\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\) [336]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 351 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {\left (d (A d (3+m)+B (2 c+d m))-2 (2+m) \left (A c d (3+m)+B \left (c^2+d^2+c d m\right )\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {2^{\frac {1}{2}+m} \left (A (3+m) \left (2 c d m (2+m)+d^2 \left (1+m+m^2\right )+c^2 \left (2+3 m+m^2\right )\right )+B \left (d^2 m \left (5+3 m+m^2\right )+c^2 m \left (6+5 m+m^2\right )+2 c d \left (3+4 m+4 m^2+m^3\right )\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m) (2+m) (3+m)}-\frac {d (A d (3+m)+B (2 c+d m)) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (2+m) (3+m)}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2}{f (3+m)} \] Output:

(d*(A*d*(3+m)+B*(d*m+2*c))-2*(2+m)*(A*c*d*(3+m)+B*(c*d*m+c^2+d^2)))*cos(f* 
x+e)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m)/(3+m)-2^(1/2+m)*(A*(3+m)*(2*c*d*m*(2 
+m)+d^2*(m^2+m+1)+c^2*(m^2+3*m+2))+B*(d^2*m*(m^2+3*m+5)+c^2*m*(m^2+5*m+6)+ 
2*c*d*(m^3+4*m^2+4*m+3)))*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2* 
sin(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m/f/(1+m)/(2+m)/(3+m) 
-d*(A*d*(3+m)+B*(d*m+2*c))*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(2+m)/(3+ 
m)-B*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^2/f/(3+m)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 27.68 (sec) , antiderivative size = 854, normalized size of antiderivative = 2.43 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {i (a (1+\sin (e+f x)))^m (1-i \cos (e+f x)+\sin (e+f x))^{-2 m} \left (\frac {8 A c^2 \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {8 B c d \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {4 A d^2 \operatorname {Hypergeometric2F1}(-2 m,-m,1-m,i \cos (e+f x)-\sin (e+f x))}{m}+\frac {4 B c^2 \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {8 A c d \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {3 B d^2 \operatorname {Hypergeometric2F1}(1-m,-2 m,2-m,i \cos (e+f x)-\sin (e+f x)) (-i \cos (e+f x)+\sin (e+f x))}{-1+m}+\frac {4 B c^2 \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}+\frac {8 A c d \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}+\frac {3 B d^2 \operatorname {Hypergeometric2F1}(-1-m,-2 m,-m,i \cos (e+f x)-\sin (e+f x)) (i \cos (e+f x)+\sin (e+f x))}{1+m}-\frac {2 d (2 B c+A d) \operatorname {Hypergeometric2F1}(-2-m,-2 m,-1-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))-i \sin (2 (e+f x)))}{2+m}-\frac {4 B c d \operatorname {Hypergeometric2F1}(2-m,-2 m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}-\frac {2 A d^2 \operatorname {Hypergeometric2F1}(2-m,-2 m,3-m,i \cos (e+f x)-\sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x)))}{-2+m}-\frac {i B d^2 \operatorname {Hypergeometric2F1}(-3-m,-2 m,-2-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))-i \sin (3 (e+f x)))}{3+m}+\frac {i B d^2 \operatorname {Hypergeometric2F1}(3-m,-2 m,4-m,i \cos (e+f x)-\sin (e+f x)) (\cos (3 (e+f x))+i \sin (3 (e+f x)))}{-3+m}\right )}{8 f} \] Input:

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

Output:

((I/8)*(a*(1 + Sin[e + f*x]))^m*((8*A*c^2*Hypergeometric2F1[-2*m, -m, 1 - 
m, I*Cos[e + f*x] - Sin[e + f*x]])/m + (8*B*c*d*Hypergeometric2F1[-2*m, -m 
, 1 - m, I*Cos[e + f*x] - Sin[e + f*x]])/m + (4*A*d^2*Hypergeometric2F1[-2 
*m, -m, 1 - m, I*Cos[e + f*x] - Sin[e + f*x]])/m + (4*B*c^2*Hypergeometric 
2F1[1 - m, -2*m, 2 - m, I*Cos[e + f*x] - Sin[e + f*x]]*((-I)*Cos[e + f*x] 
+ Sin[e + f*x]))/(-1 + m) + (8*A*c*d*Hypergeometric2F1[1 - m, -2*m, 2 - m, 
 I*Cos[e + f*x] - Sin[e + f*x]]*((-I)*Cos[e + f*x] + Sin[e + f*x]))/(-1 + 
m) + (3*B*d^2*Hypergeometric2F1[1 - m, -2*m, 2 - m, I*Cos[e + f*x] - Sin[e 
 + f*x]]*((-I)*Cos[e + f*x] + Sin[e + f*x]))/(-1 + m) + (4*B*c^2*Hypergeom 
etric2F1[-1 - m, -2*m, -m, I*Cos[e + f*x] - Sin[e + f*x]]*(I*Cos[e + f*x] 
+ Sin[e + f*x]))/(1 + m) + (8*A*c*d*Hypergeometric2F1[-1 - m, -2*m, -m, I* 
Cos[e + f*x] - Sin[e + f*x]]*(I*Cos[e + f*x] + Sin[e + f*x]))/(1 + m) + (3 
*B*d^2*Hypergeometric2F1[-1 - m, -2*m, -m, I*Cos[e + f*x] - Sin[e + f*x]]* 
(I*Cos[e + f*x] + Sin[e + f*x]))/(1 + m) - (2*d*(2*B*c + A*d)*Hypergeometr 
ic2F1[-2 - m, -2*m, -1 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x 
)] - I*Sin[2*(e + f*x)]))/(2 + m) - (4*B*c*d*Hypergeometric2F1[2 - m, -2*m 
, 3 - m, I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f 
*x)]))/(-2 + m) - (2*A*d^2*Hypergeometric2F1[2 - m, -2*m, 3 - m, I*Cos[e + 
 f*x] - Sin[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]))/(-2 + m) - 
(I*B*d^2*Hypergeometric2F1[-3 - m, -2*m, -2 - m, I*Cos[e + f*x] - Sin[e...
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 349, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3462, 3042, 3447, 3042, 3502, 3042, 3230, 3042, 3131, 3042, 3130}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2dx\)

