3.227 \(\int \cos (c+d x) (a+a \cos (c+d x)) (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=97 \[ -\frac {a (B+C) \sin ^3(c+d x)}{3 d}+\frac {a (B+C) \sin (c+d x)}{d}+\frac {a (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 B+3 C)+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3029, 2968, 3023, 2748, 2635, 8, 2633} \[ -\frac {a (B+C) \sin ^3(c+d x)}{3 d}+\frac {a (B+C) \sin (c+d x)}{d}+\frac {a (4 B+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 B+3 C)+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*B + 3*C)*x)/8 + (a*(B + C)*Sin[c + d*x])/d + (a*(4*B + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*C*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(B + C)*Sin[c + d*x]^3)/(3*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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

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 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 76, normalized size = 0.78 \[ \frac {a (72 (B+C) \sin (c+d x)+24 (B+C) \sin (2 (c+d x))+8 B \sin (3 (c+d x))+48 B d x+8 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+36 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(48*B*d*x + 36*C*d*x + 72*(B + C)*Sin[c + d*x] + 24*(B + C)*Sin[2*(c + d*x)] + 8*B*Sin[3*(c + d*x)] + 8*C*S
in[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.42, size = 74, normalized size = 0.76 \[ \frac {3 \, {\left (4 \, B + 3 \, C\right )} a d x + {\left (6 \, C a \cos \left (d x + c\right )^{3} + 8 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (B + C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 89, normalized size = 0.92 \[ \frac {1}{8} \, {\left (4 \, B a + 3 \, C a\right )} x + \frac {C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (B a + C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, {\left (B a + C a\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 107, normalized size = 1.10 \[ \frac {a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*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*a*B*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
a*C*(2+cos(d*x+c)^2)*sin(d*x+c)+a*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.65, size = 101, normalized size = 1.04 \[ -\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a + 32*(sin(d*x + c)^3
- 3*sin(d*x + c))*C*a - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a)/d

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mupad [B]  time = 2.12, size = 212, normalized size = 2.19 \[ \frac {\left (B\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,B\,a}{3}+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {13\,B\,a}{3}+\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,B+3\,C\right )}{4\,\left (B\,a+\frac {3\,C\,a}{4}\right )}\right )\,\left (4\,B+3\,C\right )}{4\,d}-\frac {a\,\left (4\,B+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*B*a + (13*C*a)/4) + tan(c/2 + (d*x)/2)^7*(B*a + (3*C*a)/4) + tan(c/2 + (d*x)/2)^3*((13*
B*a)/3 + (31*C*a)/12) + tan(c/2 + (d*x)/2)^5*((7*B*a)/3 + (49*C*a)/12))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2
+ (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*atan((a*tan(c/2 + (d*x)/2)*(4*B + 3*C
))/(4*(B*a + (3*C*a)/4)))*(4*B + 3*C))/(4*d) - (a*(4*B + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)

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sympy [A]  time = 1.03, size = 255, normalized size = 2.63 \[ \begin {cases} \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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