\(\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\) [628]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 109 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d} \]

[Out]

2*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*(
a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2895} \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d} \]

[In]

Int[1/(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]),x]

[Out]

(2*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a
+ b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d)

Rule 2895

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(
Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqrt[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]
*EllipticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2]], -(a + b)/(a - b)], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.56 \[ \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx=-\frac {4 (a+b) \cos ^{\frac {3}{2}}(c+d x) \sqrt {-\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{a-b}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {a+b \cos (c+d x)}{a (-1+\cos (c+d x))}}\right ),\frac {2 a}{a-b}\right )}{a d \sqrt {a+b \cos (c+d x)} \left (-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}\right )^{3/2}} \]

[In]

Integrate[1/(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]),x]

[Out]

(-4*(a + b)*Cos[c + d*x]^(3/2)*Sqrt[-(((a + b)*Cot[(c + d*x)/2]^2)/(a - b))]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c
 + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[-((a + b*Cos[c + d*x])/(a*(-1 + Cos[c + d*x])))]], (2*a)/(
a - b)])/(a*d*Sqrt[a + b*Cos[c + d*x]]*(-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a))^(3/2))

Maple [A] (verified)

Time = 8.49 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.02

method result size
default \(-\frac {2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, F\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}}{d \sqrt {a +\cos \left (d x +c \right ) b}\, \sqrt {\cos \left (d x +c \right )}}\) \(111\)

[In]

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

[Out]

-2/d*(1+cos(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(a+cos(d*x+c)*b)^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(
-(a-b)/(a+b))^(1/2))*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)/cos(d*x+c)^(1/2)

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(b*cos(d*x + c)^2 + a*cos(d*x + c)), x)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(sqrt(a + b*cos(c + d*x))*sqrt(cos(c + d*x))), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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