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

Optimal. Leaf size=77 \[ \frac{a (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a x (A+B)+\frac{a B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0783281, antiderivative size = 77, normalized size of antiderivative = 1., 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} \[ \frac{a (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a x (A+B)+\frac{a B \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A + B)*x)/2 + (a*(3*A + 2*B)*Sin[c + d*x])/(3*d) + (a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*B*Cos[
c + d*x]^2*Sin[c + d*x])/(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 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+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx &=\int \cos (c+d x) \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a B \cos ^2(c+d x) \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos (c+d x) (a (3 A+2 B)+3 a (A+B) \cos (c+d x)) \, dx\\ &=\frac{1}{2} a (A+B) x+\frac{a (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a (A+B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a B \cos ^2(c+d x) \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.166033, size = 65, normalized size = 0.84 \[ \frac{a (3 (4 A+3 B) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+6 A c+6 A d x+B \sin (3 (c+d x))+6 B c+6 B d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 85, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\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 ) +aB \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +aA\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+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 = 0.984232, size = 107, normalized size = 1.39 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 12 \, A a \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.37541, size = 146, normalized size = 1.9 \begin{align*} \frac{3 \,{\left (A + B\right )} a d x +{\left (2 \, B a \cos \left (d x + c\right )^{2} + 3 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \,{\left (3 \, A + 2 \, B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.928453, size = 168, normalized size = 2.18 \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{A a \sin{\left (c + d x \right )}}{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{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{B a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\right ) \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+a*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) + A*a*sin(c
+ d*x)/d + B*a*x*sin(c + d*x)**2/2 + B*a*x*cos(c + d*x)**2/2 + 2*B*a*sin(c + d*x)**3/(3*d) + B*a*sin(c + d*x)
*cos(c + d*x)**2/d + B*a*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)*cos(c),
True))

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Giac [A]  time = 1.10459, size = 92, normalized size = 1.19 \begin{align*} \frac{1}{2} \,{\left (A a + B a\right )} x + \frac{B a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, A a + 3 \, B a\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+a*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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