\(\int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [988]

Optimal result

Integrand size = 31, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^5}{5 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^6}{6 a^3 d}-\frac {B (a+a \sin (c+d x))^7}{7 a^4 d} \]

[Out]

2/5*(A-B)*(a+a*sin(d*x+c))^5/a^2/d-1/6*(A-3*B)*(a+a*sin(d*x+c))^6/a^3/d-1/7*B*(a+a*sin(d*x+c))^7/a^4/d

Rubi [A] (verified)

Time = 0.10 (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} \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {B (a \sin (c+d x)+a)^7}{7 a^4 d}-\frac {(A-3 B) (a \sin (c+d x)+a)^6}{6 a^3 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^5}{5 a^2 d} \]

[In]

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

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^6)/(6*a^3*d) - (B*(a + a*Sin[c
+ d*x])^7)/(7*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)^4 \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)^4+(-A+3 B) (a+x)^5-\frac {B (a+x)^6}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {2 (A-B) (a+a \sin (c+d x))^5}{5 a^2 d}-\frac {(A-3 B) (a+a \sin (c+d x))^6}{6 a^3 d}-\frac {B (a+a \sin (c+d x))^7}{7 a^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.68 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 (1+\sin (c+d x))^5 \left (-49 A+9 B+5 (7 A-9 B) \sin (c+d x)+30 B \sin ^2(c+d x)\right )}{210 d} \]

[In]

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

[Out]

-1/210*(a^3*(1 + Sin[c + d*x])^5*(-49*A + 9*B + 5*(7*A - 9*B)*Sin[c + d*x] + 30*B*Sin[c + d*x]^2))/d

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.45

method	result	size
derivativedivides	\(-\frac {a^{3} \left (\frac {\left (\sin ^{7}\left (d x +c \right )\right ) B}{7}+\frac {\left (A +3 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (3 A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 B +2 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-3 B -2 A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(113\)
default	\(-\frac {a^{3} \left (\frac {\left (\sin ^{7}\left (d x +c \right )\right ) B}{7}+\frac {\left (A +3 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (3 A +2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 B +2 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-3 B -2 A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\)	\(113\)
parallelrisch	\(\frac {9 \left (\left (-\frac {3 A}{8}-\frac {17 B}{72}\right ) \cos \left (2 d x +2 c \right )+\left (-\frac {A}{12}-\frac {B}{36}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {A}{216}+\frac {B}{72}\right ) \cos \left (6 d x +6 c \right )+\left (\frac {A}{54}-\frac {5 B}{72}\right ) \sin \left (3 d x +3 c \right )+\left (-\frac {A}{30}-\frac {13 B}{360}\right ) \sin \left (5 d x +5 c \right )+\frac {B \sin \left (7 d x +7 c \right )}{504}+\left (A +\frac {3 B}{8}\right ) \sin \left (d x +c \right )+\frac {49 A}{108}+\frac {B}{4}\right ) a^{3}}{8 d}\)	\(125\)
risch	\(\frac {9 \sin \left (d x +c \right ) A \,a^{3}}{8 d}+\frac {27 a^{3} B \sin \left (d x +c \right )}{64 d}+\frac {\sin \left (7 d x +7 c \right ) B \,a^{3}}{448 d}+\frac {a^{3} \cos \left (6 d x +6 c \right ) A}{192 d}+\frac {a^{3} \cos \left (6 d x +6 c \right ) B}{64 d}-\frac {3 \sin \left (5 d x +5 c \right ) A \,a^{3}}{80 d}-\frac {13 \sin \left (5 d x +5 c \right ) B \,a^{3}}{320 d}-\frac {3 a^{3} \cos \left (4 d x +4 c \right ) A}{32 d}-\frac {a^{3} \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{3}}{48 d}-\frac {5 \sin \left (3 d x +3 c \right ) B \,a^{3}}{64 d}-\frac {27 a^{3} \cos \left (2 d x +2 c \right ) A}{64 d}-\frac {17 a^{3} \cos \left (2 d x +2 c \right ) B}{64 d}\)	\(230\)
norman	\(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (22 A \,a^{3}+18 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (22 A \,a^{3}+18 B \,a^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (76 A \,a^{3}+36 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (76 A \,a^{3}+36 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (13 A +6 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a^{3} \left (13 A +6 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {24 a^{3} \left (49 A +6 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {2 a^{3} \left (241 A +144 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a^{3} \left (241 A +144 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\)	\(356\)

[In]

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

[Out]

