)^(1/2)*InverseJacobiAM(1/2*b*x+1/2*a,2^(1/2
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(c*cos(b*x + a)), x)

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mupad [B]  time = 0.15, size = 33, normalized size = 0.87 \[ \frac {2\,\sqrt {\cos \left (a+b\,x\right )}\,\mathrm {F}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{b\,\sqrt {c\,\cos \left (a+b\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cos(a + b*x))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c*cos(a + b*x)), x)

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