3.290 \(\int \cos ^4(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{(6 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{(6 A+5 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x (6 A+5 C)+\frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

[Out]

((6*A + 5*C)*x)/16 + (B*Sin[c + d*x])/d + ((6*A + 5*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*A + 5*C)*Cos[c
+ d*x]^3*Sin[c + d*x])/(24*d) + (C*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*B*Sin[c + d*x]^3)/(3*d) + (B*Sin[c
+ d*x]^5)/(5*d)

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Rubi [A]  time = 0.114, antiderivative size = 132, 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, 2635, 8, 2633} \[ \frac{(6 A+5 C) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{(6 A+5 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x (6 A+5 C)+\frac{B \sin ^5(c+d x)}{5 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

((6*A + 5*C)*x)/16 + (B*Sin[c + d*x])/d + ((6*A + 5*C)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*A + 5*C)*Cos[c
+ d*x]^3*Sin[c + d*x])/(24*d) + (C*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*B*Sin[c + d*x]^3)/(3*d) + (B*Sin[c
+ d*x]^5)/(5*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 ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^4(c+d x) (6 A+5 C+6 B \cos (c+d x)) \, dx\\ &=\frac{C \cos ^5(c+d x) \sin (c+d x)}{6 d}+B \int \cos ^5(c+d x) \, dx+\frac{1}{6} (6 A+5 C) \int \cos ^4(c+d x) \, dx\\ &=\frac{(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{C \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{8} (6 A+5 C) \int \cos ^2(c+d x) \, dx-\frac{B \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{B \sin (c+d x)}{d}+\frac{(6 A+5 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{C \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}+\frac{1}{16} (6 A+5 C) \int 1 \, dx\\ &=\frac{1}{16} (6 A+5 C) x+\frac{B \sin (c+d x)}{d}+\frac{(6 A+5 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{(6 A+5 C) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{C \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{2 B \sin ^3(c+d x)}{3 d}+\frac{B \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.275709, size = 102, normalized size = 0.77 \[ \frac{5 ((48 A+45 C) \sin (2 (c+d x))+(6 A+9 C) \sin (4 (c+d x))+72 A c+72 A d x+C \sin (6 (c+d x))+60 c C+60 C d x)+192 B \sin ^5(c+d x)-640 B \sin ^3(c+d x)+960 B \sin (c+d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(960*B*Sin[c + d*x] - 640*B*Sin[c + d*x]^3 + 192*B*Sin[c + d*x]^5 + 5*(72*A*c + 60*c*C + 72*A*d*x + 60*C*d*x +
(48*A + 45*C)*Sin[2*(c + d*x)] + (6*A + 9*C)*Sin[4*(c + d*x)] + C*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.015, size = 115, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( C \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{B\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 ) }+A \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/d*(C*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/5*B*(8/3+cos(d*x+c)^
4+4/3*cos(d*x+c)^2)*sin(d*x+c)+A*(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 = 1.01, size = 155, normalized size = 1.17 \begin{align*} \frac{30 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A + 64 \,{\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 )} B - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/960*(30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3
+ 15*sin(d*x + c))*B - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*C)
/d

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Fricas [A]  time = 1.88558, size = 244, normalized size = 1.85 \begin{align*} \frac{15 \,{\left (6 \, A + 5 \, C\right )} d x +{\left (40 \, C \cos \left (d x + c\right )^{5} + 48 \, B \cos \left (d x + c\right )^{4} + 10 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 64 \, B \cos \left (d x + c\right )^{2} + 15 \,{\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 128 \, B\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(6*A + 5*C)*d*x + (40*C*cos(d*x + c)^5 + 48*B*cos(d*x + c)^4 + 10*(6*A + 5*C)*cos(d*x + c)^3 + 64*B*
cos(d*x + c)^2 + 15*(6*A + 5*C)*cos(d*x + c) + 128*B)*sin(d*x + c))/d

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Sympy [A]  time = 4.59171, size = 321, normalized size = 2.43 \begin{align*} \begin{cases} \frac{3 A x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 A x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 A x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 A \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 A \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 B \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 B \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{B \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{5 C x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 C x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 C x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 C x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 C \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 C \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 C \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{4}{\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)**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((3*A*x*sin(c + d*x)**4/8 + 3*A*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*x*cos(c + d*x)**4/8 + 3*A*s
in(c + d*x)**3*cos(c + d*x)/(8*d) + 5*A*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 8*B*sin(c + d*x)**5/(15*d) + 4*B*
sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*sin(c + d*x)*cos(c + d*x)**4/d + 5*C*x*sin(c + d*x)**6/16 + 15*C*x*s
in(c + d*x)**4*cos(c + d*x)**2/16 + 15*C*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*C*x*cos(c + d*x)**6/16 + 5*C
*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*C*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*C*sin(c + d*x)*cos(c + d
*x)**5/(16*d), Ne(d, 0)), (x*(A + B*cos(c) + C*cos(c)**2)*cos(c)**4, True))

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Giac [A]  time = 1.20374, size = 149, normalized size = 1.13 \begin{align*} \frac{1}{16} \,{\left (6 \, A + 5 \, C\right )} x + \frac{C \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{B \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{{\left (2 \, A + 3 \, C\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{5 \, B \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (16 \, A + 15 \, C\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{5 \, B \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)^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/16*(6*A + 5*C)*x + 1/192*C*sin(6*d*x + 6*c)/d + 1/80*B*sin(5*d*x + 5*c)/d + 1/64*(2*A + 3*C)*sin(4*d*x + 4*c
)/d + 5/48*B*sin(3*d*x + 3*c)/d + 1/64*(16*A + 15*C)*sin(2*d*x + 2*c)/d + 5/8*B*sin(d*x + c)/d