\(\int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [1066]

Optimal result

Integrand size = 41, antiderivative size = 154 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A b+7 a B+5 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 (7 A b+7 a B+5 b C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 (b B+a C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:

2/5*(5*A*a+3*B*b+3*C*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(7*A* 
b+7*B*a+5*C*b)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/21*(7*A*b+7*B*a+ 
5*C*b)*cos(d*x+c)^(1/2)*sin(d*x+c)/d+2/5*(B*b+C*a)*cos(d*x+c)^(3/2)*sin(d* 
x+c)/d+2/7*b*C*cos(d*x+c)^(5/2)*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.76 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {42 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (7 A b+7 a B+5 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (70 A b+70 a B+65 b C+42 (b B+a C) \cos (c+d x)+15 b C \cos (2 (c+d x))) \sin (c+d x)}{105 d} \] Input:

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

Output:

(42*(5*a*A + 3*b*B + 3*a*C)*EllipticE[(c + d*x)/2, 2] + 10*(7*A*b + 7*a*B 
+ 5*b*C)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(70*A*b + 70*a*B + 
 65*b*C + 42*(b*B + a*C)*Cos[c + d*x] + 15*b*C*Cos[2*(c + d*x)])*Sin[c + d 
*x])/(105*d)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3512, 27, 3042, 3502, 27, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

\(\displaystyle \frac {2}{7} \int \frac {1}{2} \sqrt {\cos (c+d x)} \left (7 (b B+a C) \cos ^2(c+d x)+(7 A b+5 C b+7 a B) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \sqrt {\cos (c+d x)} \left (7 (b B+a C) \cos ^2(c+d x)+(7 A b+5 C b+7 a B) \cos (c+d x)+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 (b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(7 A b+5 C b+7 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+7 a A\right )dx+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \cos (c+d x))dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\cos (c+d x)} (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \cos (c+d x))dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 (5 a A+3 b B+3 a C)+5 (7 A b+5 C b+7 a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 (7 a B+7 A b+5 b C) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 (5 a A+3 a C+3 b B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (7 (5 a A+3 a C+3 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 (7 a B+7 A b+5 b C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{d}\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{d}+5 (7 a B+7 A b+5 b C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {14 (a C+b B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

Input:

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

Output:

(2*b*C*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + ((14*(b*B + a*C)*Cos[c + d 
*x]^(3/2)*Sin[c + d*x])/(5*d) + ((14*(5*a*A + 3*b*B + 3*a*C)*EllipticE[(c 
+ d*x)/2, 2])/d + 5*(7*A*b + 7*a*B + 5*b*C)*((2*EllipticF[(c + d*x)/2, 2]) 
/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5)/7
 

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 3115

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

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. \(514\) vs. \(2(141)=282\).

Time = 6.32 (sec) , antiderivative size = 515, normalized size of antiderivative = 3.34

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 (240 C b \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-168 B b -168 C a -360 C b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A b +140 B a +168 B b +168 C a +280 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 (-70 A b -70 B a -42 B b -42 C a -80 C b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A 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 )-105 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 ) a +35 B 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 )-63 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 ) b +25 C 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 )-63 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 ) a \right )}{105 \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}\)	\(515\)
parts	\(\text {Expression too large to display}\)	\(729\)

Input:

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

Output:

-2/105*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-168*B*b-168*C*a-360*C*b)*sin(1/2* 
d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b+140*B*a+168*B*b+168*C*a+280*C*b)* 
sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*b-70*B*a-42*B*b-42*C*a-80*C 
*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*A*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 
))-105*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)*Ell 
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a+35*B*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))-63*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(c 
os(1/2*d*x+1/2*c),2^(1/2))*b+25*C*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))-63*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))*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)/(-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 = 0.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.44 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, C b \cos \left (d x + c\right )^{2} + 35 \, B a + 5 \, {\left (7 \, A + 5 \, C\right )} b + 21 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (7 i \, B a + i \, {\left (7 \, A + 5 \, C\right )} b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-7 i \, B a - i \, {\left (7 \, A + 5 \, C\right )} b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 3 \, C\right )} a - 3 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, \sqrt {2} {\left (i \, {\left (5 \, A + 3 \, C\right )} a + 3 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{105 \, d} \] Input:

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

Output:

1/105*(2*(15*C*b*cos(d*x + c)^2 + 35*B*a + 5*(7*A + 5*C)*b + 21*(C*a + B*b 
)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 5*sqrt(2)*(7*I*B*a + I*( 
7*A + 5*C)*b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 
5*sqrt(2)*(-7*I*B*a - I*(7*A + 5*C)*b)*weierstrassPInverse(-4, 0, cos(d*x 
+ c) - I*sin(d*x + c)) - 21*sqrt(2)*(-I*(5*A + 3*C)*a - 3*I*B*b)*weierstra 
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 
 21*sqrt(2)*(I*(5*A + 3*C)*a + 3*I*B*b)*weierstrassZeta(-4, 0, weierstrass 
PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

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

integrate(cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^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 = 1.27 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.40 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2\,A\,b\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,B\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(2*A*b*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 
*d) + (2*B*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2 
)))/(3*d) + (2*A*a*ellipticE(c/2 + (d*x)/2, 2))/d - (2*B*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)) - (2*C*a*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)) - (2*C*b*cos(c + 
 d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d 
*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}d x \right ) a^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x \right ) b c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) b^{2} \] Input:

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

Output:

int(sqrt(cos(c + d*x)),x)*a**2 + 2*int(sqrt(cos(c + d*x))*cos(c + d*x),x)* 
a*b + int(sqrt(cos(c + d*x))*cos(c + d*x)**3,x)*b*c + int(sqrt(cos(c + d*x 
))*cos(c + d*x)**2,x)*a*c + int(sqrt(cos(c + d*x))*cos(c + d*x)**2,x)*b**2