\(\int (a+b \cos (c+d x)) (A+C \cos ^2(c+d x)) \, dx\) [524]

Optimal result

Integrand size = 23, antiderivative size = 96 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} a (2 A+C) x-\frac {\left (a^2 C-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b d}-\frac {a C \cos (c+d x) \sin (c+d x)}{6 d}+\frac {C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3103, 2813} \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {\left (a^2 C-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b d}+\frac {1}{2} a x (2 A+C)+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 b d}-\frac {a C \sin (c+d x) \cos (c+d x)}{6 d} \]

[In]

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

[Out]

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

Rule 3103

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[
(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*
x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !Lt
Q[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 b d}+\frac {\int (a+b \cos (c+d x)) (b (3 A+2 C)-a C \cos (c+d x)) \, dx}{3 b} \\ & = \frac {1}{2} a (2 A+C) x-\frac {\left (a^2 C-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b d}-\frac {a C \cos (c+d x) \sin (c+d x)}{6 d}+\frac {C (a+b \cos (c+d x))^2 \sin (c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {6 a c C+12 a A d x+6 a C d x+3 b (4 A+3 C) \sin (c+d x)+3 a C \sin (2 (c+d x))+b C \sin (3 (c+d x))}{12 d} \]

[In]

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

[Out]

(6*a*c*C + 12*a*A*d*x + 6*a*C*d*x + 3*b*(4*A + 3*C)*Sin[c + d*x] + 3*a*C*Sin[2*(c + d*x)] + b*C*Sin[3*(c + d*x
)])/(12*d)

Maple [A] (verified)

Time = 3.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {3 a \sin \left (2 d x +2 c \right ) C +b \sin \left (3 d x +3 c \right ) C +12 b \left (A +\frac {3 C}{4}\right ) \sin \left (d x +c \right )+12 x \left (A +\frac {C}{2}\right ) d a}{12 d}\)	\(56\)
derivativedivides	\(\frac {\frac {C b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \sin \left (d x +c \right ) A +a A \left (d x +c \right )}{d}\)	\(68\)
default	\(\frac {\frac {C b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \sin \left (d x +c \right ) A +a A \left (d x +c \right )}{d}\)	\(68\)
risch	\(a x A +\frac {a C x}{2}+\frac {\sin \left (d x +c \right ) A b}{d}+\frac {3 b C \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) C b}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\)	\(68\)
parts	\(a x A +\frac {\sin \left (d x +c \right ) A b}{d}+\frac {a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {C b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(69\)
norman	\(\frac {\left (a A +\frac {1}{2} a C \right ) x +\left (a A +\frac {1}{2} a C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} a C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a A +\frac {3}{2} a C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 A b -a C +2 C b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +a C +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 b \left (3 A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(168\)

[In]

int((a+cos(d*x+c)*b)*(A+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.58 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A + C\right )} a d x + {\left (2 \, C b \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 2 \, C\right )} b\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.26 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} A a x + \frac {A b \sin {\left (c + d x \right )}}{d} + \frac {C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((A*a*x + A*b*sin(c + d*x)/d + C*a*x*sin(c + d*x)**2/2 + C*a*x*cos(c + d*x)**2/2 + C*a*sin(c + d*x)*c
os(c + d*x)/(2*d) + 2*C*b*sin(c + d*x)**3/(3*d) + C*b*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(A + C*cos
(c)**2)*(a + b*cos(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (d x + c\right )} A a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b + 12 \, A b \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} \, {\left (2 \, A a + C a\right )} x + \frac {C b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {C a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A b + 3 \, C b\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx=A\,a\,x+\frac {C\,a\,x}{2}+\frac {A\,b\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

A*a*x + (C*a*x)/2 + (A*b*sin(c + d*x))/d + (3*C*b*sin(c + d*x))/(4*d) + (C*a*sin(2*c + 2*d*x))/(4*d) + (C*b*si
n(3*c + 3*d*x))/(12*d)