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

Optimal. Leaf size=170 \[ \frac {\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (4 a^2 C+8 a b B+3 b^2 C\right )+\frac {\left (a^3 (-C)+4 a^2 b B+8 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 b d}+\frac {(4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d} \]

[Out]

1/8*(8*B*a*b+4*C*a^2+3*C*b^2)*x+1/6*(4*B*a^2*b+4*B*b^3-C*a^3+8*C*a*b^2)*sin(d*x+c)/b/d+1/24*(8*B*a*b-2*C*a^2+9
*C*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/12*(4*B*b-C*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d+1/4*C*(a+b*cos(d*x+c))^3*si
n(d*x+c)/b/d

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Rubi [A]  time = 0.19, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3023, 2753, 2734} \[ \frac {\left (4 a^2 b B+a^3 (-C)+8 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 b d}+\frac {\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (4 a^2 C+8 a b B+3 b^2 C\right )+\frac {(4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*b*B + 4*a^2*C + 3*b^2*C)*x)/8 + ((4*a^2*b*B + 4*b^3*B - a^3*C + 8*a*b^2*C)*Sin[c + d*x])/(6*b*d) + ((8*a
*b*B - 2*a^2*C + 9*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*b*B - a*C)*(a + b*Cos[c + d*x])^2*Sin[c + d*
x])/(12*b*d) + (C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*b*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 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

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 (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}+\frac {\int (a+b \cos (c+d x))^2 (3 b C+(4 b B-a C) \cos (c+d x)) \, dx}{4 b}\\ &=\frac {(4 b B-a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 b d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}+\frac {\int (a+b \cos (c+d x)) \left (b (8 b B+7 a C)+\left (8 a b B-2 a^2 C+9 b^2 C\right ) \cos (c+d x)\right ) \, dx}{12 b}\\ &=\frac {1}{8} \left (8 a b B+4 a^2 C+3 b^2 C\right ) x+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \sin (c+d x)}{6 b d}+\frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 b d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 118, normalized size = 0.69 \[ \frac {12 (c+d x) \left (4 a^2 C+8 a b B+3 b^2 C\right )+24 \left (4 a^2 B+6 a b C+3 b^2 B\right ) \sin (c+d x)+24 \left (a^2 C+2 a b B+b^2 C\right ) \sin (2 (c+d x))+8 b (2 a C+b B) \sin (3 (c+d x))+3 b^2 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a*b*B + 4*a^2*C + 3*b^2*C)*(c + d*x) + 24*(4*a^2*B + 3*b^2*B + 6*a*b*C)*Sin[c + d*x] + 24*(2*a*b*B + a^
2*C + b^2*C)*Sin[2*(c + d*x)] + 8*b*(b*B + 2*a*C)*Sin[3*(c + d*x)] + 3*b^2*C*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.50, size = 114, normalized size = 0.67 \[ \frac {3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} d x + {\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 24 \, B a^{2} + 32 \, C a b + 16 \, B b^{2} + 8 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*d*x + (6*C*b^2*cos(d*x + c)^3 + 24*B*a^2 + 32*C*a*b + 16*B*b^2 + 8*(2*C*
a*b + B*b^2)*cos(d*x + c)^2 + 3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.20, size = 124, normalized size = 0.73 \[ \frac {C b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} x + \frac {{\left (2 \, C a b + B b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (C a^{2} + 2 \, B a b + C b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*C*b^2*sin(4*d*x + 4*c)/d + 1/8*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*x + 1/12*(2*C*a*b + B*b^2)*sin(3*d*x + 3*c)/
d + 1/4*(C*a^2 + 2*B*a*b + C*b^2)*sin(2*d*x + 2*c)/d + 1/4*(4*B*a^2 + 6*C*a*b + 3*B*b^2)*sin(d*x + c)/d

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maple [A]  time = 0.24, size = 152, normalized size = 0.89 \[ \frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+\frac {2 C a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 B a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 142, normalized size = 0.84 \[ \frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 96 \, B a^{2} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(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))*C*a^2 + 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b - 64*(sin(d*x + c)
^3 - 3*sin(d*x + c))*C*a*b - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^2 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*C*b^2 + 96*B*a^2*sin(d*x + c))/d

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mupad [B]  time = 1.80, size = 169, normalized size = 0.99 \[ \frac {C\,a^2\,x}{2}+\frac {3\,C\,b^2\,x}{8}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+B\,a\,b\,x+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,C\,a\,b\,\sin \left (c+d\,x\right )}{2\,d}+\frac {B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{6\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*a^2*x)/2 + (3*C*b^2*x)/8 + (B*a^2*sin(c + d*x))/d + (3*B*b^2*sin(c + d*x))/(4*d) + B*a*b*x + (C*a^2*sin(2*c
+ 2*d*x))/(4*d) + (B*b^2*sin(3*c + 3*d*x))/(12*d) + (C*b^2*sin(2*c + 2*d*x))/(4*d) + (C*b^2*sin(4*c + 4*d*x))
/(32*d) + (3*C*a*b*sin(c + d*x))/(2*d) + (B*a*b*sin(2*c + 2*d*x))/(2*d) + (C*a*b*sin(3*c + 3*d*x))/(6*d)

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sympy [A]  time = 1.35, size = 340, normalized size = 2.00 \[ \begin {cases} \frac {B a^{2} \sin {\left (c + d x \right )}}{d} + B a b x \sin ^{2}{\left (c + d x \right )} + B a b x \cos ^{2}{\left (c + d x \right )} + \frac {B a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 B b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {C a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 C a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 C a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{2} \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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