\(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [4]

Optimal result

Integrand size = 21, antiderivative size = 47 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} a (2 A+B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) \sin (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, 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} \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a (A+B) \sin (c+d x)}{d}+\frac {1}{2} a x (2 A+B)+\frac {a B \sin (c+d x) \cos (c+d x)}{2 d} \]

[In]

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

[Out]

(a*(2*A + B)*x)/2 + (a*(A + B)*Sin[c + d*x])/d + (a*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} a (2 A+B) x+\frac {a (A+B) \sin (c+d x)}{d}+\frac {a B \cos (c+d x) \sin (c+d x)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {a (2 B c+4 A d x+2 B d x+4 (A+B) \sin (c+d x)+B \sin (2 (c+d x)))}{4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(\frac {\left (\frac {\sin \left (2 d x +2 c \right ) B}{4}+\left (A +B \right ) \sin \left (d x +c \right )+d x \left (A +\frac {B}{2}\right )\right ) a}{d}\)	\(37\)
risch	\(a x A +\frac {a B x}{2}+\frac {\sin \left (d x +c \right ) a A}{d}+\frac {a B \sin \left (d x +c \right )}{d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}\)	\(51\)
parts	\(a x A +\frac {\left (a A +B a \right ) \sin \left (d x +c \right )}{d}+\frac {B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(51\)
derivativedivides	\(\frac {B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )+B a \sin \left (d x +c \right )+a A \left (d x +c \right )}{d}\)	\(57\)
default	\(\frac {B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )+B a \sin \left (d x +c \right )+a A \left (d x +c \right )}{d}\)	\(57\)
norman	\(\frac {\frac {a \left (2 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+a \left (2 A +B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a \left (2 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a \left (2 A +B \right ) x}{2}+\frac {a \left (2 A +B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(108\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.81 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {{\left (2 \, A + B\right )} a d x + {\left (B a \cos \left (d x + c\right ) + 2 \, {\left (A + B\right )} a\right )} \sin \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

1/2*((2*A + B)*a*d*x + (B*a*cos(d*x + c) + 2*(A + B)*a)*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. \(2 (42) = 84\).

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.00 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} A a x + \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.17 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {4 \, {\left (d x + c\right )} A a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a + 4 \, A a \sin \left (d x + c\right ) + 4 \, B a \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{2} \, {\left (2 \, A a + B a\right )} x + \frac {B a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (A a + B a\right )} \sin \left (d x + c\right )}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx=A\,a\,x+\frac {B\,a\,x}{2}+\frac {A\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

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

[Out]

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