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

Optimal. Leaf size=92 \[ \frac {B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 B x}{8}+\frac {C \sin ^5(c+d x)}{5 d}-\frac {2 C \sin ^3(c+d x)}{3 d}+\frac {C \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3010, 2748, 2635, 8, 2633} \[ \frac {B \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 B \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 B x}{8}+\frac {C \sin ^5(c+d x)}{5 d}-\frac {2 C \sin ^3(c+d x)}{3 d}+\frac {C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*B*x)/8 + (C*Sin[c + d*x])/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)
- (2*C*Sin[c + d*x]^3)/(3*d) + (C*Sin[c + d*x]^5)/(5*d)

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]

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 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 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
]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 89, normalized size = 0.97 \[ \frac {3 B (c+d x)}{8 d}+\frac {B \sin (2 (c+d x))}{4 d}+\frac {B \sin (4 (c+d x))}{32 d}+\frac {C \sin ^5(c+d x)}{5 d}-\frac {2 C \sin ^3(c+d x)}{3 d}+\frac {C \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*B*(c + d*x))/(8*d) + (C*Sin[c + d*x])/d - (2*C*Sin[c + d*x]^3)/(3*d) + (C*Sin[c + d*x]^5)/(5*d) + (B*Sin[2*
(c + d*x)])/(4*d) + (B*Sin[4*(c + d*x)])/(32*d)

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fricas [A]  time = 0.60, size = 64, normalized size = 0.70 \[ \frac {45 \, B d x + {\left (24 \, C \cos \left (d x + c\right )^{4} + 30 \, B \cos \left (d x + c\right )^{3} + 32 \, C \cos \left (d x + c\right )^{2} + 45 \, B \cos \left (d x + c\right ) + 64 \, C\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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 + 32*C*cos(d*x + c)^2 + 45*B*cos(d*x + c) + 64*C)
*sin(d*x + c))/d

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giac [A]  time = 0.37, size = 77, normalized size = 0.84 \[ \frac {3}{8} \, B x + \frac {C \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {B \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {5 \, C \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {5 \, C \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(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 + 5/48*C*sin(3*d*x + 3*c)/d + 1/4*B*sin(2*d*x
+ 2*c)/d + 5/8*C*sin(d*x + c)/d

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maple [A]  time = 0.21, size = 70, normalized size = 0.76 \[ \frac {\frac {C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(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))

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maxima [A]  time = 0.44, size = 69, normalized size = 0.75 \[ \frac {15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(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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mupad [B]  time = 4.63, size = 115, normalized size = 1.25 \[ \frac {3\,B\,x}{8}+\frac {\left (2\,C-\frac {5\,B}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,C}{3}-\frac {B}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {116\,C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (\frac {B}{2}+\frac {8\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*B*x)/8 + (tan(c/2 + (d*x)/2)^3*(B/2 + (8*C)/3) - tan(c/2 + (d*x)/2)^9*((5*B)/4 - 2*C) - tan(c/2 + (d*x)/2)^
7*(B/2 - (8*C)/3) + (116*C*tan(c/2 + (d*x)/2)^5)/15 + tan(c/2 + (d*x)/2)*((5*B)/4 + 2*C))/(d*(tan(c/2 + (d*x)/
2)^2 + 1)^5)

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sympy [A]  time = 1.79, size = 173, normalized size = 1.88 \[ \begin {cases} \frac {3 B x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 C \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((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*s
in(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*(B*cos(c) + C*cos(c)**
2)*cos(c)**3, True))

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