\(\Big \downarrow \) 3462

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {\frac {a^2 \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a^2 \cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{f (m+2)}}{a (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^2 \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) \int (\sin (e+f x) a+a)^mdx}{m+1}+\frac {a^2 \cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{f (m+2)}}{a (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {\frac {a^2 \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a^2 \cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{f (m+2)}}{a (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^2 \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (\sin (e+f x)+1)^mdx}{m+1}+\frac {a^2 \cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1)}}{a (m+2)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{f (m+2)}}{a (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {\frac {a^2 \cos (e+f x) \left (d (A d (m+3)+B (2 c+d m))-2 (m+2) \left (A c d (m+3)+B \left (c^2+c d m+d^2\right )\right )\right ) (a \sin (e+f x)+a)^m}{f (m+1)}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (A (m+3) \left (c^2 \left (m^2+3 m+2\right )+2 c d m (m+2)+d^2 \left (m^2+m+1\right )\right )+B \left (c^2 m \left (m^2+5 m+6\right )+2 c d \left (m^3+4 m^2+4 m+3\right )+d^2 m \left (m^2+3 m+5\right )\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1)}}{a (m+2)}-\frac {d \cos (e+f x) (A d (m+3)+B (2 c+d m)) (a \sin (e+f x)+a)^{m+1}}{f (m+2)}}{a (m+3)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^2}{f (m+3)}\)

Input:

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

Output:

-((B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^2)/(f*(3 + m 
))) + (-((d*(A*d*(3 + m) + B*(2*c + d*m))*Cos[e + f*x]*(a + a*Sin[e + f*x] 
)^(1 + m))/(f*(2 + m))) + ((a^2*(d*(A*d*(3 + m) + B*(2*c + d*m)) - 2*(2 + 
m)*(A*c*d*(3 + m) + B*(c^2 + d^2 + c*d*m)))*Cos[e + f*x]*(a + a*Sin[e + f* 
x])^m)/(f*(1 + m)) - (2^(1/2 + m)*a^2*(A*(3 + m)*(2*c*d*m*(2 + m) + d^2*(1 
 + m + m^2) + c^2*(2 + 3*m + m^2)) + B*(d^2*m*(5 + 3*m + m^2) + c^2*m*(6 + 
 5*m + m^2) + 2*c*d*(3 + 4*m + 4*m^2 + m^3)))*Cos[e + f*x]*Hypergeometric2 
F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)* 
(a + a*Sin[e + f*x])^m)/(f*(1 + m)))/(a*(2 + m)))/(a*(3 + m))
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3130
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3230
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

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 3462
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[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 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 [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{2}d x\]

Input:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)
 

Output:

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)
 

Fricas [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algori 
thm="fricas")
 

Output:

integral((A*c^2 + 2*B*c*d + A*d^2 - (2*B*c*d + A*d^2)*cos(f*x + e)^2 - (B* 
d^2*cos(f*x + e)^2 - B*c^2 - 2*A*c*d - B*d^2)*sin(f*x + e))*(a*sin(f*x + e 
) + a)^m, x)
 

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{2}\, dx \] Input:

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2,x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))*(c + d*sin(e + f*x 
))**2, x)
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algori 
thm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2*(a*sin(f*x + e) + a) 
^m, x)
 

Giac [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algori 
thm="giac")
 

Output:

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^2*(a*sin(f*x + e) + a) 
^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \] Input:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^2,x)
 

Output:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^2, x)
 

Reduce [F]

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

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)
 

Output:

int((sin(e + f*x)*a + a)**m,x)*a*c**2 + int((sin(e + f*x)*a + a)**m*sin(e 
+ f*x)**3,x)*b*d**2 + int((sin(e + f*x)*a + a)**m*sin(e + f*x)**2,x)*a*d** 
2 + 2*int((sin(e + f*x)*a + a)**m*sin(e + f*x)**2,x)*b*c*d + 2*int((sin(e 
+ f*x)*a + a)**m*sin(e + f*x),x)*a*c*d + int((sin(e + f*x)*a + a)**m*sin(e 
 + f*x),x)*b*c**2