3.215 \(\int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac{(a B+A b) \sin ^3(c+d x)}{3 d}+\frac{(a B+A b) \sin (c+d x)}{d}+\frac{(4 a A+3 b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a A+3 b B)+\frac{b B \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

((4*a*A + 3*b*B)*x)/8 + ((A*b + a*B)*Sin[c + d*x])/d + ((4*a*A + 3*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*
B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - ((A*b + a*B)*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.169751, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2968, 3023, 2748, 2635, 8, 2633} \[ -\frac{(a B+A b) \sin ^3(c+d x)}{3 d}+\frac{(a B+A b) \sin (c+d x)}{d}+\frac{(4 a A+3 b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 a A+3 b B)+\frac{b B \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*A + 3*b*B)*x)/8 + ((A*b + a*B)*Sin[c + d*x])/d + ((4*a*A + 3*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*
B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - ((A*b + a*B)*Sin[c + d*x]^3)/(3*d)

Rule 2968

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 3023

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]

Rule 2748

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 2635

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] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx &=\int \cos ^2(c+d x) \left (a A+(A b+a B) \cos (c+d x)+b B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{b B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^2(c+d x) (4 a A+3 b B+4 (A b+a B) \cos (c+d x)) \, dx\\ &=\frac{b B \cos ^3(c+d x) \sin (c+d x)}{4 d}+(A b+a B) \int \cos ^3(c+d x) \, dx+\frac{1}{4} (4 a A+3 b B) \int \cos ^2(c+d x) \, dx\\ &=\frac{(4 a A+3 b B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{8} (4 a A+3 b B) \int 1 \, dx-\frac{(A b+a B) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{1}{8} (4 a A+3 b B) x+\frac{(A b+a B) \sin (c+d x)}{d}+\frac{(4 a A+3 b B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{(A b+a B) \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.219568, size = 91, normalized size = 0.87 \[ \frac{-32 (a B+A b) \sin ^3(c+d x)+96 (a B+A b) \sin (c+d x)+24 (a A+b B) \sin (2 (c+d x))+48 a A c+48 a A d x+3 b B \sin (4 (c+d x))+36 b B c+36 b B d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a*A*c + 36*b*B*c + 48*a*A*d*x + 36*b*B*d*x + 96*(A*b + a*B)*Sin[c + d*x] - 32*(A*b + a*B)*Sin[c + d*x]^3 +
24*(a*A + b*B)*Sin[2*(c + d*x)] + 3*b*B*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.042, size = 107, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( Bb \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{Ab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{aB \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+aA \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(B*b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*A*b*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
a*B*(2+cos(d*x+c)^2)*sin(d*x+c)+a*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.998019, size = 136, normalized size = 1.3 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a - 32*(sin(d*x + c)^3
- 3*sin(d*x + c))*A*b + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b)/d

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Fricas [A]  time = 1.39391, size = 205, normalized size = 1.95 \begin{align*} \frac{3 \,{\left (4 \, A a + 3 \, B b\right )} d x +{\left (6 \, B b \cos \left (d x + c\right )^{3} + 8 \,{\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 16 \, B a + 16 \, A b + 3 \,{\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*A*a + 3*B*b)*d*x + (6*B*b*cos(d*x + c)^3 + 8*(B*a + A*b)*cos(d*x + c)^2 + 16*B*a + 16*A*b + 3*(4*A*
a + 3*B*b)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.30528, size = 252, normalized size = 2.4 \begin{align*} \begin{cases} \frac{A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 A b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{B a \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 B b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a*x*sin(c + d*x)**2/2 + A*a*x*cos(c + d*x)**2/2 + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*A*b*sin
(c + d*x)**3/(3*d) + A*b*sin(c + d*x)*cos(c + d*x)**2/d + 2*B*a*sin(c + d*x)**3/(3*d) + B*a*sin(c + d*x)*cos(c
+ d*x)**2/d + 3*B*b*x*sin(c + d*x)**4/8 + 3*B*b*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*B*b*x*cos(c + d*x)**4
/8 + 3*B*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*B*b*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(A + B
*cos(c))*(a + b*cos(c))*cos(c)**2, True))

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Giac [A]  time = 1.43524, size = 120, normalized size = 1.14 \begin{align*} \frac{1}{8} \,{\left (4 \, A a + 3 \, B b\right )} x + \frac{B b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (B a + A b\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (A a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{3 \,{\left (B a + A b\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*A*a + 3*B*b)*x + 1/32*B*b*sin(4*d*x + 4*c)/d + 1/12*(B*a + A*b)*sin(3*d*x + 3*c)/d + 1/4*(A*a + B*b)*si
n(2*d*x + 2*c)/d + 3/4*(B*a + A*b)*sin(d*x + c)/d