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

Optimal. Leaf size=120 \[ \frac {2 b \left (-2 a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (-2 a^3 C+2 a^2 b B+a b^2 C+b^3 B\right )+\frac {b^2 (3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

1/2*(2*B*a^2*b+B*b^3-2*C*a^3+C*a*b^2)*x+2/3*b*(3*B*a*b-2*C*a^2+C*b^2)*sin(d*x+c)/d+1/6*b^2*(3*B*b-C*a)*cos(d*x
+c)*sin(d*x+c)/d+1/3*b*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.21, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3015, 2753, 2734} \[ \frac {2 b \left (-2 a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (2 a^2 b B-2 a^3 C+a b^2 C+b^3 B\right )+\frac {b^2 (3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*b*B + b^3*B - 2*a^3*C + a*b^2*C)*x)/2 + (2*b*(3*a*b*B - 2*a^2*C + b^2*C)*Sin[c + d*x])/(3*d) + (b^2*(3
*b*B - a*C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (b*C*(a + b*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 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)) \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))^2 \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))^2 \sin (c+d x)}{3 d}+\frac {\int (a+b \cos (c+d x)) \left (b^2 \left (2 b^2 C+3 a (b B-a C)\right )+b^3 (3 b B-a C) \cos (c+d x)\right ) \, dx}{3 b^2}\\ &=\frac {1}{2} \left (2 a^2 b B+b^3 B-2 a^3 C+a b^2 C\right ) x+\frac {2 b \left (3 a b B-2 a^2 C+b^2 C\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 b B-a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 102, normalized size = 0.85 \[ \frac {3 b \left (-4 a^2 C+8 a b B+3 b^2 C\right ) \sin (c+d x)-6 (c+d x) \left (2 a^3 C-2 a^2 b B-a b^2 C-b^3 B\right )+3 b^2 (a C+b B) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 100, normalized size = 0.83 \[ -\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} d x - {\left (2 \, C b^{3} \cos \left (d x + c\right )^{2} - 6 \, C a^{2} b + 12 \, B a b^{2} + 4 \, C b^{3} + 3 \, {\left (C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(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]

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

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giac [A]  time = 0.18, size = 107, normalized size = 0.89 \[ \frac {C b^{3} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {1}{2} \, {\left (2 \, C a^{3} - 2 \, B a^{2} b - C a b^{2} - B b^{3}\right )} x + \frac {{\left (C a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (4 \, C a^{2} b - 8 \, B a b^{2} - 3 \, C b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(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]

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

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maple [A]  time = 0.19, size = 131, normalized size = 1.09 \[ \frac {\frac {b^{3} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a \,b^{2} \sin \left (d x +c \right )-C \,a^{2} b \sin \left (d x +c \right )+B \left (d x +c \right ) a^{2} b -C \,a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.35, size = 125, normalized size = 1.04 \[ -\frac {12 \, {\left (d x + c\right )} C a^{3} - 12 \, {\left (d x + c\right )} B a^{2} b - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} + 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 12 \, C a^{2} b \sin \left (d x + c\right ) - 24 \, B a b^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(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]

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

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mupad [B]  time = 1.93, size = 132, normalized size = 1.10 \[ \frac {B\,b^3\,x}{2}-C\,a^3\,x+B\,a^2\,b\,x+\frac {C\,a\,b^2\,x}{2}+\frac {3\,C\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{d}-\frac {C\,a^2\,b\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.67, size = 241, normalized size = 2.01 \[ \begin {cases} B a^{2} b x + \frac {2 B a b^{2} \sin {\left (c + d x \right )}}{d} + \frac {B b^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B b^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - C a^{3} x - \frac {C a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {C a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right ) \left (B a b + B b^{2} \cos {\relax (c )} - C a^{2} + C b^{2} \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(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**2*b*x + 2*B*a*b**2*sin(c + d*x)/d + B*b**3*x*sin(c + d*x)**2/2 + B*b**3*x*cos(c + d*x)**2/2 +
B*b**3*sin(c + d*x)*cos(c + d*x)/(2*d) - C*a**3*x - C*a**2*b*sin(c + d*x)/d + C*a*b**2*x*sin(c + d*x)**2/2 + C
*a*b**2*x*cos(c + d*x)**2/2 + C*a*b**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*C*b**3*sin(c + d*x)**3/(3*d) + C*b*
*3*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(a + b*cos(c))*(B*a*b + B*b**2*cos(c) - C*a**2 + C*b**2*cos(c
)**2), True))

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