\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [995]

Optimal result

Integrand size = 31, antiderivative size = 91 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {3}{2} a^3 (2 A+3 B) x+\frac {2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac {a^3 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(A+B) \sec (c+d x) (a+a \sin (c+d x))^3}{d} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\sec (c+d x) \left (4 \sqrt {2} a^3 (2 A+3 B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)}-B (a+a \sin (c+d x))^3\right )}{2 d} \] Input:

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

Output:

(Sec[c + d*x]*(4*Sqrt[2]*a^3*(2*A + 3*B)*Hypergeometric2F1[-3/2, -1/2, 1/2 
, (1 - Sin[c + d*x])/2]*Sqrt[1 + Sin[c + d*x]] - B*(a + a*Sin[c + d*x])^3) 
)/(2*d)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 3334, 3042, 3123}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^2(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3 (A+B \sin (c+d x))}{\cos (c+d x)^2}dx\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d}-a (2 A+3 B) \int (\sin (c+d x) a+a)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d}-a (2 A+3 B) \int (\sin (c+d x) a+a)^2dx\)

\(\Big \downarrow \) 3123

\(\displaystyle \frac {(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d}-a (2 A+3 B) \left (-\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2}\right )\)

Input:

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

Output:

((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^3)/d - a*(2*A + 3*B)*((3*a^2*x) 
/2 - (2*a^2*Cos[c + d*x])/d - (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3123

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(2*a^2 + b^ 
2)*(x/2), x] + (-Simp[2*a*b*(Cos[c + d*x]/d), x] - Simp[b^2*Cos[c + d*x]*(S 
in[c + d*x]/(2*d)), x]) /; FreeQ[{a, b, c, d}, x]
 

rule 3334

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* 
c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) 
, x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1)))   Int[(g*Cos[e + 
f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, 
 f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29

method	result	size
parallelrisch	\(\frac {a^{3} \left (36 A +44 B +B \sin \left (3 d x +3 c \right )+33 B \sin \left (d x +c \right )+4 A \cos \left (2 d x +2 c \right )+40 A \cos \left (d x +c \right )-24 d x A \cos \left (d x +c \right )+12 B \cos \left (2 d x +2 c \right )+56 B \cos \left (d x +c \right )-36 d x B \cos \left (d x +c \right )+32 A \sin \left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )}\)	\(117\)
risch	\(-3 a^{3} x A -\frac {9 a^{3} B x}{2}+\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )} A}{2 d}+\frac {3 a^{3} {\mathrm e}^{i \left (d x +c \right )} B}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )} A}{2 d}+\frac {3 a^{3} {\mathrm e}^{-i \left (d x +c \right )} B}{2 d}+\frac {8 a^{3} A}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {8 a^{3} B}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {\sin \left (2 d x +2 c \right ) a^{3} B}{4 d}\)	\(152\)
derivativedivides	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} A \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+\frac {3 a^{3} A}{\cos \left (d x +c \right )}+3 a^{3} B \left (\tan \left (d x +c \right )-d x -c \right )+a^{3} A \tan \left (d x +c \right )+\frac {a^{3} B}{\cos \left (d x +c \right )}}{d}\)	\(219\)
default	\(\frac {a^{3} A \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+a^{3} B \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} A \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{3} B \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+\frac {3 a^{3} A}{\cos \left (d x +c \right )}+3 a^{3} B \left (\tan \left (d x +c \right )-d x -c \right )+a^{3} A \tan \left (d x +c \right )+\frac {a^{3} B}{\cos \left (d x +c \right )}}{d}\)	\(219\)
norman	\(\frac {\left (3 a^{3} A +\frac {9}{2} a^{3} B \right ) x +\left (-9 a^{3} A -\frac {27}{2} a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-3 a^{3} A -\frac {9}{2} a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (9 a^{3} A +\frac {27}{2} a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-6 a^{3} A -9 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (6 a^{3} A +9 a^{3} B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {10 a^{3} A +14 a^{3} B}{d}-\frac {\left (6 a^{3} A +2 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {2 \left (14 a^{3} A +10 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {2 \left (24 a^{3} A +24 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {\left (36 a^{3} A +44 a^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {32 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {32 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{3} \left (8 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{3} \left (8 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {2 a^{3} \left (24 A +23 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(432\)

Input:

int(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

1/8/d*a^3*(36*A+44*B+B*sin(3*d*x+3*c)+33*B*sin(d*x+c)+4*A*cos(2*d*x+2*c)+4 
0*A*cos(d*x+c)-24*d*x*A*cos(d*x+c)+12*B*cos(2*d*x+2*c)+56*B*cos(d*x+c)-36* 
d*x*B*cos(d*x+c)+32*A*sin(d*x+c))/cos(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + 2 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a^{3} - {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x - {\left (10 \, A + 13 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) + {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + B a^{3} \cos \left (d x + c\right )^{2} - {\left (2 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/2*(B*a^3*cos(d*x + c)^3 - 3*(2*A + 3*B)*a^3*d*x + 2*(A + 3*B)*a^3*cos(d* 
x + c)^2 + 8*(A + B)*a^3 - (3*(2*A + 3*B)*a^3*d*x - (10*A + 13*B)*a^3)*cos 
(d*x + c) + (3*(2*A + 3*B)*a^3*d*x + B*a^3*cos(d*x + c)^2 - (2*A + 5*B)*a^ 
3*cos(d*x + c) + 8*(A + B)*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x 
+ c) + d)
 

