3.2.31 \(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [131]

3.2.31.1 Optimal result
3.2.31.2 Mathematica [C] (warning: unable to verify)
3.2.31.3 Rubi [A] (verified)
3.2.31.4 Maple [A] (verified)
3.2.31.5 Fricas [C] (verification not implemented)
3.2.31.6 Sympy [F(-1)]
3.2.31.7 Maxima [F]
3.2.31.8 Giac [F]
3.2.31.9 Mupad [B] (verification not implemented)

3.2.31.1 Optimal result

Integrand size = 33, antiderivative size = 161 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {4 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (7 A+6 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^2 (7 A+6 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a^2 (7 A+9 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{7 d} \]

output
4/5*a^2*(4*A+3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic 
E(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^2*(7*A+6*B)*(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+2/35*a^2* 
(7*A+9*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*B*cos(d*x+c)^(3/2)*(a^2+a^2*co 
s(d*x+c))*sin(d*x+c)/d+4/21*a^2*(7*A+6*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d
 
3.2.31.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.54 (sec) , antiderivative size = 898, normalized size of antiderivative = 5.58 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(4 A+3 B) \cot (c)}{5 d}+\frac {(56 A+51 B) \cos (d x) \sin (c)}{168 d}+\frac {(A+2 B) \cos (2 d x) \sin (2 c)}{20 d}+\frac {B \cos (3 d x) \sin (3 c)}{56 d}+\frac {(56 A+51 B) \cos (c) \sin (d x)}{168 d}+\frac {(A+2 B) \cos (2 c) \sin (2 d x)}{20 d}+\frac {B \cos (3 c) \sin (3 d x)}{56 d}\right )-\frac {A (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}-\frac {2 B (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{7 d \sqrt {1+\cot ^2(c)}}-\frac {2 A (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d}-\frac {3 B (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d} \]

input
Integrate[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x]),x 
]
 
output
Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/5*((4*A 
 + 3*B)*Cot[c])/d + ((56*A + 51*B)*Cos[d*x]*Sin[c])/(168*d) + ((A + 2*B)*C 
os[2*d*x]*Sin[2*c])/(20*d) + (B*Cos[3*d*x]*Sin[3*c])/(56*d) + ((56*A + 51* 
B)*Cos[c]*Sin[d*x])/(168*d) + ((A + 2*B)*Cos[2*c]*Sin[2*d*x])/(20*d) + (B* 
Cos[3*c]*Sin[3*d*x])/(56*d)) - (A*(a + a*Cos[c + d*x])^2*Csc[c]*Hypergeome 
tricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2] 
^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sq 
rt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - Arc 
Tan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (2*B*(a + a*Cos[c + d*x])^2*Csc[ 
c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c 
/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c] 
]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + 
Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (2*A*(a + a*Cos[c + 
 d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/ 
4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 
 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos 
[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - (( 
Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x 
 + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]* 
Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d) - (3*B*(a + a*Cos...
 
3.2.31.3 Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3455, 27, 3042, 3447, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

\(\displaystyle \frac {2}{7} \int \frac {1}{2} \sqrt {\cos (c+d x)} (\cos (c+d x) a+a) (a (7 A+3 B)+a (7 A+9 B) \cos (c+d x))dx+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \sqrt {\cos (c+d x)} (\cos (c+d x) a+a) (a (7 A+3 B)+a (7 A+9 B) \cos (c+d x))dx+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (7 A+3 B)+a (7 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {1}{7} \int \sqrt {\cos (c+d x)} \left ((7 A+9 B) \cos ^2(c+d x) a^2+(7 A+3 B) a^2+\left ((7 A+3 B) a^2+(7 A+9 B) a^2\right ) \cos (c+d x)\right )dx+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left ((7 A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(7 A+3 B) a^2+\left ((7 A+3 B) a^2+(7 A+9 B) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \sqrt {\cos (c+d x)} \left (7 (4 A+3 B) a^2+5 (7 A+6 B) \cos (c+d x) a^2\right )dx+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (7 (4 A+3 B) a^2+5 (7 A+6 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 a^2 (4 A+3 B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (7 a^2 (4 A+3 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a^2 (7 A+6 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (7 a^2 (4 A+3 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (7 a^2 (4 A+3 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \left (5 a^2 (7 A+6 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {14 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\frac {2 a^2 (7 A+9 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2}{5} \left (\frac {14 a^2 (4 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 a^2 (7 A+6 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )\right )+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{7 d}\)

input
Int[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x]),x]
 
output
(2*B*Cos[c + d*x]^(3/2)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(7*d) + ((2 
*a^2*(7*A + 9*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*((14*a^2*(4*A 
 + 3*B)*EllipticE[(c + d*x)/2, 2])/d + 5*a^2*(7*A + 6*B)*((2*EllipticF[(c 
+ d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))))/5)/7
 

3.2.31.3.1 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 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 3455
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + 
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 
 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1 
] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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]
 
3.2.31.4 Maple [A] (verified)

Time = 11.96 (sec) , antiderivative size = 385, normalized size of antiderivative = 2.39

method result size
default \(-\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{2} \left (120 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-84 A -348 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (224 A +378 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-91 A -117 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-84 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+30 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(385\)
parts \(\text {Expression too large to display}\) \(743\)

input
int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(120*B* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-84*A-348*B)*sin(1/2*d*x+1/2*c)^6 
*cos(1/2*d*x+1/2*c)+(224*A+378*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+ 
(-91*A-117*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*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))-84*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))+30*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))-6 
3*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)*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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
 
3.2.31.5 Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.26 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (7 \, A + 6 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (4 \, A + 3 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (15 \, B a^{2} \cos \left (d x + c\right )^{2} + 21 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 10 \, {\left (7 \, A + 6 \, B\right )} a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d} \]

input
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorith 
m="fricas")
 
output
-2/105*(5*I*sqrt(2)*(7*A + 6*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c 
) + I*sin(d*x + c)) - 5*I*sqrt(2)*(7*A + 6*B)*a^2*weierstrassPInverse(-4, 
0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2)*(4*A + 3*B)*a^2*weierstra 
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 
 21*I*sqrt(2)*(4*A + 3*B)*a^2*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c))) - (15*B*a^2*cos(d*x + c)^2 + 21*(A + 
 2*B)*a^2*cos(d*x + c) + 10*(7*A + 6*B)*a^2)*sqrt(cos(d*x + c))*sin(d*x + 
c))/d
 
3.2.31.6 Sympy [F(-1)]

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

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

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

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

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

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

Time = 1.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.43 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {2\,A\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

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