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

Optimal. Leaf size=88 \[ \frac{(4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 A+3 C)-\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0945887, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3023, 2748, 2635, 8, 2633} \[ \frac{(4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x (4 A+3 C)-\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A + 3*C)*x)/8 + (B*Sin[c + d*x])/d + ((4*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^3*Sin
[c + d*x])/(4*d) - (B*Sin[c + d*x]^3)/(3*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 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 ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^2(c+d x) (4 A+3 C+4 B \cos (c+d x)) \, dx\\ &=\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}+B \int \cos ^3(c+d x) \, dx+\frac{1}{4} (4 A+3 C) \int \cos ^2(c+d x) \, dx\\ &=\frac{(4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{8} (4 A+3 C) \int 1 \, dx-\frac{B \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{1}{8} (4 A+3 C) x+\frac{B \sin (c+d x)}{d}+\frac{(4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{B \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.153774, size = 70, normalized size = 0.8 \[ \frac{24 (A+C) \sin (2 (c+d x))+48 A c+48 A d x-32 B \sin ^3(c+d x)+96 B \sin (c+d x)+3 C \sin (4 (c+d x))+36 c C+36 C d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 96*B*Sin[c + d*x] - 32*B*Sin[c + d*x]^3 + 24*(A + C)*Sin[2*(c + d*x)]
+ 3*C*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.013, size = 84, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( C \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{B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A \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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/d*(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*(2+cos(d*x+c)^2)*sin(d*x+c)+A*(1/2*c
os(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.967407, size = 104, normalized size = 1.18 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 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}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B + 3*(12*d*x + 12*c + sin(
4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C)/d

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Fricas [A]  time = 1.9633, size = 163, normalized size = 1.85 \begin{align*} \frac{3 \,{\left (4 \, A + 3 \, C\right )} d x +{\left (6 \, C \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \,{\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(4*A + 3*C)*d*x + (6*C*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 3*(4*A + 3*C)*cos(d*x + c) + 16*B)*sin(d*
x + c))/d

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Sympy [A]  time = 1.29187, size = 197, normalized size = 2.24 \begin{align*} \begin{cases} \frac{A x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 B \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{B \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 C x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 C x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 C x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 C \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 C \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\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)**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((A*x*sin(c + d*x)**2/2 + A*x*cos(c + d*x)**2/2 + A*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*B*sin(c + d*x
)**3/(3*d) + B*sin(c + d*x)*cos(c + d*x)**2/d + 3*C*x*sin(c + d*x)**4/8 + 3*C*x*sin(c + d*x)**2*cos(c + d*x)**
2/4 + 3*C*x*cos(c + d*x)**4/8 + 3*C*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*C*sin(c + d*x)*cos(c + d*x)**3/(8*d
), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**2, True))

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Giac [A]  time = 1.17457, size = 95, normalized size = 1.08 \begin{align*} \frac{1}{8} \,{\left (4 \, A + 3 \, C\right )} x + \frac{C \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{B \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (A + C\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{3 \, B \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)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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