\(\int \frac {1}{\sqrt {\cos (a+b x)}} \, dx\) [13]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2720} \[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{b} \]

[In]

Int[1/Sqrt[Cos[a + b*x]],x]

[Out]

(2*EllipticF[(a + b*x)/2, 2])/b

Rule 2720

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{b} \]

[In]

Integrate[1/Sqrt[Cos[a + b*x]],x]

[Out]

(2*EllipticF[(a + b*x)/2, 2])/b

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {2 \,\operatorname {am}^{-1}\left (\frac {b x}{2}+\frac {a}{2}| \sqrt {2}\right )}{b}\)	\(18\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.19 \[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\frac {-i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{b} \]

[In]

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

[Out]

(-I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + I*sqrt(2)*weierstrassPInverse(-4, 0, c
os(b*x + a) - I*sin(b*x + a)))/b

Sympy [F]

\[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\int \frac {1}{\sqrt {\cos {\left (a + b x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(cos(a + b*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\int { \frac {1}{\sqrt {\cos \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(cos(b*x + a)), x)

Giac [F]

\[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\int { \frac {1}{\sqrt {\cos \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(cos(b*x + a)), x)

Mupad [B] (verification not implemented)

Time = 13.59 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx=\frac {2\,\mathrm {F}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{b} \]

[In]

int(1/cos(a + b*x)^(1/2),x)

[Out]

(2*ellipticF(a/2 + (b*x)/2, 2))/b