\(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [125]

Optimal result

Integrand size = 31, antiderivative size = 101 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {2 a (5 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/5*a*(5*A+3*B)*(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+2/3*a*
(A+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+2/5*a*B*cos(d*x+
c)^(3/2)*sin(d*x+c)/d+2/3*a*(A+B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3047, 3102, 2827, 2719, 2715, 2720} \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {2 a (A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (5 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 a B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d} \]

[In]

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

[Out]

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

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2719

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 2720

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3047

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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {\cos (c+d x)} \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {\cos (c+d x)} \left (\frac {1}{2} a (5 A+3 B)+\frac {5}{2} a (A+B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+(a (A+B)) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{5} (a (5 A+3 B)) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 a (5 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} (a (A+B)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a (5 A+3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 5.46 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.59 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 (5 A+3 B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (9 (5 A+3 B) \cos (c-d x-\arctan (\tan (c))) \csc (c) \sec (c)+15 A \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)+9 B \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)-60 A \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-36 B \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-20 (A+B) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)+20 A \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+20 B \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)+12 B \cos ^2(c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{60 d \sqrt {\cos (c+d x)} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \]

[In]

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

[Out]

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-6*(5*A + 3*B)*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcT
an[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]] + (9*(5*A + 3*B)*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] +
15*A*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 9*B*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] - 60*A*Cos
[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 36*B*Cos[c + d*x]*Cot[c]*Sqrt[Sec[c]^2] - 20*(A + B)*Cos[c + d*x]*Sqrt[Cos[d
*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sqrt[
Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 20*A*Cos[c + d*x]*Sqrt[Sec[c]^2]*Sin[c + d*x] + 20*B*Cos[c + d*x]
*Sqrt[Sec[c]^2]*Sin[c + d*x] + 12*B*Cos[c + d*x]^2*Sqrt[Sec[c]^2]*Sin[c + d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]
^2]))/(60*d*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(354\) vs. \(2(141)=282\).

Time = 8.57 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.51

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (20 A +44 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A -16 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(355\)
parts	\(-\frac {2 \left (a A +B a \right ) \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 (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\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 )\right )}{3 \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}+\frac {2 a A \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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}{\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}-\frac {2 B a \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 (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \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 )}{5 \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}\)	\(524\)

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(20*A+44*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A-16*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*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))+5*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))-9*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)))/(-2*sin(1/2*d*x+1/2*c)^4+s
in(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 higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.59 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (A + B\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, B a \cos \left (d x + c\right ) + 5 \, {\left (A + B\right )} a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,B\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*A*a*(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*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))