3.233 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=171 \[ \frac{\left (16 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d}+\frac{b \left (6 a^2 B+20 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right )+\frac{(3 a B+4 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

((8*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 3*b^3*B)*x)/8 + ((16*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Sin[c + d
*x])/(6*d) + (b*(20*a*A*b + 6*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*A*b + 3*a*B)*(a + b*Cos
[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.196937, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ \frac{\left (16 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d}+\frac{b \left (6 a^2 B+20 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right )+\frac{(3 a B+4 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]

[Out]

((8*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 3*b^3*B)*x)/8 + ((16*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Sin[c + d
*x])/(6*d) + (b*(20*a*A*b + 6*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*A*b + 3*a*B)*(a + b*Cos
[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

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 (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \int (a+b \cos (c+d x))^2 (4 a A+3 b B+(4 A b+3 a B) \cos (c+d x)) \, dx\\ &=\frac{(4 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (12 a^2 A+8 A b^2+15 a b B+\left (20 a A b+6 a^2 B+9 b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{1}{8} \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) x+\frac{\left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac{b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.381848, size = 140, normalized size = 0.82 \[ \frac{12 (c+d x) \left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right )+24 b \left (3 a^2 B+3 a A b+b^2 B\right ) \sin (2 (c+d x))+24 \left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right ) \sin (c+d x)+8 b^2 (3 a B+A b) \sin (3 (c+d x))+3 b^3 B \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]

[Out]

(12*(8*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 3*b^3*B)*(c + d*x) + 24*(12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*
Sin[c + d*x] + 24*b*(3*a*A*b + 3*a^2*B + b^2*B)*Sin[2*(c + d*x)] + 8*b^2*(A*b + 3*a*B)*Sin[3*(c + d*x)] + 3*b^
3*B*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.045, size = 180, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( B{b}^{3} \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{A{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Ba{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,Aa{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{2}bB \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,A{a}^{2}b\sin \left ( dx+c \right ) +{a}^{3}B\sin \left ( dx+c \right ) +A{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.16311, size = 231, normalized size = 1.35 \begin{align*} \frac{96 \,{\left (d x + c\right )} A a^{3} + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b + 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} - 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} + 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^{3} + 96 \, B a^{3} \sin \left (d x + c\right ) + 288 \, A a^{2} b \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/96*(96*(d*x + c)*A*a^3 + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2*b + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*A
*a*b^2 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b^2 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^3 + 3*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b^3 + 96*B*a^3*sin(d*x + c) + 288*A*a^2*b*sin(d*x + c))/d

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Fricas [A]  time = 1.34661, size = 321, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} d x +{\left (6 \, B b^{3} \cos \left (d x + c\right )^{3} + 24 \, B a^{3} + 72 \, A a^{2} b + 48 \, B a b^{2} + 16 \, A b^{3} + 8 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{2} + 9 \,{\left (4 \, B a^{2} b + 4 \, A a b^{2} + B b^{3}\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((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(3*(8*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 3*B*b^3)*d*x + (6*B*b^3*cos(d*x + c)^3 + 24*B*a^3 + 72*A*a^2*b +
48*B*a*b^2 + 16*A*b^3 + 8*(3*B*a*b^2 + A*b^3)*cos(d*x + c)^2 + 9*(4*B*a^2*b + 4*A*a*b^2 + B*b^3)*cos(d*x + c))
*sin(d*x + c))/d

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Sympy [A]  time = 1.76105, size = 386, normalized size = 2.26 \begin{align*} \begin{cases} A a^{3} x + \frac{3 A a^{2} b \sin{\left (c + d x \right )}}{d} + \frac{3 A a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 A a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 A a b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 A b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{B a^{3} \sin{\left (c + d x \right )}}{d} + \frac{3 B a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{2} b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 B a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac{3 B a b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 B b^{3} \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 )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.36552, size = 200, normalized size = 1.17 \begin{align*} \frac{B b^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} x + \frac{{\left (3 \, B a b^{2} + A b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (3 \, B a^{2} b + 3 \, A a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/32*B*b^3*sin(4*d*x + 4*c)/d + 1/8*(8*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 3*B*b^3)*x + 1/12*(3*B*a*b^2 + A*b^3)
*sin(3*d*x + 3*c)/d + 1/4*(3*B*a^2*b + 3*A*a*b^2 + B*b^3)*sin(2*d*x + 2*c)/d + 1/4*(4*B*a^3 + 12*A*a^2*b + 9*B
*a*b^2 + 3*A*b^3)*sin(d*x + c)/d