\(\int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [802]

Optimal result

Integrand size = 29, antiderivative size = 80 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {2 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)}}+\frac {2 A b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {b \cos (c+d x)}} \] Output:

2*B*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c 
)^(1/2)+2*A*b*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d/(b 
*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.69 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {2 b \sqrt {\cos (c+d x)} \left (B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )\right )}{d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 2030, 3227, 3042, 3121, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\sqrt {b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b \left (A \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {B \int \sqrt {b \cos (c+d x)}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (A \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {B \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b \left (\frac {A \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b \left (\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {b \cos (c+d x)}}+\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )\)

Input:

Int[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x],x]
 

Output:

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

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3119

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

rule 3120

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]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(160\) vs. \(2(75)=150\).

Time = 7.86 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.01

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\)	\(161\)
parts	\(-\frac {2 A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}+\frac {2 B \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\)	\(286\)
risch	\(-\frac {i B \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}}{d}-\frac {i \left (\frac {i A \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} b}}+B \left (-\frac {2 \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}{b \sqrt {{\mathrm e}^{i \left (d x +c \right )} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} b}}\right )\right ) \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{-i \left (d x +c \right )}}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \,{\mathrm e}^{i \left (d x +c \right )}}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(408\)

Input:

int((cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c),x,method=_RETURNVERBO 
SE)
 

Output:

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(A*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(2*sin(1/2 
*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/sin(1/2*d*x+1/2*c)/((2*cos(1/2* 
d*x+1/2*c)^2-1)*b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.50 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {\frac {1}{2}} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {\frac {1}{2}} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )\right )}}{d} \] Input:

integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="f 
ricas")
 

Output:

-2*(I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin( 
d*x + c)) - I*sqrt(1/2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) 
- I*sin(d*x + c)) - I*sqrt(1/2)*B*sqrt(b)*weierstrassZeta(-4, 0, weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt(1/2)*B*sqrt(b)* 
weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c))))/d
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="m 
axima")
                                                                                    
                                                                                    
 

Output:

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

Giac [F]

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

integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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