3.291 \(\int \cos ^3(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=113 \[ -\frac{(5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac{(5 A+4 C) \sin (c+d x)}{5 d}+\frac{B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 B x}{8}+\frac{C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

(3*B*x)/8 + ((5*A + 4*C)*Sin[c + d*x])/(5*d) + (3*B*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (B*Cos[c + d*x]^3*Sin[c
+ d*x])/(4*d) + (C*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - ((5*A + 4*C)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.107283, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3023, 2748, 2633, 2635, 8} \[ -\frac{(5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac{(5 A+4 C) \sin (c+d x)}{5 d}+\frac{B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 B x}{8}+\frac{C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(3*B*x)/8 + ((5*A + 4*C)*Sin[c + d*x])/(5*d) + (3*B*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (B*Cos[c + d*x]^3*Sin[c
+ d*x])/(4*d) + (C*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - ((5*A + 4*C)*Sin[c + d*x]^3)/(15*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 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]

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]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^3(c+d x) (5 A+4 C+5 B \cos (c+d x)) \, dx\\ &=\frac{C \cos ^4(c+d x) \sin (c+d x)}{5 d}+B \int \cos ^4(c+d x) \, dx+\frac{1}{5} (5 A+4 C) \int \cos ^3(c+d x) \, dx\\ &=\frac{B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac{1}{4} (3 B) \int \cos ^2(c+d x) \, dx-\frac{(5 A+4 C) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{(5 A+4 C) \sin (c+d x)}{5 d}+\frac{3 B \cos (c+d x) \sin (c+d x)}{8 d}+\frac{B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{C \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{(5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac{1}{8} (3 B) \int 1 \, dx\\ &=\frac{3 B x}{8}+\frac{(5 A+4 C) \sin (c+d x)}{5 d}+\frac{3 B \cos (c+d x) \sin (c+d x)}{8 d}+\frac{B \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{C \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{(5 A+4 C) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.176897, size = 87, normalized size = 0.77 \[ \frac{60 (6 A+5 C) \sin (c+d x)+40 A \sin (3 (c+d x))+120 B \sin (2 (c+d x))+15 B \sin (4 (c+d x))+180 B c+180 B d x+50 C \sin (3 (c+d x))+6 C \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(180*B*c + 180*B*d*x + 60*(6*A + 5*C)*Sin[c + d*x] + 120*B*Sin[2*(c + d*x)] + 40*A*Sin[3*(c + d*x)] + 50*C*Sin
[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.013, size = 89, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{C\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+B \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{A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/d*(1/5*C*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+B*(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*A*(2+cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.10748, size = 120, normalized size = 1.06 \begin{align*} -\frac{160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A - 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B - 32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B
- 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C)/d

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Fricas [A]  time = 1.96333, size = 194, normalized size = 1.72 \begin{align*} \frac{45 \, B d x +{\left (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 80 \, A + 64 \, C\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(45*B*d*x + (24*C*cos(d*x + c)^4 + 30*B*cos(d*x + c)^3 + 8*(5*A + 4*C)*cos(d*x + c)^2 + 45*B*cos(d*x + c
) + 80*A + 64*C)*sin(d*x + c))/d

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Sympy [A]  time = 2.39523, size = 209, normalized size = 1.85 \begin{align*} \begin{cases} \frac{2 A \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 B \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 C \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 C \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{C \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{3}{\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)**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((2*A*sin(c + d*x)**3/(3*d) + A*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*x*sin(c + d*x)**4/8 + 3*B*x*sin(
c + d*x)**2*cos(c + d*x)**2/4 + 3*B*x*cos(c + d*x)**4/8 + 3*B*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*B*sin(c +
d*x)*cos(c + d*x)**3/(8*d) + 8*C*sin(c + d*x)**5/(15*d) + 4*C*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + C*sin(c
+ d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**3, True))

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Giac [A]  time = 1.2193, size = 120, normalized size = 1.06 \begin{align*} \frac{3}{8} \, B x + \frac{C \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{B \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (4 \, A + 5 \, C\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (6 \, A + 5 \, C\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

3/8*B*x + 1/80*C*sin(5*d*x + 5*c)/d + 1/32*B*sin(4*d*x + 4*c)/d + 1/48*(4*A + 5*C)*sin(3*d*x + 3*c)/d + 1/4*B*
sin(2*d*x + 2*c)/d + 1/8*(6*A + 5*C)*sin(d*x + c)/d