3.7.76 \(\int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) (A+C \cos ^2(c+d x)) \, dx\) [676]

3.7.76.1 Optimal result
3.7.76.2 Mathematica [A] (verified)
3.7.76.3 Rubi [A] (verified)
3.7.76.4 Maple [B] (verified)
3.7.76.5 Fricas [C] (verification not implemented)
3.7.76.6 Sympy [F(-1)]
3.7.76.7 Maxima [F]
3.7.76.8 Giac [F]
3.7.76.9 Mupad [B] (verification not implemented)

3.7.76.1 Optimal result

Integrand size = 33, antiderivative size = 134 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 b C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \]

output
2/5*a*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE( 
sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*b*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2 
)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*a*C*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+2/21*b*(7*A 
+5*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
 
3.7.76.2 Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.73 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {42 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 b (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (70 A b+65 b C+42 a C \cos (c+d x)+15 b C \cos (2 (c+d x))) \sin (c+d x)}{105 d} \]

input
Integrate[Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2),x 
]
 
output
(42*a*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2] + 10*b*(7*A + 5*C)*EllipticF[( 
c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(70*A*b + 65*b*C + 42*a*C*Cos[c + d*x] 
 + 15*b*C*Cos[2*(c + d*x)])*Sin[c + d*x])/(105*d)
 
3.7.76.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3513, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3513

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.7.76.3.1 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 3513
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3) 
)), x] + Simp[1/(b*(m + 3))   Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c 
*(m + 3) + b*d*(C*(m + 2) + A*(m + 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 
 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 
3.7.76.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(170)=340\).

Time = 19.55 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.99

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (240 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (-168 a C -360 C b \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (140 A b +168 a C +280 C b \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A b -42 a C -80 C b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +25 C b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(401\)
parts \(\text {Expression too large to display}\) \(719\)

input
int((a+cos(d*x+c)*b)*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8*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+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-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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a+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)*Ellipti 
cF(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)/(2*cos(1/2* 
d*x+1/2*c)^2-1)^(1/2)/d
 
3.7.76.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.38 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (15 \, C b \cos \left (d x + c\right )^{2} + 21 \, C a \cos \left (d x + c\right ) + 5 \, {\left (7 \, A + 5 \, C\right )} b\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d} \]

input
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorith 
m="fricas")
 
output
1/105*(-5*I*sqrt(2)*(7*A + 5*C)*b*weierstrassPInverse(-4, 0, cos(d*x + c) 
+ I*sin(d*x + c)) + 5*I*sqrt(2)*(7*A + 5*C)*b*weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c)) + 21*I*sqrt(2)*(5*A + 3*C)*a*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I* 
sqrt(2)*(5*A + 3*C)*a*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c))) + 2*(15*C*b*cos(d*x + c)^2 + 21*C*a*cos(d*x 
+ c) + 5*(7*A + 5*C)*b)*sqrt(cos(d*x + c))*sin(d*x + c))/d
 
3.7.76.6 Sympy [F(-1)]

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

input
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)
 
output
Timed out
 
3.7.76.7 Maxima [F]

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

input
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorith 
m="maxima")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)*sqrt(cos(d*x + c)), 
x)
 
3.7.76.8 Giac [F]

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

input
integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorith 
m="giac")
 
output
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)*sqrt(cos(d*x + c)), 
x)
 
3.7.76.9 Mupad [B] (verification not implemented)

Time = 2.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04 \[ \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2\,A\,b\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,C\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(cos(c + d*x)^(1/2)*(A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)),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*A*a*ellipticE(c/2 + (d*x)/2, 2))/d - (2*C*a*cos(c + d*x)^(7/2)*si 
n(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))