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

Optimal. Leaf size=85 \[ \frac{a (3 B+C) \sin (c+d x)}{3 d}+\frac{a (3 B-C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a x (B+C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 a d} \]

[Out]

(a*(B + C)*x)/2 + (a*(3*B + C)*Sin[c + d*x])/(3*d) + (a*(3*B - C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (C*(a + a
*Cos[c + d*x])^2*Sin[c + d*x])/(3*a*d)

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Rubi [A]  time = 0.0774457, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3023, 2734} \[ \frac{a (3 B+C) \sin (c+d x)}{3 d}+\frac{a (3 B-C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a x (B+C)+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(B + C)*x)/2 + (a*(3*B + C)*Sin[c + d*x])/(3*d) + (a*(3*B - C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (C*(a + a
*Cos[c + d*x])^2*Sin[c + d*x])/(3*a*d)

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 (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 a d}+\frac{\int (a+a \cos (c+d x)) (2 a C+a (3 B-C) \cos (c+d x)) \, dx}{3 a}\\ &=\frac{1}{2} a (B+C) x+\frac{a (3 B+C) \sin (c+d x)}{3 d}+\frac{a (3 B-C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.174457, size = 65, normalized size = 0.76 \[ \frac{a (3 (4 B+3 C) \sin (c+d x)+3 (B+C) \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x))+6 c C+6 C d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 85, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{aC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Ba \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +aC \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Ba\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10515, size = 107, normalized size = 1.26 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 12 \, B a \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58984, size = 146, normalized size = 1.72 \begin{align*} \frac{3 \,{\left (B + C\right )} a d x +{\left (2 \, C a \cos \left (d x + c\right )^{2} + 3 \,{\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \,{\left (3 \, B + 2 \, C\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((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.698973, size = 170, normalized size = 2. \begin{align*} \begin{cases} \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{B a \sin{\left (c + d x \right )}}{d} + \frac{C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{C a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{C a \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{C a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((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) + B*a*sin(c
+ d*x)/d + C*a*x*sin(c + d*x)**2/2 + C*a*x*cos(c + d*x)**2/2 + 2*C*a*sin(c + d*x)**3/(3*d) + C*a*sin(c + d*x)
*cos(c + d*x)**2/d + C*a*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*(a*cos(c) + a
), True))

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Giac [A]  time = 1.33715, size = 92, normalized size = 1.08 \begin{align*} \frac{1}{2} \,{\left (B a + C a\right )} x + \frac{C a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, 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((a+a*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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