\(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx\) [279]

Optimal result

Integrand size = 35, antiderivative size = 225 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx=\frac {d^2 (3 B (c-d)+A d) x}{a^3}+\frac {d^2 (3 B (c-9 d)+A (2 c+7 d)) \cos (e+f x)}{15 a^3 f}-\frac {(c-d) \left (3 B \left (c^2+6 c d-15 d^2\right )+A \left (2 c^2+7 c d+15 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(3 B (c-3 d)+2 A (c+2 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{15 a f (a+a \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a+a \sin (e+f x))^3} \] Output:

d^2*(3*B*(c-d)+A*d)*x/a^3+1/15*d^2*(3*B*(c-9*d)+A*(2*c+7*d))*cos(f*x+e)/a^ 
3/f-1/15*(c-d)*(3*B*(c^2+6*c*d-15*d^2)+A*(2*c^2+7*c*d+15*d^2))*cos(f*x+e)/ 
f/(a^3+a^3*sin(f*x+e))-1/15*(3*B*(c-3*d)+2*A*(c+2*d))*cos(f*x+e)*(c+d*sin( 
f*x+e))^2/a/f/(a+a*sin(f*x+e))^2-1/5*(A-B)*cos(f*x+e)*(c+d*sin(f*x+e))^3/f 
/(a+a*sin(f*x+e))^3
 

Mathematica [A] (verified)

Time = 5.46 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (A-B) (c-d)^3 \sin \left (\frac {1}{2} (e+f x)\right )-3 (A-B) (c-d)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (c-d)^2 (3 B (c-6 d)+A (2 c+13 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(c-d)^2 (3 B (c-6 d)+A (2 c+13 d)) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 (c-d) \left (3 B \left (c^2+8 c d-24 d^2\right )+A \left (2 c^2+11 c d+32 d^2\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-15 d^2 (-3 B c-A d+3 B d) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-15 B d^3 \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{15 a^3 f (1+\sin (e+f x))^3} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(A - B)*(c - d)^3*Sin[(e + f*x)/ 
2] - 3*(A - B)*(c - d)^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - d) 
^2*(3*B*(c - 6*d) + A*(2*c + 13*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + S 
in[(e + f*x)/2])^2 - (c - d)^2*(3*B*(c - 6*d) + A*(2*c + 13*d))*(Cos[(e + 
f*x)/2] + Sin[(e + f*x)/2])^3 + 2*(c - d)*(3*B*(c^2 + 8*c*d - 24*d^2) + A* 
(2*c^2 + 11*c*d + 32*d^2))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f 
*x)/2])^4 - 15*d^2*(-3*B*c - A*d + 3*B*d)*(e + f*x)*(Cos[(e + f*x)/2] + Si 
n[(e + f*x)/2])^5 - 15*B*d^3*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x 
)/2])^5))/(15*a^3*f*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3456, 3042, 3456, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 3127}

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.

