\(\int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx\) [733]

Optimal result

Integrand size = 25, antiderivative size = 155 \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (1+\cos (c+d x))}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \csc (c+d x) \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right ),-\frac {a-b}{a+b}\right ) \sqrt {\sec (c+d x)}}{\sqrt {a+b} d} \] Output:

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

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.94 \[ \int \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \, dx=\frac {2 \sqrt {a+b \cos (c+d x)} \left ((a-b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )+2 b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)}}{(a+b) d \sqrt {\frac {(a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}} \] Input:

Integrate[Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4710, 3042, 3290}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3290

\(\displaystyle -\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\sec (c+d x)} \sqrt {\frac {a (1-\cos (c+d x))}{a+b \cos (c+d x)}} \sqrt {\frac {a (\cos (c+d x)+1)}{a+b \cos (c+d x)}} (a+b \cos (c+d x)) \operatorname {EllipticPi}\left (\frac {b}{a+b},\arcsin \left (\frac {\sqrt {a+b} \sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}\right ),-\frac {a-b}{a+b}\right )}{d \sqrt {a+b}}\)

Input:

Int[Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]],x]
 

Output:

(-2*Sqrt[Cos[c + d*x]]*Sqrt[(a*(1 - Cos[c + d*x]))/(a + b*Cos[c + d*x])]*S 
qrt[(a*(1 + Cos[c + d*x]))/(a + b*Cos[c + d*x])]*(a + b*Cos[c + d*x])*Csc[ 
c + d*x]*EllipticPi[b/(a + b), ArcSin[(Sqrt[a + b]*Sqrt[Cos[c + d*x]])/Sqr 
t[a + b*Cos[c + d*x]]], -((a - b)/(a + b))]*Sqrt[Sec[c + d*x]])/(Sqrt[a + 
b]*d)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3290

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

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 14.20 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {2 \left (\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) a -\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) b +2 b \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \cos \left (d x +c \right )}{d \sqrt {a +\cos \left (d x +c \right ) b}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(179\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(c + d*x))*sqrt(cos(c + d*x)*b + a),x)