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

Optimal. Leaf size=170 \[ \frac{\left (4 a^2 A b+a^3 (-B)+8 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 b d}+\frac{\left (-2 a^2 B+8 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (4 a^2 B+8 a A b+3 b^2 B\right )+\frac{(4 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d} \]

[Out]

((8*a*A*b + 4*a^2*B + 3*b^2*B)*x)/8 + ((4*a^2*A*b + 4*A*b^3 - a^3*B + 8*a*b^2*B)*Sin[c + d*x])/(6*b*d) + ((8*a
*A*b - 2*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*A*b - a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*
x])/(12*b*d) + (B*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*b*d)

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Rubi [A]  time = 0.233517, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2968, 3023, 2753, 2734} \[ \frac{\left (4 a^2 A b+a^3 (-B)+8 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 b d}+\frac{\left (-2 a^2 B+8 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (4 a^2 B+8 a A b+3 b^2 B\right )+\frac{(4 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A*b + 4*a^2*B + 3*b^2*B)*x)/8 + ((4*a^2*A*b + 4*A*b^3 - a^3*B + 8*a*b^2*B)*Sin[c + d*x])/(6*b*d) + ((8*a
*A*b - 2*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*A*b - a*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*
x])/(12*b*d) + (B*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*b*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 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx &=\int (a+b \cos (c+d x))^2 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}+\frac{\int (a+b \cos (c+d x))^2 (3 b B+(4 A b-a B) \cos (c+d x)) \, dx}{4 b}\\ &=\frac{(4 A b-a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 b d}+\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}+\frac{\int (a+b \cos (c+d x)) \left (b (8 A b+7 a B)+\left (8 a A b-2 a^2 B+9 b^2 B\right ) \cos (c+d x)\right ) \, dx}{12 b}\\ &=\frac{1}{8} \left (8 a A b+4 a^2 B+3 b^2 B\right ) x+\frac{\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \sin (c+d x)}{6 b d}+\frac{\left (8 a A b-2 a^2 B+9 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 A b-a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 b d}+\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}\\ \end{align*}

Mathematica [A]  time = 0.430768, size = 118, normalized size = 0.69 \[ \frac{12 (c+d x) \left (4 a^2 B+8 a A b+3 b^2 B\right )+24 \left (4 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)+24 \left (a^2 B+2 a A b+b^2 B\right ) \sin (2 (c+d x))+8 b (2 a B+A b) \sin (3 (c+d x))+3 b^2 B \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a*A*b + 4*a^2*B + 3*b^2*B)*(c + d*x) + 24*(4*a^2*A + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x] + 24*(2*a*A*b + a^
2*B + b^2*B)*Sin[2*(c + d*x)] + 8*b*(A*b + 2*a*B)*Sin[3*(c + d*x)] + 3*b^2*B*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.047, size = 152, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{2}A\sin \left ( dx+c \right ) +B{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,Aab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{2\,Bab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{A{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{2}B \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09274, size = 192, normalized size = 1.13 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 48 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 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^{2} + 96 \, A a^{2} \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b - 64*(sin(d*x + c)
^3 - 3*sin(d*x + c))*B*a*b - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*B*b^2 + 96*A*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.42785, size = 274, normalized size = 1.61 \begin{align*} \frac{3 \,{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} d x +{\left (6 \, B b^{2} \cos \left (d x + c\right )^{3} + 24 \, A a^{2} + 32 \, B a b + 16 \, A b^{2} + 8 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\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)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*d*x + (6*B*b^2*cos(d*x + c)^3 + 24*A*a^2 + 32*B*a*b + 16*A*b^2 + 8*(2*B*
a*b + A*b^2)*cos(d*x + c)^2 + 3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.7511, size = 338, normalized size = 1.99 \begin{align*} \begin{cases} \frac{A a^{2} \sin{\left (c + d x \right )}}{d} + A a b x \sin ^{2}{\left (c + d x \right )} + A a b x \cos ^{2}{\left (c + d x \right )} + \frac{A a b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{2 A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{B a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{4 B a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{2 B a b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 B b^{2} \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 )^{2} \cos{\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)*(a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.56091, size = 167, normalized size = 0.98 \begin{align*} \frac{B b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} x + \frac{{\left (2 \, B a b + A b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\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)*(a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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