\(\displaystyle \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a \sin (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x))^2 (a (2 A c+3 B c+3 A d-3 B d)-a (A-6 B) d \sin (e+f x))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x))^2 (a (2 A c+3 B c+3 A d-3 B d)-a (A-6 B) d \sin (e+f x))}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {(c+d \sin (e+f x)) \left (a^2 \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right )-a^2 d (3 B (c-9 d)+A (2 c+7 d)) \sin (e+f x)\right )}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(c+d \sin (e+f x)) \left (a^2 \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right )-a^2 d (3 B (c-9 d)+A (2 c+7 d)) \sin (e+f x)\right )}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {\int \frac {-d^2 (3 B (c-9 d)+A (2 c+7 d)) \sin ^2(e+f x) a^2+c \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right ) a^2+\left (a^2 d \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right )-a^2 c d (3 B (c-9 d)+A (2 c+7 d))\right ) \sin (e+f x)}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {-d^2 (3 B (c-9 d)+A (2 c+7 d)) \sin (e+f x)^2 a^2+c \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right ) a^2+\left (a^2 d \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right )-a^2 c d (3 B (c-9 d)+A (2 c+7 d))\right ) \sin (e+f x)}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {\frac {\int \frac {c \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right ) a^3+15 d^2 (3 B (c-d)+A d) \sin (e+f x) a^3}{\sin (e+f x) a+a}dx}{a}+\frac {a d^2 (A (2 c+7 d)+3 B (c-9 d)) \cos (e+f x)}{f}}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {c \left (3 B \left (c^2+5 d c-6 d^2\right )+A \left (2 c^2+5 d c+8 d^2\right )\right ) a^3+15 d^2 (3 B (c-d)+A d) \sin (e+f x) a^3}{\sin (e+f x) a+a}dx}{a}+\frac {a d^2 (A (2 c+7 d)+3 B (c-9 d)) \cos (e+f x)}{f}}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {\frac {a^3 (c-d) \left (A \left (2 c^2+7 c d+15 d^2\right )+3 B \left (c^2+6 c d-15 d^2\right )\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a^2 d^2 x (A d+3 B (c-d))}{a}+\frac {a d^2 (A (2 c+7 d)+3 B (c-9 d)) \cos (e+f x)}{f}}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^3 (c-d) \left (A \left (2 c^2+7 c d+15 d^2\right )+3 B \left (c^2+6 c d-15 d^2\right )\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a^2 d^2 x (A d+3 B (c-d))}{a}+\frac {a d^2 (A (2 c+7 d)+3 B (c-9 d)) \cos (e+f x)}{f}}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3127

\(\displaystyle \frac {\frac {\frac {15 a^2 d^2 x (A d+3 B (c-d))-\frac {a^3 (c-d) \left (A \left (2 c^2+7 c d+15 d^2\right )+3 B \left (c^2+6 c d-15 d^2\right )\right ) \cos (e+f x)}{f (a \sin (e+f x)+a)}}{a}+\frac {a d^2 (A (2 c+7 d)+3 B (c-9 d)) \cos (e+f x)}{f}}{3 a^2}-\frac {a (2 A (c+2 d)+3 B (c-3 d)) \cos (e+f x) (c+d \sin (e+f x))^2}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^3}{5 f (a \sin (e+f x)+a)^3}\)

Input:

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

Output:

-1/5*((A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(f*(a + a*Sin[e + f*x]) 
^3) + (-1/3*(a*(3*B*(c - 3*d) + 2*A*(c + 2*d))*Cos[e + f*x]*(c + d*Sin[e + 
 f*x])^2)/(f*(a + a*Sin[e + f*x])^2) + ((a*d^2*(3*B*(c - 9*d) + A*(2*c + 7 
*d))*Cos[e + f*x])/f + (15*a^2*d^2*(3*B*(c - d) + A*d)*x - (a^3*(c - d)*(3 
*B*(c^2 + 6*c*d - 15*d^2) + A*(2*c^2 + 7*c*d + 15*d^2))*Cos[e + f*x])/(f*( 
a + a*Sin[e + f*x])))/a)/(3*a^2))/(5*a^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3127

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + 
 d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b 
^2, 0]
 

rule 3214

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d   Int[1/(c + d 
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3456

