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]

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

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, 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 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]

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*}

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Mathematica [A]  time = 0.23, 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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fricas [A]  time = 0.66, size = 85, normalized size = 0.79 \[ \frac {3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} d x + {\left (2 \, B b^{2} \cos \left (d x + c\right )^{2} + 6 \, B a^{2} + 12 \, A a b + 4 \, B b^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

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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giac [A]  time = 0.47, size = 93, normalized size = 0.87 \[ \frac {B b^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {1}{2} \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} x + \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

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

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maple [A]  time = 0.05, size = 114, normalized size = 1.07 \[ \frac {\frac {b^{2} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A a b \sin \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(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 = 0.31, size = 108, normalized size = 1.01 \[ \frac {12 \, {\left (d x + c\right )} A a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 12 \, B a^{2} \sin \left (d x + c\right ) + 24 \, A a b \sin \left (d x + c\right )}{12 \, d} \]

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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mupad [B]  time = 0.45, size = 115, normalized size = 1.07 \[ A\,a^2\,x+\frac {A\,b^2\,x}{2}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+B\,a\,b\,x+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,A\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.59, size = 199, normalized size = 1.86 \[ \begin {cases} A a^{2} x + \frac {2 A a b \sin {\left (c + d x \right )}}{d} + \frac {A b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a^{2} \sin {\left (c + d x \right )}}{d} + B a b x \sin ^{2}{\left (c + d x \right )} + B a b x \cos ^{2}{\left (c + d x \right )} + \frac {B a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 B b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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