\(\int \frac {(a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1067]

Optimal result

Integrand size = 41, antiderivative size = 116 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 (5 A b+5 a B+3 b C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (b B+a (3 A+C)) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 (b B+a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 b C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

2/5*(5*A*b+5*B*a+3*C*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(B*b+a 
*(3*A+C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*(B*b+C*a)*cos(d*x+c 
)^(1/2)*sin(d*x+c)/d+2/5*b*C*cos(d*x+c)^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.79 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (3 (5 A b+5 a B+3 b C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (3 a A+b B+a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 b B+5 a C+3 b C \cos (c+d x)) \sin (c+d x)\right )}{15 d} \] Input:

Integrate[((a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/S 
qrt[Cos[c + d*x]],x]
 

Output:

(2*(3*(5*A*b + 5*a*B + 3*b*C)*EllipticE[(c + d*x)/2, 2] + 5*(3*a*A + b*B + 
 a*C)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*b*B + 5*a*C + 3*b* 
C*Cos[c + d*x])*Sin[c + d*x]))/(15*d)
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 3512, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {2}{5} \int \frac {5 (b B+a C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 a A}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {5 (b B+a C) \cos ^2(c+d x)+(5 A b+3 C b+5 a B) \cos (c+d x)+5 a A}{\sqrt {\cos (c+d x)}}dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {5 (b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(5 A b+3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+5 a A}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {5 (b B+a (3 A+C))+3 (5 A b+3 C b+5 a B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 (b B+a (3 A+C))+3 (5 A b+3 C b+5 a B) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {5 (b B+a (3 A+C))+3 (5 A b+3 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 (a (3 A+C)+b B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 (5 a B+5 A b+3 b C) \int \sqrt {\cos (c+d x)}dx\right )+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 (a (3 A+C)+b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 (5 a B+5 A b+3 b C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (5 (a (3 A+C)+b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (a (3 A+C)+b B)}{d}+\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a B+5 A b+3 b C)}{d}\right )+\frac {10 (a C+b B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\)

Input:

Int[((a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[Co 
s[c + d*x]],x]
 

Output:

(2*b*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (((6*(5*A*b + 5*a*B + 3*b* 
C)*EllipticE[(c + d*x)/2, 2])/d + (10*(b*B + a*(3*A + C))*EllipticF[(c + d 
*x)/2, 2])/d)/3 + (10*(b*B + a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))/ 
5
 

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 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]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(107)=214\).

Time = 4.17 (sec) , antiderivative size = 465, normalized size of antiderivative = 4.01

method	result	size
default	\(-\frac {2 \sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b +\left (20 B b +20 C a +24 C b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 B b -10 C a -6 C b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 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 )-15 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +5 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 )-15 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +5 C 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 )-9 C \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b \right )}{15 \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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(465\)
parts	\(\frac {2 \left (A b +B a \right ) \sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}-\frac {2 \left (B b +C a \right ) \sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\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 \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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}-\frac {2 C b \sqrt {\left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \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 {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}+\frac {2 a A \,\operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d}\)	\(548\)

Input:

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-2/15*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*C*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b+(20*B*b+20*C*a+24*C*b)*sin(1/2*d*x+1 
/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*B*b-10*C*a-6*C*b)*sin(1/2*d*x+1/2*c)^2*cos 
(1/2*d*x+1/2*c)+15*a*A*(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(cos(1/2*d*x+1/2*c),2^(1/2))-15*A*(sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2))*b+5*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(cos(1/2*d*x+1/2*c),2^(1/2))-15*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a+5 
*C*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))-9*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*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)/(-1+2*cos(1/ 
2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.68 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \, {\left (3 \, C b \cos \left (d x + c\right ) + 5 \, C a + 5 \, B b\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (i \, {\left (3 \, A + C\right )} a + i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-i \, {\left (3 \, A + C\right )} a - i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-5 i \, B a - i \, {\left (5 \, A + 3 \, C\right )} b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (5 i \, B a + i \, {\left (5 \, A + 3 \, C\right )} b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \] Input:

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

Output:

1/15*(2*(3*C*b*cos(d*x + c) + 5*C*a + 5*B*b)*sqrt(cos(d*x + c))*sin(d*x + 
c) - 5*sqrt(2)*(I*(3*A + C)*a + I*B*b)*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) - 5*sqrt(2)*(-I*(3*A + C)*a - I*B*b)*weierstrassPIn 
verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-5*I*B*a - I*(5*A 
 + 3*C)*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
+ I*sin(d*x + c))) - 3*sqrt(2)*(5*I*B*a + I*(5*A + 3*C)*b)*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)/sqr 
t(cos(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)/sqr 
t(cos(d*x + c)), x)
 

Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,B\,b\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,C\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,C\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(((a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/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*C*a*(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 - (2*C*b*cos(c + d 
*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*( 
sin(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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