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.55

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A \,c^{3}-A \,d^{3}-3 B c \,d^{2}+3 B \,d^{3}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A \,c^{3}+6 A \,c^{2} d -2 A \,d^{3}+2 B \,c^{3}-6 B c \,d^{2}+4 B \,d^{3}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A \,c^{3}-18 A \,c^{2} d +12 A c \,d^{2}-2 A \,d^{3}-6 B \,c^{3}+12 B \,c^{2} d -6 B c \,d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 A \,c^{3}+24 A \,c^{2} d -24 A c \,d^{2}+8 A \,d^{3}+8 B \,c^{3}-24 B \,c^{2} d +24 B c \,d^{2}-8 B \,d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A \,c^{3}-12 A \,c^{2} d +12 A c \,d^{2}-4 A \,d^{3}-4 B \,c^{3}+12 B \,c^{2} d -12 B c \,d^{2}+4 B \,d^{3}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+2 d^{2} \left (-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d +3 B c -3 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{a^{3} f}\)	\(349\)
default	\(\frac {-\frac {2 \left (A \,c^{3}-A \,d^{3}-3 B c \,d^{2}+3 B \,d^{3}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 A \,c^{3}+6 A \,c^{2} d -2 A \,d^{3}+2 B \,c^{3}-6 B c \,d^{2}+4 B \,d^{3}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A \,c^{3}-18 A \,c^{2} d +12 A c \,d^{2}-2 A \,d^{3}-6 B \,c^{3}+12 B \,c^{2} d -6 B c \,d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 A \,c^{3}+24 A \,c^{2} d -24 A c \,d^{2}+8 A \,d^{3}+8 B \,c^{3}-24 B \,c^{2} d +24 B c \,d^{2}-8 B \,d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A \,c^{3}-12 A \,c^{2} d +12 A c \,d^{2}-4 A \,d^{3}-4 B \,c^{3}+12 B \,c^{2} d -12 B c \,d^{2}+4 B \,d^{3}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+2 d^{2} \left (-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d +3 B c -3 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{a^{3} f}\)	\(349\)
parallelrisch	\(\frac {\left (\left (-300 f x A +900 f x B -540 A +1755 B \right ) d^{3}+180 c \left (-5 f x B +A -9 B \right ) d^{2}+180 c^{2} \left (A +B \right ) d +180 \left (A +\frac {B}{3}\right ) c^{3}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\left (30 f x A -90 f x B +108 A -363 B \right ) d^{3}-54 c \left (-\frac {5}{3} f x B +A -6 B \right ) d^{2}-36 c^{2} \left (A +\frac {3 B}{2}\right ) d -18 c^{3} \left (A +\frac {2 B}{3}\right )\right ) \cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\left (\left (-300 f x A +900 f x B -340 A +1125 B \right ) d^{3}+60 c \left (-15 f x B +A -17 B \right ) d^{2}+180 \left (A +\frac {B}{3}\right ) c^{2} d +100 c^{3} \left (A +\frac {3 B}{5}\right )\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\left (-150 f x A +450 f x B -360 A +1215 B \right ) d^{3}+90 c \left (-5 f x B +A -12 B \right ) d^{2}+180 c^{2} \left (A +\frac {B}{2}\right ) d +90 c^{3} \left (A +\frac {2 B}{3}\right )\right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (\left (150 f x A -450 f x B +80 A -225 B \right ) d^{3}-30 c \left (-15 f x B +A -8 B \right ) d^{2}-30 B \,c^{2} d -50 A \,c^{3}\right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (\left (30 f x A -90 f x B -20 A +75 B \right ) d^{3}+30 c \left (3 f x B +A -2 B \right ) d^{2}+30 B \,c^{2} d -10 A \,c^{3}\right ) \sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )-15 B \,d^{3} \left (\cos \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )+\sin \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )\right )}{30 f \,a^{3} \left (-10 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+5 \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )-5 \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )-10 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )\right )}\)	\(496\)
risch	\(\frac {d^{3} x A}{a^{3}}+\frac {3 d^{2} x B c}{a^{3}}-\frac {3 d^{3} x B}{a^{3}}-\frac {B \,d^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}-\frac {B \,d^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {2 \left (2 A \,c^{3}+3 B \,c^{3}-96 B c \,d^{2}+9 A \,c^{2} d +21 A c \,d^{2}+21 B \,c^{2} d -32 A \,d^{3}+72 B \,d^{3}-420 B \,d^{3} {\mathrm e}^{2 i \left (f x +e \right )}-20 A \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}-15 B \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+90 B \,d^{3} {\mathrm e}^{4 i \left (f x +e \right )}-45 A \,d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+90 i A c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+90 i B \,c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}-405 i B c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-60 i B \,c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}-135 B c \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-45 A \,c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+555 B c \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-120 A c \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-120 B \,c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i B \,c^{3} {\mathrm e}^{3 i \left (f x +e \right )}+45 A c \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+45 B \,c^{2} d \,{\mathrm e}^{4 i \left (f x +e \right )}-270 i B \,d^{3} {\mathrm e}^{i \left (f x +e \right )}+115 i A \,d^{3} {\mathrm e}^{i \left (f x +e \right )}-135 i A \,d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+300 i B \,d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+45 i A \,c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}+185 A \,d^{3} {\mathrm e}^{2 i \left (f x +e \right )}-45 i A \,c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}-60 i A c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+345 i B c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}-15 i B \,c^{3} {\mathrm e}^{i \left (f x +e \right )}-10 i A \,c^{3} {\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\)	\(599\)
norman	\(\text {Expression too large to display}\)	\(1348\)

