\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [13]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {1}{2} a^2 (3 A+2 B) x+\frac {2 a^2 (3 A+2 B) \sin (c+d x)}{3 d}+\frac {a^2 (3 A+2 B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, 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 = {2830, 2723} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {2 a^2 (3 A+2 B) \sin (c+d x)}{3 d}+\frac {a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a^2 x (3 A+2 B)+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[In]

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

[Out]

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

Rule 2723

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(2*a^2 + b^2)*(x/2), x] + (-Simp[2*a*b*(Cos[c
+ d*x]/d), x] - Simp[b^2*Cos[c + d*x]*(Sin[c + d*x]/(2*d)), x]) /; FreeQ[{a, b, c, d}, x]

Rule 2830

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.89 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {a^2 \sin (c+d x) \left (12 A+11 B+3 (A+2 B) \cos (c+d x)+B \cos (2 (c+d x))+\frac {6 (3 A+2 B) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{6 d} \]

[In]

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

[Out]

(a^2*Sin[c + d*x]*(12*A + 11*B + 3*(A + 2*B)*Cos[c + d*x] + B*Cos[2*(c + d*x)] + (6*(3*A + 2*B)*ArcSin[Sqrt[Si
n[(c + d*x)/2]^2]])/Sqrt[Sin[c + d*x]^2]))/(6*d)

Maple [A] (verified)

Time = 1.93 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(\frac {3 \left (\left (\frac {A}{6}+\frac {B}{3}\right ) \sin \left (2 d x +2 c \right )+\frac {B \sin \left (3 d x +3 c \right )}{18}+\left (\frac {4 A}{3}+\frac {7 B}{6}\right ) \sin \left (d x +c \right )+d x \left (A +\frac {2 B}{3}\right )\right ) a^{2}}{2 d}\)	\(61\)
parts	\(a^{2} x A +\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(93\)
risch	\(\frac {3 a^{2} x A}{2}+a^{2} B x +\frac {2 \sin \left (d x +c \right ) A \,a^{2}}{d}+\frac {7 \sin \left (d x +c \right ) B \,a^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}\)	\(99\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A \,a^{2} \sin \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )}{d}\)	\(116\)
default	\(\frac {A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A \,a^{2} \sin \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )}{d}\)	\(116\)
norman	\(\frac {\frac {a^{2} \left (3 A +2 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (5 A +6 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (3 A +2 B \right ) x}{2}+\frac {8 a^{2} \left (3 A +2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {3 a^{2} \left (3 A +2 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a^{2} \left (3 A +2 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a^{2} \left (3 A +2 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(177\)

[In]

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

[Out]

3/2*((1/6*A+1/3*B)*sin(2*d*x+2*c)+1/18*B*sin(3*d*x+3*c)+(4/3*A+7/6*B)*sin(d*x+c)+d*x*(A+2/3*B))*a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (3 \, A + 2 \, B\right )} a^{2} d x + {\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (6 \, A + 5 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(3*A + 2*B)*a^2*d*x + (2*B*a^2*cos(d*x + c)^2 + 3*(A + 2*B)*a^2*cos(d*x + c) + 2*(6*A + 5*B)*a^2)*sin(d
*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (85) = 170\).

Time = 0.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.12 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + A a^{2} x + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a^{2} \sin {\left (c + d x \right )}}{d} + B a^{2} x \sin ^{2}{\left (c + d x \right )} + B a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right )^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.17 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 12 \, {\left (d x + c\right )} A a^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 24 \, A a^{2} \sin \left (d x + c\right ) + 12 \, B a^{2} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 12*(d*x + c)*A*a^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^
2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 24*A*a^2*sin(d*x + c) + 12*B*a^2*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {B a^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {1}{2} \, {\left (3 \, A a^{2} + 2 \, B a^{2}\right )} x + \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/12*B*a^2*sin(3*d*x + 3*c)/d + 1/2*(3*A*a^2 + 2*B*a^2)*x + 1/4*(A*a^2 + 2*B*a^2)*sin(2*d*x + 2*c)/d + 1/4*(8*
A*a^2 + 7*B*a^2)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {3\,A\,a^2\,x}{2}+B\,a^2\,x+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

(3*A*a^2*x)/2 + B*a^2*x + (2*A*a^2*sin(c + d*x))/d + (7*B*a^2*sin(c + d*x))/(4*d) + (A*a^2*sin(2*c + 2*d*x))/(
4*d) + (B*a^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/(12*d)