\(\int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx\) [225]

Optimal result

Integrand size = 32, antiderivative size = 121 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3}{16} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {2 a^3 A \cos ^5(c+d x)}{5 d}-\frac {3 a^3 A \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d} \]

[Out]

3/16*a^3*A*x-2/3*a^3*A*cos(d*x+c)^3/d+2/5*a^3*A*cos(d*x+c)^5/d-3/16*a^3*A*cos(d*x+c)*sin(d*x+c)/d+5/24*a^3*A*c
os(d*x+c)*sin(d*x+c)^3/d+1/6*a^3*A*cos(d*x+c)*sin(d*x+c)^5/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3045, 2715, 8, 2713} \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {2 a^3 A \cos ^5(c+d x)}{5 d}-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {a^3 A \sin ^5(c+d x) \cos (c+d x)}{6 d}+\frac {5 a^3 A \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {3 a^3 A \sin (c+d x) \cos (c+d x)}{16 d}+\frac {3}{16} a^3 A x \]

[In]

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

[Out]

(3*a^3*A*x)/16 - (2*a^3*A*Cos[c + d*x]^3)/(3*d) + (2*a^3*A*Cos[c + d*x]^5)/(5*d) - (3*a^3*A*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (5*a^3*A*Cos[c + d*x]*Sin[c + d*x]^3)/(24*d) + (a^3*A*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 A \sin ^2(c+d x)+2 a^3 A \sin ^3(c+d x)-2 a^3 A \sin ^5(c+d x)-a^3 A \sin ^6(c+d x)\right ) \, dx \\ & = \left (a^3 A\right ) \int \sin ^2(c+d x) \, dx-\left (a^3 A\right ) \int \sin ^6(c+d x) \, dx+\left (2 a^3 A\right ) \int \sin ^3(c+d x) \, dx-\left (2 a^3 A\right ) \int \sin ^5(c+d x) \, dx \\ & = -\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}+\frac {1}{2} \left (a^3 A\right ) \int 1 \, dx-\frac {1}{6} \left (5 a^3 A\right ) \int \sin ^4(c+d x) \, dx-\frac {\left (2 a^3 A\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {\left (2 a^3 A\right ) \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {1}{2} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {2 a^3 A \cos ^5(c+d x)}{5 d}-\frac {a^3 A \cos (c+d x) \sin (c+d x)}{2 d}+\frac {5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{8} \left (5 a^3 A\right ) \int \sin ^2(c+d x) \, dx \\ & = \frac {1}{2} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {2 a^3 A \cos ^5(c+d x)}{5 d}-\frac {3 a^3 A \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d}-\frac {1}{16} \left (5 a^3 A\right ) \int 1 \, dx \\ & = \frac {3}{16} a^3 A x-\frac {2 a^3 A \cos ^3(c+d x)}{3 d}+\frac {2 a^3 A \cos ^5(c+d x)}{5 d}-\frac {3 a^3 A \cos (c+d x) \sin (c+d x)}{16 d}+\frac {5 a^3 A \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^3 A \cos (c+d x) \sin ^5(c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.64 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {a^3 A (180 c+180 d x-240 \cos (c+d x)-40 \cos (3 (c+d x))+24 \cos (5 (c+d x))-15 \sin (2 (c+d x))-45 \sin (4 (c+d x))+5 \sin (6 (c+d x)))}{960 d} \]

[In]

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

[Out]

(a^3*A*(180*c + 180*d*x - 240*Cos[c + d*x] - 40*Cos[3*(c + d*x)] + 24*Cos[5*(c + d*x)] - 15*Sin[2*(c + d*x)] -
45*Sin[4*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {A \,a^{3} \left (-180 d x +240 \cos \left (d x +c \right )-5 \sin \left (6 d x +6 c \right )-24 \cos \left (5 d x +5 c \right )+45 \sin \left (4 d x +4 c \right )+40 \cos \left (3 d x +3 c \right )+15 \sin \left (2 d x +2 c \right )+256\right )}{960 d}\)	\(79\)
risch	\(\frac {3 a^{3} A x}{16}-\frac {a^{3} A \cos \left (d x +c \right )}{4 d}+\frac {A \,a^{3} \sin \left (6 d x +6 c \right )}{192 d}+\frac {A \,a^{3} \cos \left (5 d x +5 c \right )}{40 d}-\frac {3 A \,a^{3} \sin \left (4 d x +4 c \right )}{64 d}-\frac {A \,a^{3} \cos \left (3 d x +3 c \right )}{24 d}-\frac {A \,a^{3} \sin \left (2 d x +2 c \right )}{64 d}\)	\(114\)
derivativedivides	\(\frac {-A \,a^{3} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {2 A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(136\)
default	\(\frac {-A \,a^{3} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {2 A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}-\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(136\)
parts	\(\frac {A \,a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 A \,a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3 d}+\frac {2 A \,a^{3} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5 d}-\frac {A \,a^{3} \left (-\frac {\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(144\)
norman	\(\frac {-\frac {8 A \,a^{3}}{15 d}+\frac {3 a^{3} A x}{16}-\frac {16 A \,a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 A \,a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {8 A \,a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {13 A \,a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {25 A \,a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {25 A \,a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {13 A \,a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {3 A \,a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {9 a^{3} A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {45 a^{3} A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {15 a^{3} A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {45 a^{3} A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {9 a^{3} A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a^{3} A x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(320\)