Input:

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

2/f/a^3*(-(A*c^3-A*d^3-3*B*c*d^2+3*B*d^3)/(tan(1/2*f*x+1/2*e)+1)-1/2*(-4*A 
*c^3+6*A*c^2*d-2*A*d^3+2*B*c^3-6*B*c*d^2+4*B*d^3)/(tan(1/2*f*x+1/2*e)+1)^2 
-1/3*(8*A*c^3-18*A*c^2*d+12*A*c*d^2-2*A*d^3-6*B*c^3+12*B*c^2*d-6*B*c*d^2)/ 
(tan(1/2*f*x+1/2*e)+1)^3-1/4*(-8*A*c^3+24*A*c^2*d-24*A*c*d^2+8*A*d^3+8*B*c 
^3-24*B*c^2*d+24*B*c*d^2-8*B*d^3)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(4*A*c^3-12 
*A*c^2*d+12*A*c*d^2-4*A*d^3-4*B*c^3+12*B*c^2*d-12*B*c*d^2+4*B*d^3)/(tan(1/ 
2*f*x+1/2*e)+1)^5+d^2*(-B*d/(1+tan(1/2*f*x+1/2*e)^2)+(A*d+3*B*c-3*B*d)*arc 
tan(tan(1/2*f*x+1/2*e))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 649 vs. \(2 (217) = 434\).

Time = 0.11 (sec) , antiderivative size = 649, normalized size of antiderivative = 2.88 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algori 
thm="fricas")
 

Output:

-1/15*(15*B*d^3*cos(f*x + e)^4 - 3*(A - B)*c^3 + 9*(A - B)*c^2*d - 9*(A - 
B)*c*d^2 + 3*(A - B)*d^3 + ((2*A + 3*B)*c^3 + 3*(3*A + 7*B)*c^2*d + 3*(7*A 
 - 32*B)*c*d^2 - (32*A - 117*B)*d^3 - 15*(3*B*c*d^2 + (A - 3*B)*d^3)*f*x)* 
cos(f*x + e)^3 + 60*(3*B*c*d^2 + (A - 3*B)*d^3)*f*x - (2*(2*A + 3*B)*c^3 + 
 3*(6*A - B)*c^2*d - 3*(A + 19*B)*c*d^2 - (19*A - 84*B)*d^3 + 45*(3*B*c*d^ 
2 + (A - 3*B)*d^3)*f*x)*cos(f*x + e)^2 - 3*((3*A + 2*B)*c^3 + 3*(2*A + 3*B 
)*c^2*d + 9*(A - 6*B)*c*d^2 - 9*(2*A - 7*B)*d^3 - 10*(3*B*c*d^2 + (A - 3*B 
)*d^3)*f*x)*cos(f*x + e) + (15*B*d^3*cos(f*x + e)^3 + 3*(A - B)*c^3 - 9*(A 
 - B)*c^2*d + 9*(A - B)*c*d^2 - 3*(A - B)*d^3 + 60*(3*B*c*d^2 + (A - 3*B)* 
d^3)*f*x - ((2*A + 3*B)*c^3 + 3*(3*A + 7*B)*c^2*d + 3*(7*A - 32*B)*c*d^2 - 
 2*(16*A - 51*B)*d^3 + 15*(3*B*c*d^2 + (A - 3*B)*d^3)*f*x)*cos(f*x + e)^2 
- 3*((2*A + 3*B)*c^3 + 3*(3*A + 2*B)*c^2*d + 3*(2*A - 17*B)*c*d^2 - (17*A 
- 62*B)*d^3 - 10*(3*B*c*d^2 + (A - 3*B)*d^3)*f*x)*cos(f*x + e))*sin(f*x + 
e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) 
- 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f* 
x + e))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11456 vs. \(2 (206) = 412\).

