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

Optimal. Leaf size=107 \[ \frac{2 \left (a^2 B+3 a A b+b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (2 a^2 A+2 a b B+A b^2\right )+\frac{b (2 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

((2*a^2*A + A*b^2 + 2*a*b*B)*x)/2 + (2*(3*a*A*b + a^2*B + b^2*B)*Sin[c + d*x])/(3*d) + (b*(3*A*b + 2*a*B)*Cos[
c + d*x]*Sin[c + d*x])/(6*d) + (B*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.0935754, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ \frac{2 \left (a^2 B+3 a A b+b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (2 a^2 A+2 a b B+A b^2\right )+\frac{b (2 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*A + A*b^2 + 2*a*b*B)*x)/2 + (2*(3*a*A*b + a^2*B + b^2*B)*Sin[c + d*x])/(3*d) + (b*(3*A*b + 2*a*B)*Cos[
c + d*x]*Sin[c + d*x])/(6*d) + (B*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

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))^2 (A+B \cos (c+d x)) \, dx &=\frac{B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int (a+b \cos (c+d x)) (3 a A+2 b B+(3 A b+2 a B) \cos (c+d x)) \, dx\\ &=\frac{1}{2} \left (2 a^2 A+A b^2+2 a b B\right ) x+\frac{2 \left (3 a A b+a^2 B+b^2 B\right ) \sin (c+d x)}{3 d}+\frac{b (3 A b+2 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.213408, size = 90, normalized size = 0.84 \[ \frac{6 (c+d x) \left (2 a^2 A+2 a b B+A b^2\right )+3 \left (4 a^2 B+8 a A b+3 b^2 B\right ) \sin (c+d x)+3 b (2 a B+A b) \sin (2 (c+d x))+b^2 B \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 114, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,Bab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,Aab\sin \left ( dx+c \right ) +B{a}^{2}\sin \left ( dx+c \right ) +{a}^{2}A \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))^2*(A+B*cos(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.0408, size = 146, normalized size = 1.36 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a^{2} + 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 12 \, B a^{2} \sin \left (d x + c\right ) + 24 \, A a b \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))^2*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*A*a^2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^2
 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^2 + 12*B*a^2*sin(d*x + c) + 24*A*a*b*sin(d*x + c))/d

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Fricas [A]  time = 1.33801, size = 201, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} d x +{\left (2 \, B b^{2} \cos \left (d x + c\right )^{2} + 6 \, B a^{2} + 12 \, A a b + 4 \, B b^{2} + 3 \,{\left (2 \, B a b + A b^{2}\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))^2*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.839749, size = 199, normalized size = 1.86 \begin{align*} \begin{cases} A a^{2} x + \frac{2 A a b \sin{\left (c + d x \right )}}{d} + \frac{A b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \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} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3853, size = 126, normalized size = 1.18 \begin{align*} \frac{B b^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{1}{2} \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} x + \frac{{\left (2 \, B a b + A b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\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))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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