\(\int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [12]

Optimal result

Integrand size = 29, antiderivative size = 129 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {1}{8} a^2 (8 A+7 B) x+\frac {a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac {a^2 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 A-B) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d} \]

[Out]

1/8*a^2*(8*A+7*B)*x+1/6*a^2*(8*A+7*B)*sin(d*x+c)/d+1/24*a^2*(8*A+7*B)*cos(d*x+c)*sin(d*x+c)/d+1/12*(4*A-B)*(a+
a*cos(d*x+c))^2*sin(d*x+c)/d+1/4*B*(a+a*cos(d*x+c))^3*sin(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3047, 3102, 2830, 2723} \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac {a^2 (8 A+7 B) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} a^2 x (8 A+7 B)+\frac {(4 A-B) \sin (c+d x) (a \cos (c+d x)+a)^2}{12 d}+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d} \]

[In]

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

[Out]

(a^2*(8*A + 7*B)*x)/8 + (a^2*(8*A + 7*B)*Sin[c + d*x])/(6*d) + (a^2*(8*A + 7*B)*Cos[c + d*x]*Sin[c + d*x])/(24
*d) + ((4*A - B)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*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)]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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}& = \int (a+a \cos (c+d x))^2 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx \\ & = \frac {B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {\int (a+a \cos (c+d x))^2 (3 a B+a (4 A-B) \cos (c+d x)) \, dx}{4 a} \\ & = \frac {(4 A-B) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac {1}{12} (8 A+7 B) \int (a+a \cos (c+d x))^2 \, dx \\ & = \frac {1}{8} a^2 (8 A+7 B) x+\frac {a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac {a^2 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 A-B) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.67 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {a^2 (84 B c+96 A d x+84 B d x+24 (7 A+6 B) \sin (c+d x)+48 (A+B) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+16 B \sin (3 (c+d x))+3 B \sin (4 (c+d x)))}{96 d} \]

[In]

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

[Out]

(a^2*(84*B*c + 96*A*d*x + 84*B*d*x + 24*(7*A + 6*B)*Sin[c + d*x] + 48*(A + B)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c
+ d*x)] + 16*B*Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.57

method	result	size
parallelrisch	\(\frac {\left (\frac {\left (A +B \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (\frac {A}{2}+B \right ) \sin \left (3 d x +3 c \right )}{6}+\frac {\sin \left (4 d x +4 c \right ) B}{32}+\frac {\left (\frac {7 A}{2}+3 B \right ) \sin \left (d x +c \right )}{2}+d x \left (A +\frac {7 B}{8}\right )\right ) a^{2}}{d}\)	\(74\)
parts	\(\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\sin \left (d x +c \right ) A \,a^{2}}{d}+\frac {B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(128\)
risch	\(a^{2} x A +\frac {7 a^{2} B x}{8}+\frac {7 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) B \,a^{2}}{2 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{6 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}\)	\(135\)
derivativedivides	\(\frac {\frac {A \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(154\)
default	\(\frac {\frac {A \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(154\)
norman	\(\frac {\frac {a^{2} \left (8 A +7 B \right ) x}{8}+\frac {11 a^{2} \left (8 A +7 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{2} \left (8 A +7 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (8 A +7 B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {3 a^{2} \left (8 A +7 B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{2} \left (8 A +7 B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a^{2} \left (8 A +7 B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a^{2} \left (24 A +25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (136 A +83 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(229\)

[In]

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

[Out]

(1/2*(A+B)*sin(2*d*x+2*c)+1/6*(1/2*A+B)*sin(3*d*x+3*c)+1/32*sin(4*d*x+4*c)*B+1/2*(7/2*A+3*B)*sin(d*x+c)+d*x*(A
+7/8*B))*a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.70 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (8 \, A + 7 \, B\right )} a^{2} d x + {\left (6 \, B a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (8 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (5 \, A + 4 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(8*A + 7*B)*a^2*d*x + (6*B*a^2*cos(d*x + c)^3 + 8*(A + 2*B)*a^2*cos(d*x + c)^2 + 3*(8*A + 7*B)*a^2*cos
(d*x + c) + 8*(5*A + 4*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. 338 vs. \(2 (112) = 224\).

Time = 0.19 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.62 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\begin {cases} A a^{2} x \sin ^{2}{\left (c + d x \right )} + A a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*(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 + A*a**2*x*cos(c + d*x)**2 + 2*A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c
+ d*x)*cos(c + d*x)**2/d + A*a**2*sin(c + d*x)*cos(c + d*x)/d + A*a**2*sin(c + d*x)/d + 3*B*a**2*x*sin(c + d*
x)**4/8 + 3*B*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + B*a**2*x*sin(c + d*x)**2/2 + 3*B*a**2*x*cos(c + d*x)*
*4/8 + B*a**2*x*cos(c + d*x)**2/2 + 3*B*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 4*B*a**2*sin(c + d*x)**3/(3*
d) + 5*B*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 2*B*a**2*sin(c + d*x)*cos(c + d*x)**2/d + B*a**2*sin(c + d*
x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)**2*cos(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 96 \, A a^{2} \sin \left (d x + c\right )}{96 \, d} \]

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 64*(sin(d*x +
c)^3 - 3*sin(d*x + c))*B*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 24*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a^2 - 96*A*a^2*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {B a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (8 \, A a^{2} + 7 \, B a^{2}\right )} x + \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a^{2} + B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/32*B*a^2*sin(4*d*x + 4*c)/d + 1/8*(8*A*a^2 + 7*B*a^2)*x + 1/12*(A*a^2 + 2*B*a^2)*sin(3*d*x + 3*c)/d + 1/2*(A
*a^2 + B*a^2)*sin(2*d*x + 2*c)/d + 1/4*(7*A*a^2 + 6*B*a^2)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.04 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=A\,a^2\,x+\frac {7\,B\,a^2\,x}{8}+\frac {7\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \]

[In]

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

[Out]

A*a^2*x + (7*B*a^2*x)/8 + (7*A*a^2*sin(c + d*x))/(4*d) + (3*B*a^2*sin(c + d*x))/(2*d) + (A*a^2*sin(2*c + 2*d*x
))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/(12*d) + (B*a^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/(6*d) +
(B*a^2*sin(4*c + 4*d*x))/(32*d)