\(\int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [1530]

Optimal result

Integrand size = 29, antiderivative size = 97 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a A \sin (c+d x)}{d}+\frac {(A b+a B) \sin ^2(c+d x)}{2 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}-\frac {(A b+a B) \sin ^4(c+d x)}{4 d}-\frac {b B \sin ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 97, 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 = {2916, 786} \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {(a B+A b) \sin ^4(c+d x)}{4 d}-\frac {(a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}-\frac {b B \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Rule 786

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2916

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {\sin (c+d x) \left (60 a A+30 (A b+a B) \sin (c+d x)-20 (a A-b B) \sin ^2(c+d x)-15 (A b+a B) \sin ^3(c+d x)-12 b B \sin ^4(c+d x)\right )}{60 d} \]

[In]

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

[Out]

(Sin[c + d*x]*(60*a*A + 30*(A*b + a*B)*Sin[c + d*x] - 20*(a*A - b*B)*Sin[c + d*x]^2 - 15*(A*b + a*B)*Sin[c + d
*x]^3 - 12*b*B*Sin[c + d*x]^4))/(60*d)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(-\frac {\frac {B b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A b +B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a A -B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A b -B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a}{d}\)	\(83\)
default	\(-\frac {\frac {B b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A b +B a \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (a A -B b \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A b -B a \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a}{d}\)	\(83\)
parallelrisch	\(\frac {-60 A \cos \left (2 d x +2 c \right ) b -15 A \cos \left (4 d x +4 c \right ) b +40 a A \sin \left (3 d x +3 c \right )+360 A \sin \left (d x +c \right ) a -60 B \cos \left (2 d x +2 c \right ) a -6 B \sin \left (5 d x +5 c \right ) b -15 B \cos \left (4 d x +4 c \right ) a -10 B \sin \left (3 d x +3 c \right ) b +60 B b \sin \left (d x +c \right )+75 A b +75 B a}{480 d}\)	\(126\)
risch	\(\frac {3 a A \sin \left (d x +c \right )}{4 d}+\frac {b B \sin \left (d x +c \right )}{8 d}-\frac {\sin \left (5 d x +5 c \right ) B b}{80 d}-\frac {\cos \left (4 d x +4 c \right ) A b}{32 d}-\frac {\cos \left (4 d x +4 c \right ) B a}{32 d}+\frac {a A \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) B b}{48 d}-\frac {\cos \left (2 d x +2 c \right ) A b}{8 d}-\frac {\cos \left (2 d x +2 c \right ) B a}{8 d}\)	\(140\)
norman	\(\frac {\frac {\left (2 A b +2 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (2 A b +2 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 \left (2 a A +B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (2 a A +B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (25 a A -4 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A b +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(219\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {15 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B b \cos \left (d x + c\right )^{4} - {\left (5 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 10 \, A a - 2 \, B b\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

-1/60*(15*(B*a + A*b)*cos(d*x + c)^4 + 4*(3*B*b*cos(d*x + c)^4 - (5*A*a + B*b)*cos(d*x + c)^2 - 10*A*a - 2*B*b
)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.32 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (A a - B b\right )} \sin \left (d x + c\right )^{3} - 60 \, A a \sin \left (d x + c\right ) - 30 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {12 \, B b \sin \left (d x + c\right )^{5} + 15 \, B a \sin \left (d x + c\right )^{4} + 15 \, A b \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B b \sin \left (d x + c\right )^{3} - 30 \, B a \sin \left (d x + c\right )^{2} - 30 \, A b \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^5}{5}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {A\,a}{3}-\frac {B\,b}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2-A\,a\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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