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]

1/16*(6*A+5*C)*x+B*sin(d*x+c)/d+1/16*(6*A+5*C)*cos(d*x+c)*sin(d*x+c)/d+1/24*(6*A+5*C)*cos(d*x+c)^3*sin(d*x+c)/
d+1/6*C*cos(d*x+c)^5*sin(d*x+c)/d-2/3*B*sin(d*x+c)^3/d+1/5*B*sin(d*x+c)^5/d

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.28, 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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fricas [A]  time = 0.58, size = 93, normalized size = 0.70 \[ \frac {15 \, {\left (6 \, A + 5 \, C\right )} d x + {\left (40 \, C \cos \left (d x + c\right )^{5} + 48 \, B \cos \left (d x + c\right )^{4} + 10 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 64 \, B \cos \left (d x + c\right )^{2} + 15 \, {\left (6 \, A + 5 \, C\right )} \cos \left (d x + c\right ) + 128 \, B\right )} \sin \left (d x + c\right )}{240 \, d} \]

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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giac [A]  time = 0.42, size = 110, normalized size = 0.83 \[ \frac {1}{16} \, {\left (6 \, A + 5 \, C\right )} x + \frac {C \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {B \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (2 \, A + 3 \, C\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {5 \, B \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, A + 15 \, C\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {5 \, B \sin \left (d x + c\right )}{8 \, d} \]

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

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maple [A]  time = 0.26, size = 115, normalized size = 0.87 \[ \frac {C \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^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 = 0.33, size = 115, normalized size = 0.87 \[ \frac {30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C}{960 \, d} \]

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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mupad [B]  time = 1.19, size = 126, normalized size = 0.95 \[ \frac {3\,A\,x}{8}+\frac {5\,C\,x}{16}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,B\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,C\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,C\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {C\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {5\,B\,\sin \left (c+d\,x\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*x)/8 + (5*C*x)/16 + (A*sin(2*c + 2*d*x))/(4*d) + (A*sin(4*c + 4*d*x))/(32*d) + (5*B*sin(3*c + 3*d*x))/(48
*d) + (B*sin(5*c + 5*d*x))/(80*d) + (15*C*sin(2*c + 2*d*x))/(64*d) + (3*C*sin(4*c + 4*d*x))/(64*d) + (C*sin(6*
c + 6*d*x))/(192*d) + (5*B*sin(c + d*x))/(8*d)

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sympy [A]  time = 3.45, size = 321, normalized size = 2.43 \[ \begin {cases} \frac {3 A x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 B \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {B \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 C x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 C x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 C x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 C \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 C \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 C \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

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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