\(\int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [971]

Optimal result

Integrand size = 31, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^4}{2 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {B (a+a \sin (c+d x))^6}{6 a^4 d} \]

[Out]

1/2*(A-B)*(a+a*sin(d*x+c))^4/a^2/d-1/5*(A-3*B)*(a+a*sin(d*x+c))^5/a^3/d-1/6*B*(a+a*sin(d*x+c))^6/a^4/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 78} \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {B (a \sin (c+d x)+a)^6}{6 a^4 d}-\frac {(A-3 B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac {(A-B) (a \sin (c+d x)+a)^4}{2 a^2 d} \]

[In]

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

[Out]

((A - B)*(a + a*Sin[c + d*x])^4)/(2*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^5)/(5*a^3*d) - (B*(a + a*Sin[c +
d*x])^6)/(6*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)^3 \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)^3+(-A+3 B) (a+x)^4-\frac {B (a+x)^5}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {(A-B) (a+a \sin (c+d x))^4}{2 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^5}{5 a^3 d}-\frac {B (a+a \sin (c+d x))^6}{6 a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^8 (18 A-9 B+5 B \cos (2 (c+d x))-4 (3 A-4 B) \sin (c+d x))}{60 d} \]

[In]

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

[Out]

(a^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8*(18*A - 9*B + 5*B*Cos[2*(c + d*x)] - 4*(3*A - 4*B)*Sin[c + d*x]))
/(60*d)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(-\frac {a^{2} \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right ) B}{6}+\frac {\left (A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) A}{2}-\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-B -2 A \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(84\)
default	\(-\frac {a^{2} \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right ) B}{6}+\frac {\left (A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right ) A}{2}-\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-B -2 A \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(84\)
parallelrisch	\(\frac {a^{2} \left (60 A \sin \left (3 d x +3 c \right )-60 A \cos \left (4 d x +4 c \right )-12 A \sin \left (5 d x +5 c \right )-240 A \cos \left (2 d x +2 c \right )+840 A \sin \left (d x +c \right )-40 B \sin \left (3 d x +3 c \right )-30 B \cos \left (4 d x +4 c \right )-24 B \sin \left (5 d x +5 c \right )-165 B \cos \left (2 d x +2 c \right )+240 B \sin \left (d x +c \right )+5 B \cos \left (6 d x +6 c \right )+300 A +190 B \right )}{960 d}\)	\(142\)
risch	\(\frac {7 \sin \left (d x +c \right ) A \,a^{2}}{8 d}+\frac {\sin \left (d x +c \right ) B \,a^{2}}{4 d}+\frac {a^{2} \cos \left (6 d x +6 c \right ) B}{192 d}-\frac {\sin \left (5 d x +5 c \right ) A \,a^{2}}{80 d}-\frac {\sin \left (5 d x +5 c \right ) B \,a^{2}}{40 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) A}{16 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {A \,a^{2} \sin \left (3 d x +3 c \right )}{16 d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{24 d}-\frac {a^{2} \cos \left (2 d x +2 c \right ) A}{4 d}-\frac {11 a^{2} \cos \left (2 d x +2 c \right ) B}{64 d}\)	\(194\)
norman	\(\frac {\frac {\left (8 A \,a^{2}+8 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (8 A \,a^{2}+8 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (6 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (15 A +8 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (15 A +8 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a^{2} \left (17 A +4 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {4 a^{2} \left (17 A +4 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(300\)

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-a^2/d*(1/6*sin(d*x+c)^6*B+1/5*(A+2*B)*sin(d*x+c)^5+1/2*sin(d*x+c)^4*A-2/3*B*sin(d*x+c)^3+1/2*(-B-2*A)*sin(d*x
+c)^2-A*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.17 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {5 \, B a^{2} \cos \left (d x + c\right )^{6} - 15 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (3 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(5*B*a^2*cos(d*x + c)^6 - 15*(A + B)*a^2*cos(d*x + c)^4 - 2*(3*(A + 2*B)*a^2*cos(d*x + c)^4 - 2*(3*A + B)
*a^2*cos(d*x + c)^2 - 4*(3*A + 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. 228 vs. \(2 (68) = 136\).

Time = 0.34 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.92 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {4 B a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a^{2} \cos ^{6}{\left (c + d x \right )}}{12 d} - \frac {B a^{2} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((2*A*a**2*sin(c + d*x)**5/(15*d) + A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 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*cos(c + d*x)**4/(2*d) + 4*B*a**2*sin(c + d*x)**5
/(15*d) + 2*B*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a**2*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) - B*a*
*2*cos(c + d*x)**6/(12*d) - B*a**2*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c
)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 15 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/30*(5*B*a^2*sin(d*x + c)^6 + 6*(A + 2*B)*a^2*sin(d*x + c)^5 + 15*A*a^2*sin(d*x + c)^4 - 20*B*a^2*sin(d*x +
c)^3 - 15*(2*A + B)*a^2*sin(d*x + c)^2 - 30*A*a^2*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \, A a^{2} \sin \left (d x + c\right )^{5} + 12 \, B a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 30 \, A a^{2} \sin \left (d x + c\right )^{2} - 15 \, B a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/30*(5*B*a^2*sin(d*x + c)^6 + 6*A*a^2*sin(d*x + c)^5 + 12*B*a^2*sin(d*x + c)^5 + 15*A*a^2*sin(d*x + c)^4 - 2
0*B*a^2*sin(d*x + c)^3 - 30*A*a^2*sin(d*x + c)^2 - 15*B*a^2*sin(d*x + c)^2 - 30*A*a^2*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 10.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\frac {A\,a^2\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^5\,\left (A+2\,B\right )}{5}-\frac {2\,B\,a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^6}{6}-A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

[In]

int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2,x)

[Out]

-((A*a^2*sin(c + d*x)^4)/2 - (a^2*sin(c + d*x)^2*(2*A + B))/2 + (a^2*sin(c + d*x)^5*(A + 2*B))/5 - (2*B*a^2*si
n(c + d*x)^3)/3 + (B*a^2*sin(c + d*x)^6)/6 - A*a^2*sin(c + d*x))/d