\(\int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [348]

Optimal result

Integrand size = 31, antiderivative size = 75 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 (A b+a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 (3 a A+b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

2*(A*b+B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(3*A*a+B*b)*Invers 
eJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*b*B*cos(d*x+c)^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (3 (A b+a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(3 a A+b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+b B \sqrt {\cos (c+d x)} \sin (c+d x)\right )}{3 d} \] Input:

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

Output:

(2*(3*(A*b + a*B)*EllipticE[(c + d*x)/2, 2] + (3*a*A + b*B)*EllipticF[(c + 
 d*x)/2, 2] + b*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x]))/(3*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

\(\displaystyle \int \frac {(a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {2}{3} \int \frac {3 a A+b B+3 (A b+a B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 a A+b B+3 (A b+a B) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 a A+b B+3 (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{3} \left ((3 a A+b B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 (a B+A b) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left ((3 a A+b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 (a B+A b) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{3} \left ((3 a A+b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 (a B+A b) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{3} \left (\frac {2 (3 a A+b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 (a B+A b) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b B \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

Input:

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

Output:

((6*(A*b + a*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(3*a*A + b*B)*EllipticF[ 
(c + d*x)/2, 2])/d)/3 + (2*b*B*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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]
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(72)=144\).

Time = 7.00 (sec) , antiderivative size = 326, normalized size of antiderivative = 4.35

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +3 A a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b -2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +B b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a \right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(326\)
parts	\(\frac {2 \left (A b +B a \right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}+\frac {2 A a \,\operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d}-\frac {2 B b \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(340\)

Input:

int((a+cos(d*x+c)*b)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*B*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b+3*A*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(2*s 
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*b-2*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b+B*b* 
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co 
s(1/2*d*x+1/2*c),2^(1/2))-3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(c 
os(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a)/(-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)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, B b \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A a - i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, A a + i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a - i \, A b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a + i \, A b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \] Input:

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

Output:

1/3*(2*B*b*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(2)*(-3*I*A*a - I*B*b)*we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*A*a 
 + I*B*b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sq 
rt(2)*(-I*B*a - I*A*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c 
os(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(I*B*a + I*A*b)*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 24.63 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,B\,b\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d} \] Input:

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

Output:

(2*B*b*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 
*d) + (2*A*a*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*b*ellipticE(c/2 + (d*x) 
/2, 2))/d + (2*B*a*ellipticE(c/2 + (d*x)/2, 2))/d
 

Reduce [F]

\[ \int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) b^{2} \] Input:

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

Output:

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