\(\int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [989]

Optimal result

Integrand size = 29, antiderivative size = 51 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {(A-B) (a+a \sin (c+d x))^4}{4 a d}+\frac {B (a+a \sin (c+d x))^5}{5 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, 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 = {2912, 45} \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B (a \sin (c+d x)+a)^5}{5 a^2 d}+\frac {(A-B) (a \sin (c+d x)+a)^4}{4 a d} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (1+\sin (c+d x))^4 (5 A-B+4 B \sin (c+d x))}{20 d} \]

[In]

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

[Out]

(a^3*(1 + Sin[c + d*x])^4*(5*A - B + 4*B*Sin[c + d*x]))/(20*d)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84

method	result	size
parallelrisch	\(-\frac {\left (\frac {\left (7 A +5 B \right ) \cos \left (2 d x +2 c \right )}{2}+\frac {\left (-A -3 B \right ) \cos \left (4 d x +4 c \right )}{8}+\left (A +\frac {5 B}{4}\right ) \sin \left (3 d x +3 c \right )-\frac {B \sin \left (5 d x +5 c \right )}{20}+7 \left (-A -\frac {B}{2}\right ) \sin \left (d x +c \right )-\frac {27 A}{8}-\frac {17 B}{8}\right ) a^{3}}{4 d}\)	\(94\)
derivativedivides	\(\frac {\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{3}}{d}\)	\(98\)
default	\(\frac {\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{3}}{d}\)	\(98\)
risch	\(\frac {7 \sin \left (d x +c \right ) A \,a^{3}}{4 d}+\frac {7 a^{3} B \sin \left (d x +c \right )}{8 d}+\frac {\sin \left (5 d x +5 c \right ) B \,a^{3}}{80 d}+\frac {a^{3} \cos \left (4 d x +4 c \right ) A}{32 d}+\frac {3 a^{3} \cos \left (4 d x +4 c \right ) B}{32 d}-\frac {\sin \left (3 d x +3 c \right ) A \,a^{3}}{4 d}-\frac {5 \sin \left (3 d x +3 c \right ) B \,a^{3}}{16 d}-\frac {7 a^{3} \cos \left (2 d x +2 c \right ) A}{8 d}-\frac {5 a^{3} \cos \left (2 d x +2 c \right ) B}{8 d}\)	\(158\)
norman	\(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (5 A +4 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(244\)

[In]

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

[Out]

-1/4*(1/2*(7*A+5*B)*cos(2*d*x+2*c)+1/8*(-A-3*B)*cos(4*d*x+4*c)+(A+5/4*B)*sin(3*d*x+3*c)-1/20*B*sin(5*d*x+5*c)+
7*(-A-1/2*B)*sin(d*x+c)-27/8*A-17/8*B)*a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 4 \, {\left (B a^{3} \cos \left (d x + c\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{20 \, d} \]

[In]

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

[Out]

1/20*(5*(A + 3*B)*a^3*cos(d*x + c)^4 - 40*(A + B)*a^3*cos(d*x + c)^2 + 4*(B*a^3*cos(d*x + c)^4 - (5*A + 7*B)*a
^3*cos(d*x + c)^2 + 2*(5*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. 151 vs. \(2 (41) = 82\).

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.96 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 A a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.65 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{3} + 10 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]

[In]

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

[Out]

1/20*(4*B*a^3*sin(d*x + c)^5 + 5*(A + 3*B)*a^3*sin(d*x + c)^4 + 20*(A + B)*a^3*sin(d*x + c)^3 + 10*(3*A + B)*a
^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (47) = 94\).

Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.27 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {4 \, B a^{3} \sin \left (d x + c\right )^{5} + 5 \, A a^{3} \sin \left (d x + c\right )^{4} + 15 \, B a^{3} \sin \left (d x + c\right )^{4} + 20 \, A a^{3} \sin \left (d x + c\right )^{3} + 20 \, B a^{3} \sin \left (d x + c\right )^{3} + 30 \, A a^{3} \sin \left (d x + c\right )^{2} + 10 \, B a^{3} \sin \left (d x + c\right )^{2} + 20 \, A a^{3} \sin \left (d x + c\right )}{20 \, d} \]

[In]

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

[Out]

1/20*(4*B*a^3*sin(d*x + c)^5 + 5*A*a^3*sin(d*x + c)^4 + 15*B*a^3*sin(d*x + c)^4 + 20*A*a^3*sin(d*x + c)^3 + 20
*B*a^3*sin(d*x + c)^3 + 30*A*a^3*sin(d*x + c)^2 + 10*B*a^3*sin(d*x + c)^2 + 20*A*a^3*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 11.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.59 \[ \int \cos (c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A+3\,B\right )}{4}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^5}{5}+A\,a^3\,\sin \left (c+d\,x\right )+a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{d} \]

[In]

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

[Out]

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