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

Optimal. Leaf size=54 \[ \frac{B \sin (c+d x) \cos (c+d x)}{2 d}+\frac{B x}{2}-\frac{C \sin ^3(c+d x)}{3 d}+\frac{C \sin (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0631004, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3010, 2748, 2635, 8, 2633} \[ \frac{B \sin (c+d x) \cos (c+d x)}{2 d}+\frac{B x}{2}-\frac{C \sin ^3(c+d x)}{3 d}+\frac{C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.06226, size = 57, normalized size = 1.06 \[ \frac{B (c+d x)}{2 d}+\frac{B \sin (2 (c+d x))}{4 d}-\frac{C \sin ^3(c+d x)}{3 d}+\frac{C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(c + d*x))/(2*d) + (C*Sin[c + d*x])/d - (C*Sin[c + d*x]^3)/(3*d) + (B*Sin[2*(c + d*x)])/(4*d)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 49, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+B \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.21927, size = 62, normalized size = 1.15 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C)/d

________________________________________________________________________________________

Fricas [A]  time = 1.5606, size = 105, normalized size = 1.94 \begin{align*} \frac{3 \, B d x +{\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 4 \, C\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)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.611654, size = 95, normalized size = 1.76 \begin{align*} \begin{cases} \frac{B x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{B x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{B \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 C \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{C \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (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)*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.27177, size = 63, normalized size = 1.17 \begin{align*} \frac{1}{2} \, B x + \frac{C \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{3 \, C \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)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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