-a^3/d*(1/7*sin(d*x+c)^7*B+1/6*(A+3*B)*sin(d*x+c)^6+1/5*(3*A+2*B)*sin(d*x+c)^5+1/4*(-2*B+2*A)*sin(d*x+c)^4+1/3
*(-3*B-2*A)*sin(d*x+c)^3+1/2*(-3*A-B)*sin(d*x+c)^2-A*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.47 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {35 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} - 210 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, B a^{3} \cos \left (d x + c\right )^{6} - 3 \, {\left (21 \, A + 29 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (7 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{210 \, d} \]

[In]

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

[Out]

1/210*(35*(A + 3*B)*a^3*cos(d*x + c)^6 - 210*(A + B)*a^3*cos(d*x + c)^4 + 2*(15*B*a^3*cos(d*x + c)^6 - 3*(21*A
+ 29*B)*a^3*cos(d*x + c)^4 + 8*(7*A + 3*B)*a^3*cos(d*x + c)^2 + 16*(7*A + 3*B)*a^3)*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (70) = 140\).

Time = 0.48 (sec) , antiderivative size = 313, normalized size of antiderivative = 4.01 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{6}{\left (c + d x \right )}}{12 d} - \frac {3 A a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {3 B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {B a^{3} \cos ^{6}{\left (c + d x \right )}}{4 d} - \frac {B a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*A*a**3*sin(c + d*x)**5/(5*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 2*A*a**3*sin(c + d*x)**
3/(3*d) - A*a**3*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)*cos(c + d*x)**2/d - A*a**3*cos(c
+ d*x)**6/(12*d) - 3*A*a**3*cos(c + d*x)**4/(4*d) + 2*B*a**3*sin(c + d*x)**7/(35*d) + B*a**3*sin(c + d*x)**5*c
os(c + d*x)**2/(5*d) + 2*B*a**3*sin(c + d*x)**5/(5*d) + B*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d - 3*B*a**3*si
n(c + d*x)**2*cos(c + d*x)**4/(4*d) - B*a**3*cos(c + d*x)**6/(4*d) - B*a**3*cos(c + d*x)**4/(4*d), Ne(d, 0)),
(x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} + 42 \, {\left (3 \, A + 2 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 105 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} - 105 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]

[In]

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

[Out]

-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*(A + 3*B)*a^3*sin(d*x + c)^6 + 42*(3*A + 2*B)*a^3*sin(d*x + c)^5 + 105*(A
- B)*a^3*sin(d*x + c)^4 - 70*(2*A + 3*B)*a^3*sin(d*x + c)^3 - 105*(3*A + B)*a^3*sin(d*x + c)^2 - 210*A*a^3*si
n(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (72) = 144\).

Time = 0.41 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.21 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, A a^{3} \sin \left (d x + c\right )^{6} + 105 \, B a^{3} \sin \left (d x + c\right )^{6} + 126 \, A a^{3} \sin \left (d x + c\right )^{5} + 84 \, B a^{3} \sin \left (d x + c\right )^{5} + 105 \, A a^{3} \sin \left (d x + c\right )^{4} - 105 \, B a^{3} \sin \left (d x + c\right )^{4} - 140 \, A a^{3} \sin \left (d x + c\right )^{3} - 210 \, B a^{3} \sin \left (d x + c\right )^{3} - 315 \, A a^{3} \sin \left (d x + c\right )^{2} - 105 \, B a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]

[In]

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

[Out]

-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*A*a^3*sin(d*x + c)^6 + 105*B*a^3*sin(d*x + c)^6 + 126*A*a^3*sin(d*x + c)^
5 + 84*B*a^3*sin(d*x + c)^5 + 105*A*a^3*sin(d*x + c)^4 - 105*B*a^3*sin(d*x + c)^4 - 140*A*a^3*sin(d*x + c)^3 -
210*B*a^3*sin(d*x + c)^3 - 315*A*a^3*sin(d*x + c)^2 - 105*B*a^3*sin(d*x + c)^2 - 210*A*a^3*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{2}-\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A+3\,B\right )}{6}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (2\,A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+2\,B\right )}{5}-A\,a^3\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

-((a^3*sin(c + d*x)^4*(A - B))/2 - (a^3*sin(c + d*x)^2*(3*A + B))/2 + (a^3*sin(c + d*x)^6*(A + 3*B))/6 + (B*a^
3*sin(c + d*x)^7)/7 - (a^3*sin(c + d*x)^3*(2*A + 3*B))/3 + (a^3*sin(c + d*x)^5*(3*A + 2*B))/5 - A*a^3*sin(c +
d*x))/d