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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.214151, antiderivative size = 120, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.387372, size = 102, normalized size = 0.85 \[ \frac{-6 (c+d x) \left (-2 a^2 b B+2 a^3 C-a b^2 C-b^3 B\right )+3 b \left (-4 a^2 C+8 a b B+3 b^2 C\right ) \sin (c+d x)+3 b^2 (a C+b B) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.018, size = 131, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{C{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{3}B \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Ca{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,a{b}^{2}B\sin \left ( dx+c \right ) -{a}^{2}bC\sin \left ( dx+c \right ) +{a}^{2}bB \left ( dx+c \right ) -{a}^{3}C \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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]

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

________________________________________________________________________________________

Maxima [A]  time = 1.03424, size = 169, normalized size = 1.41 \begin{align*} -\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} \end{align*}

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

________________________________________________________________________________________

Fricas [A]  time = 1.64934, size = 224, normalized size = 1.87 \begin{align*} -\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} \end{align*}

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

________________________________________________________________________________________

Sympy [A]  time = 0.86374, size = 241, normalized size = 2.01 \begin{align*} \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{\left (c \right )}\right ) \left (B a b + B b^{2} \cos{\left (c \right )} - C a^{2} + C b^{2} \cos ^{2}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

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))

________________________________________________________________________________________

Giac [A]  time = 1.13797, size = 144, normalized size = 1.2 \begin{align*} \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} \end{align*}

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