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

Optimal. Leaf size=118 \[ \frac{a (3 A+2 (B+C)) \sin (c+d x)}{3 d}+\frac{a (4 A+4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x (4 A+4 B+3 C)+\frac{a (B+C) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac{a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

(a*(4*A + 4*B + 3*C)*x)/8 + (a*(3*A + 2*(B + C))*Sin[c + d*x])/(3*d) + (a*(4*A + 4*B + 3*C)*Cos[c + d*x]*Sin[c
+ d*x])/(8*d) + (a*(B + C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + (a*C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.138148, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.081, Rules used = {3033, 3023, 2734} \[ \frac{a (3 A+2 (B+C)) \sin (c+d x)}{3 d}+\frac{a (4 A+4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x (4 A+4 B+3 C)+\frac{a (B+C) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac{a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*A + 4*B + 3*C)*x)/8 + (a*(3*A + 2*(B + C))*Sin[c + d*x])/(3*d) + (a*(4*A + 4*B + 3*C)*Cos[c + d*x]*Sin[c
+ d*x])/(8*d) + (a*(B + C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + (a*C*Cos[c + d*x]^3*Sin[c + d*x])/(4*d)

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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

Mathematica [A]  time = 0.395352, size = 96, normalized size = 0.81 \[ \frac{a (24 (4 A+3 (B+C)) \sin (c+d x)+24 (A+B+C) \sin (2 (c+d x))+48 A d x+8 B \sin (3 (c+d x))+48 B c+48 B d x+8 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+24 c C+36 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(48*B*c + 24*c*C + 48*A*d*x + 48*B*d*x + 36*C*d*x + 24*(4*A + 3*(B + C))*Sin[c + d*x] + 24*(A + B + C)*Sin[
2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 8*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.024, size = 141, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( aC \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{Ba \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{aC \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 ) +Ba \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+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/d*(a*C*(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*B*a*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
a*C*(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)+B*a*(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 = 0.987671, size = 178, normalized size = 1.51 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 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 )} C a + 96 \, A a \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+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 24*(2*d*x + 2*c + s
in(2*d*x + 2*c))*B*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a + 96*A*a*sin(d*x + c))/d

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Fricas [A]  time = 1.89475, size = 221, normalized size = 1.87 \begin{align*} \frac{3 \,{\left (4 \, A + 4 \, B + 3 \, C\right )} a d x +{\left (6 \, C a \cos \left (d x + c\right )^{3} + 8 \,{\left (B + C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, A + 4 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + 8 \,{\left (3 \, A + 2 \, B + 2 \, C\right )} a\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+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.42431, size = 320, normalized size = 2.71 \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} + \frac{3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 C a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{5 C a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{C a \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\right ) \left (A + B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\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)+C*cos(d*x+c)**2),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) + 3*C*a*x*sin(c + d*x)**4/8 + 3*C*a*x*sin(c + d*x)**2
*cos(c + d*x)**2/4 + 3*C*a*x*cos(c + d*x)**4/8 + 3*C*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*C*a*sin(c + d*x)
**3/(3*d) + 5*C*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) + C*a*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a*co
s(c) + a)*(A + B*cos(c) + C*cos(c)**2)*cos(c), True))

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Giac [A]  time = 1.23747, size = 138, normalized size = 1.17 \begin{align*} \frac{1}{8} \,{\left (4 \, A a + 4 \, B a + 3 \, C a\right )} x + \frac{C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (B a + C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (A a + B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, A a + 3 \, B a + 3 \, C 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)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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