\(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [217]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} (2 a A+b B) x+\frac {(A b+a B) \sin (c+d x)}{d}+\frac {b B \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2813} \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {(a B+A b) \sin (c+d x)}{d}+\frac {1}{2} x (2 a A+b B)+\frac {b B \sin (c+d x) \cos (c+d x)}{2 d} \]

[In]

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

[Out]

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

Rule 2813

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*} \text {integral}& = \frac {1}{2} (2 a A+b B) x+\frac {(A b+a B) \sin (c+d x)}{d}+\frac {b B \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {2 b B c+4 a A d x+2 b B d x+4 (A b+a B) \sin (c+d x)+b B \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(2*b*B*c + 4*a*A*d*x + 2*b*B*d*x + 4*(A*b + a*B)*Sin[c + d*x] + b*B*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.98

method	result	size
risch	\(x a A +\frac {b B x}{2}+\frac {\sin \left (d x +c \right ) A b}{d}+\frac {a B \sin \left (d x +c \right )}{d}+\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}\)	\(51\)
parallelrisch	\(\frac {4 A a d x +2 B b d x +4 A \sin \left (d x +c \right ) b +4 B \sin \left (d x +c \right ) a +B \sin \left (2 d x +2 c \right ) b}{4 d}\)	\(51\)
parts	\(x a A +\frac {\left (A b +B a \right ) \sin \left (d x +c \right )}{d}+\frac {B b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(51\)
derivativedivides	\(\frac {B b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b +B \sin \left (d x +c \right ) a +a A \left (d x +c \right )}{d}\)	\(57\)
default	\(\frac {B b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b +B \sin \left (d x +c \right ) a +a A \left (d x +c \right )}{d}\)	\(57\)
norman	\(\frac {\left (a A +\frac {B b}{2}\right ) x +\left (a A +\frac {B b}{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 a A +B b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 A b +2 B a -B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(123\)

[In]

int((a+cos(d*x+c)*b)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {{\left (2 \, A a + B b\right )} d x + {\left (B b \cos \left (d x + c\right ) + 2 \, B a + 2 \, A b\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (44) = 88\).

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.81 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} A a x + \frac {A b \sin {\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )}}{d} + \frac {B b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {4 \, {\left (d x + c\right )} A a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b + 4 \, B a \sin \left (d x + c\right ) + 4 \, A b \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} \, {\left (2 \, A a + B b\right )} x + \frac {B b \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (B a + A b\right )} \sin \left (d x + c\right )}{d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=A\,a\,x+\frac {B\,b\,x}{2}+\frac {A\,b\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

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