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

Optimal. Leaf size=176 \[ \frac {b^2 \left (-14 a^2 C+20 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (-8 a^4 C+8 a^3 b B+12 a b^3 B+3 b^4 C\right )+\frac {b \left (-13 a^3 C+16 a^2 b B+8 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 d}+\frac {b (4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

[Out]

1/8*(8*B*a^3*b+12*B*a*b^3-8*C*a^4+3*C*b^4)*x+1/6*b*(16*B*a^2*b+4*B*b^3-13*C*a^3+8*C*a*b^2)*sin(d*x+c)/d+1/24*b
^2*(20*B*a*b-14*C*a^2+9*C*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/12*b*(4*B*b-C*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*
b*C*(a+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.35, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3015, 2753, 2734} \[ \frac {b \left (16 a^2 b B-13 a^3 C+8 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (-14 a^2 C+20 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} x \left (8 a^3 b B-8 a^4 C+12 a b^3 B+3 b^4 C\right )+\frac {b (4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*b*B + 12*a*b^3*B - 8*a^4*C + 3*b^4*C)*x)/8 + (b*(16*a^2*b*B + 4*b^3*B - 13*a^3*C + 8*a*b^2*C)*Sin[c +
d*x])/(6*d) + (b^2*(20*a*b*B - 14*a^2*C + 9*b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + (b*(4*b*B - a*C)*(a + b
*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (b*C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*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 3015

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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)*Simp[b*B - a*C + b*C*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx &=\frac {\int (a+b \cos (c+d x))^3 \left (b^2 (b B-a C)+b^3 C \cos (c+d x)\right ) \, dx}{b^2}\\ &=\frac {b C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {\int (a+b \cos (c+d x))^2 \left (b^2 \left (3 b^2 C+4 a (b B-a C)\right )+b^3 (4 b B-a C) \cos (c+d x)\right ) \, dx}{4 b^2}\\ &=\frac {b (4 b B-a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {\int (a+b \cos (c+d x)) \left (b^2 \left (12 a^2 b B+8 b^3 B-12 a^3 C+7 a b^2 C\right )+b^3 \left (20 a b B-14 a^2 C+9 b^2 C\right ) \cos (c+d x)\right ) \, dx}{12 b^2}\\ &=\frac {1}{8} \left (8 a^3 b B+12 a b^3 B-8 a^4 C+3 b^4 C\right ) x+\frac {b \left (16 a^2 b B+4 b^3 B-13 a^3 C+8 a b^2 C\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (20 a b B-14 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {b (4 b B-a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 134, normalized size = 0.76 \[ \frac {-12 (c+d x) \left (8 a^4 C-8 a^3 b B-12 a b^3 B-3 b^4 C\right )+24 b \left (-8 a^3 C+12 a^2 b B+6 a b^2 C+3 b^3 B\right ) \sin (c+d x)+24 b^3 (3 a B+b C) \sin (2 (c+d x))+8 b^3 (2 a C+b B) \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 133, normalized size = 0.76 \[ -\frac {3 \, {\left (8 \, C a^{4} - 8 \, B a^{3} b - 12 \, B a b^{3} - 3 \, C b^{4}\right )} d x - {\left (6 \, C b^{4} \cos \left (d x + c\right )^{3} - 48 \, C a^{3} b + 72 \, B a^{2} b^{2} + 32 \, C a b^{3} + 16 \, B b^{4} + 8 \, {\left (2 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (4 \, B a b^{3} + C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(8*C*a^4 - 8*B*a^3*b - 12*B*a*b^3 - 3*C*b^4)*d*x - (6*C*b^4*cos(d*x + c)^3 - 48*C*a^3*b + 72*B*a^2*b^
2 + 32*C*a*b^3 + 16*B*b^4 + 8*(2*C*a*b^3 + B*b^4)*cos(d*x + c)^2 + 9*(4*B*a*b^3 + C*b^4)*cos(d*x + c))*sin(d*x
+ c))/d

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giac [A]  time = 1.59, size = 144, normalized size = 0.82 \[ \frac {C b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {1}{8} \, {\left (8 \, C a^{4} - 8 \, B a^{3} b - 12 \, B a b^{3} - 3 \, C b^{4}\right )} x + \frac {{\left (2 \, C a b^{3} + B b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (3 \, B a b^{3} + C b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (8 \, C a^{3} b - 12 \, B a^{2} b^{2} - 6 \, C a b^{3} - 3 \, B b^{4}\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 168, normalized size = 0.95 \[ \frac {C \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b^{2} B \sin \left (d x +c \right )-2 a^{3} b C \sin \left (d x +c \right )+B \left (d x +c \right ) a^{3} b -a^{4} C \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(C*b^4*(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*b^4*(2+cos(d*x+c)^2)*sin(d*x+c)+
2/3*C*a*b^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*B*a*b^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a^2*b^2*B*sin(d*
x+c)-2*a^3*b*C*sin(d*x+c)+B*(d*x+c)*a^3*b-a^4*C*(d*x+c))

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maxima [A]  time = 0.36, size = 162, normalized size = 0.92 \[ -\frac {96 \, {\left (d x + c\right )} C a^{4} - 96 \, {\left (d x + c\right )} B a^{3} b - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{3} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} + 192 \, C a^{3} b \sin \left (d x + c\right ) - 288 \, B a^{2} b^{2} \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(96*(d*x + c)*C*a^4 - 96*(d*x + c)*B*a^3*b - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b^3 + 64*(sin(d*x +
c)^3 - 3*sin(d*x + c))*C*a*b^3 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^4 - 3*(12*d*x + 12*c + sin(4*d*x +
4*c) + 8*sin(2*d*x + 2*c))*C*b^4 + 192*C*a^3*b*sin(d*x + c) - 288*B*a^2*b^2*sin(d*x + c))/d

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mupad [B]  time = 2.02, size = 187, normalized size = 1.06 \[ \frac {3\,C\,b^4\,x}{8}-C\,a^4\,x+\frac {3\,B\,a\,b^3\,x}{2}+B\,a^3\,b\,x+\frac {3\,B\,b^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {3\,C\,a\,b^3\,\sin \left (c+d\,x\right )}{2\,d}-\frac {2\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.36, size = 357, normalized size = 2.03 \[ \begin {cases} B a^{3} b x + \frac {3 B a^{2} b^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 B a b^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a b^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a b^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B b^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - C a^{4} x - \frac {2 C a^{3} b \sin {\left (c + d x \right )}}{d} + \frac {4 C a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 C a b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{2} \left (B a b + B b^{2} \cos {\relax (c )} - C a^{2} + C b^{2} \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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