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

Optimal. Leaf size=69 \[ \frac {(3 A+2 C) \sin (c+d x)}{3 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {3023, 2734} \[ \frac {(3 A+2 C) \sin (c+d x)}{3 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, 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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 53, normalized size = 0.77 \[ \frac {3 (4 A+3 C) \sin (c+d x)+3 B \sin (2 (c+d x))+6 B c+6 B d x+C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*B*c + 6*B*d*x + 3*(4*A + 3*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)])/(12*d)

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fricas [A]  time = 0.44, size = 45, normalized size = 0.65 \[ \frac {3 \, B d x + {\left (2 \, C \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 6 \, A + 4 \, C\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*B*d*x + (2*C*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 6*A + 4*C)*sin(d*x + c))/d

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giac [A]  time = 0.50, size = 53, normalized size = 0.77 \[ \frac {1}{2} \, B x + \frac {C \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {B \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A + 3 \, C\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x + 1/12*C*sin(3*d*x + 3*c)/d + 1/4*B*sin(2*d*x + 2*c)/d + 1/4*(4*A + 3*C)*sin(d*x + c)/d

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maple [A]  time = 0.17, size = 57, normalized size = 0.83 \[ \frac {\frac {C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 55, normalized size = 0.80 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C + 12 \, A \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C + 12*A*sin(d*x + c))/d

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mupad [B]  time = 1.08, size = 66, normalized size = 0.96 \[ \frac {B\,x}{2}+\frac {A\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,\sin \left (c+d\,x\right )}{3\,d}+\frac {B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {C\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 107, normalized size = 1.55 \[ \begin {cases} \frac {A \sin {\left (c + d x \right )}}{d} + \frac {B x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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