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

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Rubi [A]  time = 0.0226469, antiderivative size = 38, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.02488, size = 38, normalized size = 1. \[ \frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{b \sqrt{c \cos (a+b x)}} \]

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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Maple [C]  time = 0.204, size = 54, normalized size = 1.4 \begin{align*} 2\,{\frac{\sqrt{2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it InverseJacobiAM} \left ( 1/2\,bx+a/2,\sqrt{2} \right ) }{b\sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c \cos \left (b x + a\right )}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c \cos \left (b x + a\right )}}{c \cos \left (b x + a\right )}, x\right ) \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c \cos \left (b x + a\right )}}\,{d x} \end{align*}

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)