\(\int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [972]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.05 (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} \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B (a \sin (c+d x)+a)^4}{4 a^2 d}+\frac {(A-B) (a \sin (c+d x)+a)^3}{3 a d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {1}{3} (A-B) (a+a \sin (c+d x))^3+\frac {B (a+a \sin (c+d x))^4}{4 a}}{a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45

method	result	size
parallelrisch	\(-\frac {\left (\left (6 A +\frac {9 B}{2}\right ) \cos \left (2 d x +2 c \right )+\left (A +2 B \right ) \sin \left (3 d x +3 c \right )-\frac {3 B \cos \left (4 d x +4 c \right )}{8}+\left (-15 A -6 B \right ) \sin \left (d x +c \right )-6 A -\frac {33 B}{8}\right ) a^{2}}{12 d}\)	\(74\)
derivativedivides	\(\frac {\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{2}}{d}\)	\(75\)
default	\(\frac {\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right ) A \,a^{2}}{d}\)	\(75\)
risch	\(\frac {5 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) B \,a^{2}}{2 d}+\frac {a^{2} \cos \left (4 d x +4 c \right ) B}{32 d}-\frac {A \,a^{2} \sin \left (3 d x +3 c \right )}{12 d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{6 d}-\frac {a^{2} \cos \left (2 d x +2 c \right ) A}{2 d}-\frac {3 a^{2} \cos \left (2 d x +2 c \right ) B}{8 d}\)	\(122\)
norman	\(\frac {\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (4 A \,a^{2}+4 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (13 A +8 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (13 A +8 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(193\)

[In]

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

[Out]

-1/12*((6*A+9/2*B)*cos(2*d*x+2*c)+(A+2*B)*sin(3*d*x+3*c)-3/8*B*cos(4*d*x+4*c)+(-15*A-6*B)*sin(d*x+c)-6*A-33/8*
B)*a^2/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \cos \left (d x + c\right )^{4} - 12 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*B*a^2*cos(d*x + c)^4 - 12*(A + B)*a^2*cos(d*x + c)^2 - 4*((A + 2*B)*a^2*cos(d*x + c)^2 - 2*(2*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. 117 vs. \(2 (41) = 82\).

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.29 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin ^{2}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \sin \left (d x + c\right )^{4} + 4 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 12 \, A a^{2} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*B*a^2*sin(d*x + c)^4 + 4*(A + 2*B)*a^2*sin(d*x + c)^3 + 6*(2*A + B)*a^2*sin(d*x + c)^2 + 12*A*a^2*sin(
d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {3 \, B a^{2} \sin \left (d x + c\right )^{4} + 4 \, A a^{2} \sin \left (d x + c\right )^{3} + 8 \, B a^{2} \sin \left (d x + c\right )^{3} + 12 \, A a^{2} \sin \left (d x + c\right )^{2} + 6 \, B a^{2} \sin \left (d x + c\right )^{2} + 12 \, A a^{2} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*B*a^2*sin(d*x + c)^4 + 4*A*a^2*sin(d*x + c)^3 + 8*B*a^2*sin(d*x + c)^3 + 12*A*a^2*sin(d*x + c)^2 + 6*B
*a^2*sin(d*x + c)^2 + 12*A*a^2*sin(d*x + c))/d

Mupad [B] (verification not implemented)

Time = 10.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.29 \[ \int \cos (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A+2\,B\right )}{3}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^4}{4}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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