Sympy [F]

\[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sin(c + d*x)*sec(c + d 
*x)**2, x) + Integral(3*A*sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(A 
*sin(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*sin(c + d*x)*sec(c + d*x 
)**2, x) + Integral(3*B*sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*B 
*sin(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*sin(c + d*x)**4*sec(c + 
d*x)**2, x))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.84 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} A a^{3} + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} B a^{3} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} B a^{3} - 2 \, A a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, B a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 2 \, A a^{3} \tan \left (d x + c\right ) - \frac {6 \, A a^{3}}{\cos \left (d x + c\right )} - \frac {2 \, B a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \] Input:

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/2*(6*(d*x + c - tan(d*x + c))*A*a^3 + (3*d*x + 3*c - tan(d*x + c)/(tan( 
d*x + c)^2 + 1) - 2*tan(d*x + c))*B*a^3 + 6*(d*x + c - tan(d*x + c))*B*a^3 
 - 2*A*a^3*(1/cos(d*x + c) + cos(d*x + c)) - 6*B*a^3*(1/cos(d*x + c) + cos 
(d*x + c)) - 2*A*a^3*tan(d*x + c) - 6*A*a^3/cos(d*x + c) - 2*B*a^3/cos(d*x 
 + c))/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {3 \, {\left (2 \, A a^{3} + 3 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {16 \, {\left (A a^{3} + B a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A a^{3} - 6 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \] Input:

integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/2*(3*(2*A*a^3 + 3*B*a^3)*(d*x + c) + 16*(A*a^3 + B*a^3)/(tan(1/2*d*x + 
1/2*c) - 1) + 2*(B*a^3*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3*tan(1/2*d*x + 1/2* 
c)^2 - 6*B*a^3*tan(1/2*d*x + 1/2*c)^2 - B*a^3*tan(1/2*d*x + 1/2*c) - 2*A*a 
^3 - 6*B*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
 

Mupad [B] (verification not implemented)

Time = 36.04 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.57 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {10\,A\,a^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3+5\,B\,a^3\right )+14\,B\,a^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a^3+7\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,A\,a^3+9\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (18\,A\,a^3+21\,B\,a^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-\frac {3\,a^3\,\mathrm {atan}\left (\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )}{6\,A\,a^3+9\,B\,a^3}\right )\,\left (2\,A+3\,B\right )}{d} \] Input:

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

Output:

- (10*A*a^3 - tan(c/2 + (d*x)/2)*(2*A*a^3 + 5*B*a^3) + 14*B*a^3 - tan(c/2 
+ (d*x)/2)^3*(2*A*a^3 + 7*B*a^3) + tan(c/2 + (d*x)/2)^4*(8*A*a^3 + 9*B*a^3 
) + tan(c/2 + (d*x)/2)^2*(18*A*a^3 + 21*B*a^3))/(d*(tan(c/2 + (d*x)/2) - 2 
*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^4 + ta 
n(c/2 + (d*x)/2)^5 - 1)) - (3*a^3*atan((3*a^3*tan(c/2 + (d*x)/2)*(2*A + 3* 
B))/(6*A*a^3 + 9*B*a^3))*(2*A + 3*B))/d
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.20 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -6 \cos \left (d x +c \right ) a d x -16 \cos \left (d x +c \right ) a -9 \cos \left (d x +c \right ) b d x -18 \cos \left (d x +c \right ) b -\sin \left (d x +c \right )^{3} b -2 \sin \left (d x +c \right )^{2} a -6 \sin \left (d x +c \right )^{2} b -6 \sin \left (d x +c \right ) a d x +2 \sin \left (d x +c \right ) a -9 \sin \left (d x +c \right ) b d x +5 \sin \left (d x +c \right ) b +6 a d x +16 a +9 b d x +18 b \right )}{2 d \left (\cos \left (d x +c \right )+\sin \left (d x +c \right )-1\right )} \] Input:

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

Output:

(a**3*(cos(c + d*x)*sin(c + d*x)**2*b + 2*cos(c + d*x)*sin(c + d*x)*a + 5* 
cos(c + d*x)*sin(c + d*x)*b - 6*cos(c + d*x)*a*d*x - 16*cos(c + d*x)*a - 9 
*cos(c + d*x)*b*d*x - 18*cos(c + d*x)*b - sin(c + d*x)**3*b - 2*sin(c + d* 
x)**2*a - 6*sin(c + d*x)**2*b - 6*sin(c + d*x)*a*d*x + 2*sin(c + d*x)*a - 
9*sin(c + d*x)*b*d*x + 5*sin(c + d*x)*b + 6*a*d*x + 16*a + 9*b*d*x + 18*b) 
)/(2*d*(cos(c + d*x) + sin(c + d*x) - 1))