3.3.15 \(\int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [215]

Optimal. Leaf size=105 \[ \frac {1}{8} (4 a A+3 b B) x+\frac {(A b+a B) \sin (c+d x)}{d}+\frac {(4 a A+3 b B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(A b+a B) \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3102, 2827, 2715, 8, 2713} \begin {gather*} -\frac {(a B+A b) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x)}{d}+\frac {(4 a A+3 b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (4 a A+3 b B)+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*A + 3*b*B)*x)/8 + ((A*b + a*B)*Sin[c + d*x])/d + ((4*a*A + 3*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b*
B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - ((A*b + a*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 3047

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

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Mathematica [A]
time = 0.25, size = 91, normalized size = 0.87 \begin {gather*} \frac {48 a A c+36 b B c+48 a A d x+36 b B d x+96 (A b+a B) \sin (c+d x)-32 (A b+a B) \sin ^3(c+d x)+24 (a A+b B) \sin (2 (c+d x))+3 b B \sin (4 (c+d x))}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 107, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {B b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
default	\(\frac {B b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
risch	\(\frac {a x A}{2}+\frac {3 b B x}{8}+\frac {3 \sin \left (d x +c \right ) A b}{4 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {B b \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A b}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a B}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}\)	\(118\)
norman	\(\frac {\left (\frac {a A}{2}+\frac {3 B b}{8}\right ) x +\left (2 a A +\frac {3 B b}{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a A +\frac {3 B b}{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {9 B b}{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {a A}{2}+\frac {3 B b}{8}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (4 a A -8 A b -8 a B +5 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a A +8 A b +8 a B +5 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a A -40 A b -40 a B -9 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (12 a A +40 A b +40 a B -9 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(247\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 101, normalized size = 0.96 \begin {gather*} \frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A 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 )} B b}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 81, normalized size = 0.77 \begin {gather*} \frac {3 \, {\left (4 \, A a + 3 \, B b\right )} d x + {\left (6 \, B b \cos \left (d x + c\right )^{3} + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 16 \, B a + 16 \, A b + 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (97) = 194\).
time = 0.20, size = 252, normalized size = 2.40 \begin {gather*} \begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \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 {3 B b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b \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 )}\right ) \left (a + b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 89, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, {\left (4 \, A a + 3 \, B b\right )} x + \frac {B b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (B a + A b\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, {\left (B a + A b\right )} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.47, size = 117, normalized size = 1.11 \begin {gather*} \frac {A\,a\,x}{2}+\frac {3\,B\,b\,x}{8}+\frac {3\,A\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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