3.216 \(\int \cos (c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=84 \[ \frac {(3 a A+2 b B) \sin (c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} x (a B+A b)+\frac {b B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2968, 3023, 2734} \[ \frac {(3 a A+2 b B) \sin (c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} x (a B+A b)+\frac {b B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b + a*B)*x)/2 + ((3*a*A + 2*b*B)*Sin[c + d*x])/(3*d) + ((A*b + a*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (b*
B*Cos[c + d*x]^2*Sin[c + d*x])/(3*d)

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 75, normalized size = 0.89 \[ \frac {3 (4 a A+3 b B) \sin (c+d x)+3 (a B+A b) \sin (2 (c+d x))+6 a B c+6 a B d x+6 A b c+6 A b d x+b B \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.03, size = 60, normalized size = 0.71 \[ \frac {3 \, {\left (B a + A b\right )} d x + {\left (2 \, B b \cos \left (d x + c\right )^{2} + 6 \, A a + 4 \, B b + 3 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 68, normalized size = 0.81 \[ \frac {1}{2} \, {\left (B a + A b\right )} x + \frac {B b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (B a + A b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A a + 3 \, B b\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 85, normalized size = 1.01 \[ \frac {\frac {B b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 79, normalized size = 0.94 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b + 12 \, A a \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.40, size = 84, normalized size = 1.00 \[ \frac {A\,b\,x}{2}+\frac {B\,a\,x}{2}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*b*x)/2 + (B*a*x)/2 + (A*a*sin(c + d*x))/d + (3*B*b*sin(c + d*x))/(4*d) + (A*b*sin(2*c + 2*d*x))/(4*d) + (B*
a*sin(2*c + 2*d*x))/(4*d) + (B*b*sin(3*c + 3*d*x))/(12*d)

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sympy [A]  time = 0.51, size = 168, normalized size = 2.00 \[ \begin {cases} \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {A b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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