\(\int (a+a \cos (c+d x)) (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [228]

Optimal result

Integrand size = 30, antiderivative size = 85 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} a (B+C) x+\frac {a (3 B+C) \sin (c+d x)}{3 d}+\frac {a (3 B-C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3102, 2813} \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a (3 B+C) \sin (c+d x)}{3 d}+\frac {a (3 B-C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a x (B+C)+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 a d} \]

[In]

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

[Out]

(a*(B + C)*x)/2 + (a*(3*B + C)*Sin[c + d*x])/(3*d) + (a*(3*B - C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (C*(a + a
*Cos[c + d*x])^2*Sin[c + d*x])/(3*a*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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a (6 B c+6 c C+6 B d x+6 C d x+3 (4 B+3 C) \sin (c+d x)+3 (B+C) \sin (2 (c+d x))+C \sin (3 (c+d x)))}{12 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.48 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {\left (\frac {\left (B +C \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\sin \left (3 d x +3 c \right ) C}{6}+\left (2 B +\frac {3 C}{2}\right ) \sin \left (d x +c \right )+\left (B +C \right ) x d \right ) a}{2 d}\)	\(54\)
parts	\(\frac {\left (B a +a C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a B \sin \left (d x +c \right )}{d}+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(70\)
derivativedivides	\(\frac {\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
default	\(\frac {\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B a \sin \left (d x +c \right )}{d}\)	\(85\)
risch	\(\frac {a B x}{2}+\frac {a C x}{2}+\frac {a B \sin \left (d x +c \right )}{d}+\frac {3 a C \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a C}{4 d}\)	\(85\)
norman	\(\frac {\frac {a \left (B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (B +C \right ) x}{2}+\frac {3 a \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 a \left (B +C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a \left (B +C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \left (B +C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {4 a \left (3 B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(138\)

[In]

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

[Out]

1/2*(1/2*(B+C)*sin(2*d*x+2*c)+1/6*sin(3*d*x+3*c)*C+(2*B+3/2*C)*sin(d*x+c)+(B+C)*x*d)*a/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (B + C\right )} a d x + {\left (2 \, C a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, {\left (3 \, B + 2 \, C\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (73) = 146\).

Time = 0.13 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.00 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a \sin {\left (c + d x \right )}}{d} + \frac {C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {C a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.93 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 12 \, B a \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{2} \, {\left (B a + C a\right )} x + \frac {C a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a + 3 \, C a\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99 \[ \int (a+a \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {B\,a\,x}{2}+\frac {C\,a\,x}{2}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

(B*a*x)/2 + (C*a*x)/2 + (B*a*sin(c + d*x))/d + (3*C*a*sin(c + d*x))/(4*d) + (B*a*sin(2*c + 2*d*x))/(4*d) + (C*
a*sin(2*c + 2*d*x))/(4*d) + (C*a*sin(3*c + 3*d*x))/(12*d)