[In]

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

[Out]

-1/960*A*a^3*(-180*d*x+240*cos(d*x+c)-5*sin(6*d*x+6*c)-24*cos(5*d*x+5*c)+45*sin(4*d*x+4*c)+40*cos(3*d*x+3*c)+1
5*sin(2*d*x+2*c)+256)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {96 \, A a^{3} \cos \left (d x + c\right )^{5} - 160 \, A a^{3} \cos \left (d x + c\right )^{3} + 45 \, A a^{3} d x + 5 \, {\left (8 \, A a^{3} \cos \left (d x + c\right )^{5} - 26 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

1/240*(96*A*a^3*cos(d*x + c)^5 - 160*A*a^3*cos(d*x + c)^3 + 45*A*a^3*d*x + 5*(8*A*a^3*cos(d*x + c)^5 - 26*A*a^
3*cos(d*x + c)^3 + 9*A*a^3*cos(d*x + c))*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (119) = 238\).

Time = 0.35 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.97 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\begin {cases} - \frac {5 A a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} - \frac {15 A a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} - \frac {15 A a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} - \frac {5 A a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {11 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {2 A a^{3} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {8 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 A a^{3} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {5 A a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {A a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {16 A a^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {4 A a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (- A \sin {\left (c \right )} + A\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \sin ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-5*A*a**3*x*sin(c + d*x)**6/16 - 15*A*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 - 15*A*a**3*x*sin(c
+ d*x)**2*cos(c + d*x)**4/16 + A*a**3*x*sin(c + d*x)**2/2 - 5*A*a**3*x*cos(c + d*x)**6/16 + A*a**3*x*cos(c +
d*x)**2/2 + 11*A*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 2*A*a**3*sin(c + d*x)**4*cos(c + d*x)/d + 5*A*a**3
*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 8*A*a**3*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 2*A*a**3*sin(c + d*x
)**2*cos(c + d*x)/d + 5*A*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - A*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) +
16*A*a**3*cos(c + d*x)**5/(15*d) - 4*A*a**3*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*(a*sin(c) + a
)**3*sin(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {128 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} A a^{3} + 640 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} A a^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 240 \, {\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3}}{960 \, d} \]

[In]

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

[Out]

1/960*(128*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*A*a^3 + 640*(cos(d*x + c)^3 - 3*cos(d*x +
c))*A*a^3 - 5*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*A*a^3 + 240*(2
*d*x + 2*c - sin(2*d*x + 2*c))*A*a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {3}{16} \, A a^{3} x + \frac {A a^{3} \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac {A a^{3} \cos \left (3 \, d x + 3 \, c\right )}{24 \, d} - \frac {A a^{3} \cos \left (d x + c\right )}{4 \, d} + \frac {A a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, A a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} - \frac {A a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

[In]

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

[Out]

3/16*A*a^3*x + 1/40*A*a^3*cos(5*d*x + 5*c)/d - 1/24*A*a^3*cos(3*d*x + 3*c)/d - 1/4*A*a^3*cos(d*x + c)/d + 1/19
2*A*a^3*sin(6*d*x + 6*c)/d - 3/64*A*a^3*sin(4*d*x + 4*c)/d - 1/64*A*a^3*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 14.48 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.12 \[ \int \sin ^2(c+d x) (a+a \sin (c+d x))^3 (A-A \sin (c+d x)) \, dx=\frac {A\,a^3\,\left (45\,c-90\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-768\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+130\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+1500\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-1280\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-1500\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-1920\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-130\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+90\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+45\,d\,x+270\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (c+d\,x\right )+675\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (c+d\,x\right )+900\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (c+d\,x\right )+675\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (c+d\,x\right )+270\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (c+d\,x\right )+45\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (c+d\,x\right )-128\right )}{240\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

[In]

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

[Out]

(A*a^3*(45*c - 90*tan(c/2 + (d*x)/2) - 768*tan(c/2 + (d*x)/2)^2 + 130*tan(c/2 + (d*x)/2)^3 + 1500*tan(c/2 + (d
*x)/2)^5 - 1280*tan(c/2 + (d*x)/2)^6 - 1500*tan(c/2 + (d*x)/2)^7 - 1920*tan(c/2 + (d*x)/2)^8 - 130*tan(c/2 + (
d*x)/2)^9 + 90*tan(c/2 + (d*x)/2)^11 + 45*d*x + 270*tan(c/2 + (d*x)/2)^2*(c + d*x) + 675*tan(c/2 + (d*x)/2)^4*
(c + d*x) + 900*tan(c/2 + (d*x)/2)^6*(c + d*x) + 675*tan(c/2 + (d*x)/2)^8*(c + d*x) + 270*tan(c/2 + (d*x)/2)^1
0*(c + d*x) + 45*tan(c/2 + (d*x)/2)^12*(c + d*x) - 128))/(240*d*(tan(c/2 + (d*x)/2)^2 + 1)^6)