Time = 15.43 (sec) , antiderivative size = 11456, normalized size of antiderivative = 50.92 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**3/(a+a*sin(f*x+e))**3,x)
 

Output:

Piecewise((-30*A*c**3*tan(e/2 + f*x/2)**6/(15*a**3*f*tan(e/2 + f*x/2)**7 + 
 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3 
*f*tan(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e 
/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c**3*tan(e 
/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2) 
**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x/2)**4 + 22 
5*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f* 
tan(e/2 + f*x/2) + 15*a**3*f) - 110*A*c**3*tan(e/2 + f*x/2)**4/(15*a**3*f* 
tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + 
 f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)* 
*3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3 
*f) - 100*A*c**3*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a 
**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3*f*ta 
n(e/2 + f*x/2)**4 + 225*a**3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + 
f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 94*A*c**3*tan(e/2 + 
f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 
 165*a**3*f*tan(e/2 + f*x/2)**5 + 225*a**3*f*tan(e/2 + f*x/2)**4 + 225*a** 
3*f*tan(e/2 + f*x/2)**3 + 165*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e 
/2 + f*x/2) + 15*a**3*f) - 40*A*c**3*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + 
 f*x/2)**7 + 75*a**3*f*tan(e/2 + f*x/2)**6 + 165*a**3*f*tan(e/2 + f*x/2...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1682 vs. \(2 (217) = 434\).

Time = 0.16 (sec) , antiderivative size = 1682, normalized size of antiderivative = 7.48 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algori 
thm="maxima")
 

Output:

-2/15*(3*B*d^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/ 
(cos(f*x + e) + 1)^2 + 200*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 160*sin(f 
*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 
15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(co 
s(f*x + e) + 1) + 11*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin( 
f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1) 
^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(co 
s(f*x + e) + 1)^6 + a^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(s 
in(f*x + e)/(cos(f*x + e) + 1))/a^3) - 3*B*c*d^2*((95*sin(f*x + e)/(cos(f* 
x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/ 
(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 
 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + 
e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e 
)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*a 
rctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) - A*d^3*((95*sin(f*x + e)/(cos 
(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e) 
^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^ 
3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x 
 + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x 
+ e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (217) = 434\).

Time = 0.21 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.52 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^3,x, algori 
thm="giac")
 

Output:

