\(\int \cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [292]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} (4 A+3 C) x+\frac {B \sin (c+d x)}{d}+\frac {(4 A+3 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {C \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {B \sin ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3102, 2827, 2715, 8, 2713} \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (4 A+3 C)-\frac {B \sin ^3(c+d x)}{3 d}+\frac {B \sin (c+d x)}{d}+\frac {C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

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 3102

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {48 A c+36 c C+48 A d x+36 C d x+96 B \sin (c+d x)-32 B \sin ^3(c+d x)+24 (A+C) \sin (2 (c+d x))+3 C \sin (4 (c+d x))}{96 d} \]

[In]

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

[Out]

(48*A*c + 36*c*C + 48*A*d*x + 36*C*d*x + 96*B*Sin[c + d*x] - 32*B*Sin[c + d*x]^3 + 24*(A + C)*Sin[2*(c + d*x)]
+ 3*C*Sin[4*(c + d*x)])/(96*d)

Maple [A] (verified)

Time = 3.83 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {24 \left (A +C \right ) \sin \left (2 d x +2 c \right )+8 B \sin \left (3 d x +3 c \right )+3 \sin \left (4 d x +4 c \right ) C +72 B \sin \left (d x +c \right )+48 \left (A +\frac {3 C}{4}\right ) x d}{96 d}\)	\(63\)
risch	\(\frac {x A}{2}+\frac {3 C x}{8}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (4 d x +4 c \right ) C}{32 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 d}\)	\(82\)
derivativedivides	\(\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
default	\(\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
parts	\(\frac {A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(89\)
norman	\(\frac {\left (\frac {A}{2}+\frac {3 C}{8}\right ) x +\left (2 A +\frac {3 C}{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A +\frac {3 C}{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A +\frac {9 C}{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {A}{2}+\frac {3 C}{8}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (4 A -8 B +5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 A +8 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A -40 B -9 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (12 A +40 B -9 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(209\)

[In]

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

[Out]

1/96*(24*(A+C)*sin(2*d*x+2*c)+8*B*sin(3*d*x+3*c)+3*sin(4*d*x+4*c)*C+72*B*sin(d*x+c)+48*(A+3/4*C)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.74 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 3 \, C\right )} d x + {\left (6 \, C \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (76) = 152\).

Time = 0.17 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.24 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C}{96 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} \, {\left (4 \, A + 3 \, C\right )} x + \frac {C \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {B \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A + C\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, B \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92 \[ \int \cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {A\,x}{2}+\frac {3\,C\,x}{8}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,\sin \left (c+d\,x\right )}{4\,d} \]

[In]

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

[Out]

(A*x)/2 + (3*C*x)/8 + (A*sin(2*c + 2*d*x))/(4*d) + (B*sin(3*c + 3*d*x))/(12*d) + (C*sin(2*c + 2*d*x))/(4*d) +
(C*sin(4*c + 4*d*x))/(32*d) + (3*B*sin(c + d*x))/(4*d)