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

Optimal. Leaf size=189 \[ \frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (4 a^2 A+6 a b B+3 A b^2\right )-\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin (c+d x)}{5 d}+\frac{b (6 a B+5 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{b B \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))}{5 d} \]

[Out]

((4*a^2*A + 3*A*b^2 + 6*a*b*B)*x)/8 + ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x])/(5*d) + ((4*a^2*A + 3*A*b^2
+ 6*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*(5*A*b + 6*a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (b*B*C
os[c + d*x]^3*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d) - ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.311075, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2990, 3023, 2748, 2635, 8, 2633} \[ \frac{\left (4 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (4 a^2 A+6 a b B+3 A b^2\right )-\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin ^3(c+d x)}{15 d}+\frac{\left (5 a (a B+2 A b)+4 b^2 B\right ) \sin (c+d x)}{5 d}+\frac{b (6 a B+5 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac{b B \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2*A + 3*A*b^2 + 6*a*b*B)*x)/8 + ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x])/(5*d) + ((4*a^2*A + 3*A*b^2
+ 6*a*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*(5*A*b + 6*a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (b*B*C
os[c + d*x]^3*(a + b*Cos[c + d*x])*Sin[c + d*x])/(5*d) - ((4*b^2*B + 5*a*(2*A*b + a*B))*Sin[c + d*x]^3)/(15*d)

Rule 2990

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(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] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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))^2 (A+B \cos (c+d x)) \, dx &=\frac{b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^2(c+d x) \left (a (5 a A+3 b B)+\left (4 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)+b (5 A b+6 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^2(c+d x) \left (5 \left (4 a^2 A+3 A b^2+6 a b B\right )+4 \left (4 b^2 B+5 a (2 A b+a B)\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{4} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{5} \left (4 b^2 B+5 a (2 A b+a B)\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{\left (4 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}+\frac{1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int 1 \, dx-\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) x+\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin (c+d x)}{5 d}+\frac{\left (4 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b (5 A b+6 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac{b B \cos ^3(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{5 d}-\frac{\left (4 b^2 B+5 a (2 A b+a B)\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.456626, size = 146, normalized size = 0.77 \[ \frac{60 (c+d x) \left (4 a^2 A+6 a b B+3 A b^2\right )+60 \left (6 a^2 B+12 a A b+5 b^2 B\right ) \sin (c+d x)+120 \left (a^2 A+2 a b B+A b^2\right ) \sin (2 (c+d x))+10 \left (4 a^2 B+8 a A b+5 b^2 B\right ) \sin (3 (c+d x))+15 b (2 a B+A b) \sin (4 (c+d x))+6 b^2 B \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(4*a^2*A + 3*A*b^2 + 6*a*b*B)*(c + d*x) + 60*(12*a*A*b + 6*a^2*B + 5*b^2*B)*Sin[c + d*x] + 120*(a^2*A + A*
b^2 + 2*a*b*B)*Sin[2*(c + d*x)] + 10*(8*a*A*b + 4*a^2*B + 5*b^2*B)*Sin[3*(c + d*x)] + 15*b*(A*b + 2*a*B)*Sin[4
*(c + d*x)] + 6*b^2*B*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.043, size = 184, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{\frac{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,Aab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,Bab \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +A{b}^{2} \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{{b}^{2}B\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \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))^2*(A+B*cos(d*x+c)),x)

[Out]

1/d*(a^2*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*B*a^2*(2+cos(d*x+c)^2)*sin(d*x+c)+2/3*A*a*b*(2+cos(d*
x+c)^2)*sin(d*x+c)+2*B*a*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)+A*b^2*(1/4*(cos(d*x+c)
^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/5*b^2*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.09796, size = 238, normalized size = 1.26 \begin{align*} \frac{120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 30 \,{\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 a b + 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{2}}{480 \, 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))^2*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 320*(sin(d*x
+ c)^3 - 3*sin(d*x + c))*A*a*b + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a*b + 15*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*b^2 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*B*b^2)/d

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Fricas [A]  time = 1.40327, size = 350, normalized size = 1.85 \begin{align*} \frac{15 \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} d x +{\left (24 \, B b^{2} \cos \left (d x + c\right )^{4} + 30 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} + 80 \, B a^{2} + 160 \, A a b + 64 \, B b^{2} + 8 \,{\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, 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))^2*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(15*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*d*x + (24*B*b^2*cos(d*x + c)^4 + 30*(2*B*a*b + A*b^2)*cos(d*x + c)^3 +
80*B*a^2 + 160*A*a*b + 64*B*b^2 + 8*(5*B*a^2 + 10*A*a*b + 4*B*b^2)*cos(d*x + c)^2 + 15*(4*A*a^2 + 6*B*a*b + 3
*A*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 3.44535, size = 459, normalized size = 2.43 \begin{align*} \begin{cases} \frac{A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{4 A a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{2 A a b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 A b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 A b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 A b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 A b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 A b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{B a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 B a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{3 B a b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{5 B a b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{8 B b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{B b^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{2} \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))**2*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((A*a**2*x*sin(c + d*x)**2/2 + A*a**2*x*cos(c + d*x)**2/2 + A*a**2*sin(c + d*x)*cos(c + d*x)/(2*d) +
4*A*a*b*sin(c + d*x)**3/(3*d) + 2*A*a*b*sin(c + d*x)*cos(c + d*x)**2/d + 3*A*b**2*x*sin(c + d*x)**4/8 + 3*A*b*
*2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*b**2*x*cos(c + d*x)**4/8 + 3*A*b**2*sin(c + d*x)**3*cos(c + d*x)/
(8*d) + 5*A*b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 2*B*a**2*sin(c + d*x)**3/(3*d) + B*a**2*sin(c + d*x)*cos
(c + d*x)**2/d + 3*B*a*b*x*sin(c + d*x)**4/4 + 3*B*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 3*B*a*b*x*cos(c +
d*x)**4/4 + 3*B*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 5*B*a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 8*B*b**
2*sin(c + d*x)**5/(15*d) + 4*B*b**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*b**2*sin(c + d*x)*cos(c + d*x)**
4/d, Ne(d, 0)), (x*(A + B*cos(c))*(a + b*cos(c))**2*cos(c)**2, True))

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Giac [A]  time = 1.43544, size = 211, normalized size = 1.12 \begin{align*} \frac{B b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{1}{8} \,{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} x + \frac{{\left (2 \, B a b + A b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (4 \, B a^{2} + 8 \, A a b + 5 \, B b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (6 \, B a^{2} + 12 \, A a b + 5 \, B b^{2}\right )} \sin \left (d x + c\right )}{8 \, 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))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/80*B*b^2*sin(5*d*x + 5*c)/d + 1/8*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*x + 1/32*(2*B*a*b + A*b^2)*sin(4*d*x + 4*c)/
d + 1/48*(4*B*a^2 + 8*A*a*b + 5*B*b^2)*sin(3*d*x + 3*c)/d + 1/4*(A*a^2 + 2*B*a*b + A*b^2)*sin(2*d*x + 2*c)/d +
1/8*(6*B*a^2 + 12*A*a*b + 5*B*b^2)*sin(d*x + c)/d