-1/15*(30*B*d^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^3) - 15*(3*B*c*d^2 + A*d^3 
 - 3*B*d^3)*(f*x + e)/a^3 + 2*(15*A*c^3*tan(1/2*f*x + 1/2*e)^4 - 45*B*c*d^ 
2*tan(1/2*f*x + 1/2*e)^4 - 15*A*d^3*tan(1/2*f*x + 1/2*e)^4 + 45*B*d^3*tan( 
1/2*f*x + 1/2*e)^4 + 30*A*c^3*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^3*tan(1/2*f* 
x + 1/2*e)^3 + 45*A*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 225*B*c*d^2*tan(1/2*f*x 
 + 1/2*e)^3 - 75*A*d^3*tan(1/2*f*x + 1/2*e)^3 + 210*B*d^3*tan(1/2*f*x + 1/ 
2*e)^3 + 40*A*c^3*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^3*tan(1/2*f*x + 1/2*e)^2 
 + 45*A*c^2*d*tan(1/2*f*x + 1/2*e)^2 + 60*B*c^2*d*tan(1/2*f*x + 1/2*e)^2 + 
 60*A*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 435*B*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 
145*A*d^3*tan(1/2*f*x + 1/2*e)^2 + 360*B*d^3*tan(1/2*f*x + 1/2*e)^2 + 20*A 
*c^3*tan(1/2*f*x + 1/2*e) + 15*B*c^3*tan(1/2*f*x + 1/2*e) + 45*A*c^2*d*tan 
(1/2*f*x + 1/2*e) + 30*B*c^2*d*tan(1/2*f*x + 1/2*e) + 30*A*c*d^2*tan(1/2*f 
*x + 1/2*e) - 285*B*c*d^2*tan(1/2*f*x + 1/2*e) - 95*A*d^3*tan(1/2*f*x + 1/ 
2*e) + 240*B*d^3*tan(1/2*f*x + 1/2*e) + 7*A*c^3 + 3*B*c^3 + 9*A*c^2*d + 6* 
B*c^2*d + 6*A*c*d^2 - 66*B*c*d^2 - 22*A*d^3 + 57*B*d^3)/(a^3*(tan(1/2*f*x 
+ 1/2*e) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 38.55 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.64 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,d^2\,\mathrm {atan}\left (\frac {2\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d+3\,B\,c-3\,B\,d\right )}{2\,A\,d^3-6\,B\,d^3+6\,B\,c\,d^2}\right )\,\left (A\,d+3\,B\,c-3\,B\,d\right )}{a^3\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (4\,A\,c^3-10\,A\,d^3+2\,B\,c^3+30\,B\,d^3+6\,A\,c^2\,d-30\,B\,c\,d^2\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {8\,A\,c^3}{3}-\frac {38\,A\,d^3}{3}+2\,B\,c^3+42\,B\,d^3+4\,A\,c\,d^2+6\,A\,c^2\,d-38\,B\,c\,d^2+4\,B\,c^2\,d\right )+\frac {14\,A\,c^3}{15}-\frac {44\,A\,d^3}{15}+\frac {2\,B\,c^3}{5}+\frac {48\,B\,d^3}{5}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {22\,A\,c^3}{3}-\frac {64\,A\,d^3}{3}+2\,B\,c^3+64\,B\,d^3+8\,A\,c\,d^2+6\,A\,c^2\,d-64\,B\,c\,d^2+8\,B\,c^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {20\,A\,c^3}{3}-\frac {68\,A\,d^3}{3}+4\,B\,c^3+80\,B\,d^3+4\,A\,c\,d^2+12\,A\,c^2\,d-68\,B\,c\,d^2+4\,B\,c^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {94\,A\,c^3}{15}-\frac {334\,A\,d^3}{15}+\frac {12\,B\,c^3}{5}+\frac {378\,B\,d^3}{5}+\frac {44\,A\,c\,d^2}{5}+\frac {36\,A\,c^2\,d}{5}-\frac {334\,B\,c\,d^2}{5}+\frac {44\,B\,c^2\,d}{5}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,A\,c^3-2\,A\,d^3+6\,B\,d^3-6\,B\,c\,d^2\right )+\frac {4\,A\,c\,d^2}{5}+\frac {6\,A\,c^2\,d}{5}-\frac {44\,B\,c\,d^2}{5}+\frac {4\,B\,c^2\,d}{5}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+11\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+15\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+15\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+11\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \] Input:

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

Output:

