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

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

[Out]

2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*
cos(d*x+c))^(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} \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*Cos[c + d*x]],x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.00 \[ \frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*Cos[c + d*x]],x]

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \cos \left (d x + c\right )}}{b \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*cos(d*x + c)), x)

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mupad [B]  time = 0.19, size = 33, normalized size = 0.87 \[ \frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d\,\sqrt {b\,\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*cos(c + d*x))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b*cos(c + d*x)), x)

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