\(\int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [955]

Optimal result

Integrand size = 29, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^3}{3 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^4}{4 a^3 d}-\frac {B (a+a \sin (c+d x))^5}{5 a^4 d} \]

[Out]

2/3*(A-B)*(a+a*sin(d*x+c))^3/a^2/d-1/4*(A-3*B)*(a+a*sin(d*x+c))^4/a^3/d-1/5*B*(a+a*sin(d*x+c))^5/a^4/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2915, 78} \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {B (a \sin (c+d x)+a)^5}{5 a^4 d}-\frac {(A-3 B) (a \sin (c+d x)+a)^4}{4 a^3 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^3}{3 a^2 d} \]

[In]

Int[Cos[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^3)/(3*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^4)/(4*a^3*d) - (B*(a + a*Sin[c
+ d*x])^5)/(5*a^4*d)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x) (a+x)^2 \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (2 a (A-B) (a+x)^2+(-A+3 B) (a+x)^3-\frac {B (a+x)^4}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {2 (A-B) (a+a \sin (c+d x))^3}{3 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^4}{4 a^3 d}-\frac {B (a+a \sin (c+d x))^5}{5 a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \left (A \sin (c+d x)+\frac {1}{2} (A+B) \sin ^2(c+d x)-\frac {1}{3} (A-B) \sin ^3(c+d x)-\frac {1}{4} (A+B) \sin ^4(c+d x)-\frac {1}{5} B \sin ^5(c+d x)\right )}{d} \]

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

(a*(A*Sin[c + d*x] + ((A + B)*Sin[c + d*x]^2)/2 - ((A - B)*Sin[c + d*x]^3)/3 - ((A + B)*Sin[c + d*x]^4)/4 - (B
*Sin[c + d*x]^5)/5))/d

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\sin ^{5}\left (d x +c \right )\right ) B}{5}+\frac {\left (A +B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A -B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(73\)
default	\(-\frac {a \left (\frac {\left (\sin ^{5}\left (d x +c \right )\right ) B}{5}+\frac {\left (A +B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A -B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(73\)
parallelrisch	\(\frac {a \left (\frac {3 \left (-A -B \right ) \cos \left (2 d x +2 c \right )}{2}+\frac {3 \left (-A -B \right ) \cos \left (4 d x +4 c \right )}{8}+\left (A -\frac {B}{4}\right ) \sin \left (3 d x +3 c \right )-\frac {3 B \sin \left (5 d x +5 c \right )}{20}+3 \left (3 A +\frac {B}{2}\right ) \sin \left (d x +c \right )+\frac {15 A}{8}+\frac {15 B}{8}\right )}{12 d}\)	\(92\)
risch	\(\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {a B \sin \left (d x +c \right )}{8 d}-\frac {\sin \left (5 d x +5 c \right ) B a}{80 d}-\frac {a \cos \left (4 d x +4 c \right ) A}{32 d}-\frac {a \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) B a}{48 d}-\frac {a \cos \left (2 d x +2 c \right ) A}{8 d}-\frac {a \cos \left (2 d x +2 c \right ) B}{8 d}\)	\(140\)
norman	\(\frac {\frac {\left (2 a A +2 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 a A +2 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (a A +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a \left (2 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a \left (2 A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a \left (25 A -4 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(214\)

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-a/d*(1/5*sin(d*x+c)^5*B+1/4*(A+B)*sin(d*x+c)^4+1/3*(A-B)*sin(d*x+c)^3+1/2*(-A-B)*sin(d*x+c)^2-A*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {15 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B a \cos \left (d x + c\right )^{4} - {\left (5 \, A + B\right )} a \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*(A + B)*a*cos(d*x + c)^4 + 4*(3*B*a*cos(d*x + c)^4 - (5*A + B)*a*cos(d*x + c)^2 - 2*(5*A + B)*a)*sin
(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.64 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((2*A*a*sin(c + d*x)**3/(3*d) + A*a*sin(c + d*x)*cos(c + d*x)**2/d - A*a*cos(c + d*x)**4/(4*d) + 2*B*
a*sin(c + d*x)**5/(15*d) + B*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a*cos(c + d*x)**4/(4*d), Ne(d, 0)), (
x*(A + B*sin(c))*(a*sin(c) + a)*cos(c)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{4} + 20 \, {\left (A - B\right )} a \sin \left (d x + c\right )^{3} - 30 \, {\left (A + B\right )} a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*(A + B)*a*sin(d*x + c)^4 + 20*(A - B)*a*sin(d*x + c)^3 - 30*(A + B)*a*sin(d*
x + c)^2 - 60*A*a*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.28 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B a \sin \left (d x + c\right )^{5} + 15 \, A a \sin \left (d x + c\right )^{4} + 15 \, B a \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B a \sin \left (d x + c\right )^{3} - 30 \, A a \sin \left (d x + c\right )^{2} - 30 \, B a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*A*a*sin(d*x + c)^4 + 15*B*a*sin(d*x + c)^4 + 20*A*a*sin(d*x + c)^3 - 20*B*a*
sin(d*x + c)^3 - 30*A*a*sin(d*x + c)^2 - 30*B*a*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\frac {B\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,\left (A-B\right )\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {a\,\left (A+B\right )\,{\sin \left (c+d\,x\right )}^2}{2}-A\,a\,\sin \left (c+d\,x\right )}{d} \]

[In]

int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + a*sin(c + d*x)),x)

[Out]

-((a*sin(c + d*x)^4*(A + B))/4 - (a*sin(c + d*x)^2*(A + B))/2 - A*a*sin(c + d*x) + (a*sin(c + d*x)^3*(A - B))/
3 + (B*a*sin(c + d*x)^5)/5)/d