(2*d^2*atan((2*d^2*tan(e/2 + (f*x)/2)*(A*d + 3*B*c - 3*B*d))/(2*A*d^3 - 6* 
B*d^3 + 6*B*c*d^2))*(A*d + 3*B*c - 3*B*d))/(a^3*f) - (tan(e/2 + (f*x)/2)^5 
*(4*A*c^3 - 10*A*d^3 + 2*B*c^3 + 30*B*d^3 + 6*A*c^2*d - 30*B*c*d^2) + tan( 
e/2 + (f*x)/2)*((8*A*c^3)/3 - (38*A*d^3)/3 + 2*B*c^3 + 42*B*d^3 + 4*A*c*d^ 
2 + 6*A*c^2*d - 38*B*c*d^2 + 4*B*c^2*d) + (14*A*c^3)/15 - (44*A*d^3)/15 + 
(2*B*c^3)/5 + (48*B*d^3)/5 + tan(e/2 + (f*x)/2)^4*((22*A*c^3)/3 - (64*A*d^ 
3)/3 + 2*B*c^3 + 64*B*d^3 + 8*A*c*d^2 + 6*A*c^2*d - 64*B*c*d^2 + 8*B*c^2*d 
) + tan(e/2 + (f*x)/2)^3*((20*A*c^3)/3 - (68*A*d^3)/3 + 4*B*c^3 + 80*B*d^3 
 + 4*A*c*d^2 + 12*A*c^2*d - 68*B*c*d^2 + 4*B*c^2*d) + tan(e/2 + (f*x)/2)^2 
*((94*A*c^3)/15 - (334*A*d^3)/15 + (12*B*c^3)/5 + (378*B*d^3)/5 + (44*A*c* 
d^2)/5 + (36*A*c^2*d)/5 - (334*B*c*d^2)/5 + (44*B*c^2*d)/5) + tan(e/2 + (f 
*x)/2)^6*(2*A*c^3 - 2*A*d^3 + 6*B*d^3 - 6*B*c*d^2) + (4*A*c*d^2)/5 + (6*A* 
c^2*d)/5 - (44*B*c*d^2)/5 + (4*B*c^2*d)/5)/(f*(11*a^3*tan(e/2 + (f*x)/2)^2 
 + 15*a^3*tan(e/2 + (f*x)/2)^3 + 15*a^3*tan(e/2 + (f*x)/2)^4 + 11*a^3*tan( 
e/2 + (f*x)/2)^5 + 5*a^3*tan(e/2 + (f*x)/2)^6 + a^3*tan(e/2 + (f*x)/2)^7 + 
 a^3 + 5*a^3*tan(e/2 + (f*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1123, normalized size of antiderivative = 4.99 \[ \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(15*cos(e + f*x)*sin(e + f*x)**3*b*d**3 + cos(e + f*x)*sin(e + f*x)**2*a*c 
**3 + 15*cos(e + f*x)*sin(e + f*x)**2*a*c*d**2 + 15*cos(e + f*x)*sin(e + f 
*x)**2*a*d**3*f*x - 16*cos(e + f*x)*sin(e + f*x)**2*a*d**3 + 15*cos(e + f* 
x)*sin(e + f*x)**2*b*c**2*d + 45*cos(e + f*x)*sin(e + f*x)**2*b*c*d**2*f*x 
 - 48*cos(e + f*x)*sin(e + f*x)**2*b*c*d**2 - 45*cos(e + f*x)*sin(e + f*x) 
**2*b*d**3*f*x + 63*cos(e + f*x)*sin(e + f*x)**2*b*d**3 + 4*cos(e + f*x)*s 
in(e + f*x)*a*c**3 + 9*cos(e + f*x)*sin(e + f*x)*a*c**2*d + 6*cos(e + f*x) 
*sin(e + f*x)*a*c*d**2 + 30*cos(e + f*x)*sin(e + f*x)*a*d**3*f*x - 19*cos( 
e + f*x)*sin(e + f*x)*a*d**3 + 3*cos(e + f*x)*sin(e + f*x)*b*c**3 + 6*cos( 
e + f*x)*sin(e + f*x)*b*c**2*d + 90*cos(e + f*x)*sin(e + f*x)*b*c*d**2*f*x 
 - 57*cos(e + f*x)*sin(e + f*x)*b*c*d**2 - 90*cos(e + f*x)*sin(e + f*x)*b* 
d**3*f*x + 63*cos(e + f*x)*sin(e + f*x)*b*d**3 + 6*cos(e + f*x)*a*c**3 + 1 
5*cos(e + f*x)*a*d**3*f*x - 6*cos(e + f*x)*a*d**3 + 45*cos(e + f*x)*b*c*d* 
*2*f*x - 18*cos(e + f*x)*b*c*d**2 - 45*cos(e + f*x)*b*d**3*f*x + 18*cos(e 
+ f*x)*b*d**3 + 15*sin(e + f*x)**4*b*d**3 + 3*sin(e + f*x)**3*a*c**3 + 18* 
sin(e + f*x)**3*a*c**2*d + 27*sin(e + f*x)**3*a*c*d**2 - 15*sin(e + f*x)** 
3*a*d**3*f*x - 48*sin(e + f*x)**3*a*d**3 + 6*sin(e + f*x)**3*b*c**3 + 27*s 
in(e + f*x)**3*b*c**2*d - 45*sin(e + f*x)**3*b*c*d**2*f*x - 144*sin(e + f* 
x)**3*b*c*d**2 + 45*sin(e + f*x)**3*b*d**3*f*x + 156*sin(e + f*x)**3*b*d** 
3 + 7*sin(e + f*x)**2*a*c**3 + 45*sin(e + f*x)**2*a*c**2*d + 15*sin(e +...