\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx\) [212]

Optimal result

Integrand size = 40, antiderivative size = 191 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-3-m}}{f (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{c f (3+2 m) (5+2 m)}+\frac {(2 A-B (3+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1-m}}{c^2 f (5+2 m) \left (3+8 m+4 m^2\right )} \] Output:

(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-3-m)/f/(5+2*m)+(2*A 
-B*(3+2*m))*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/c/f/(3+2 
*m)/(5+2*m)+(2*A-B*(3+2*m))*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)) 
^(-1-m)/c^2/f/(5+2*m)/(4*m^2+8*m+3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.74 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {\sec (e+f x) (a (1+\sin (e+f x)))^{1+m} (c-c \sin (e+f x))^{-m} \left (-B (3+2 m)+A \left (7+12 m+4 m^2\right )+(3+2 m) (-2 A+B (3+2 m)) \sin (e+f x)+(2 A-B (3+2 m)) \sin ^2(e+f x)\right )}{a c^3 f (1+2 m) (3+2 m) (5+2 m) (-1+\sin (e+f x))^2} \] Input:

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

Output:

(Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^(1 + m)*(-(B*(3 + 2*m)) + A*(7 + 12*m 
 + 4*m^2) + (3 + 2*m)*(-2*A + B*(3 + 2*m))*Sin[e + f*x] + (2*A - B*(3 + 2* 
m))*Sin[e + f*x]^2))/(a*c^3*f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(-1 + Sin[e + 
f*x])^2*(c - c*Sin[e + f*x])^m)
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3451, 3042, 3222, 3042, 3221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3451

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3221

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

Input:

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

Output:

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-3 - m) 
)/(f*(5 + 2*m)) + ((2*A - B*(3 + 2*m))*((Cos[e + f*x]*(a + a*Sin[e + f*x]) 
^m*(c - c*Sin[e + f*x])^(-2 - m))/(f*(3 + 2*m)) + (Cos[e + f*x]*(a + a*Sin 
[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 - m))/(c*f*(1 + 2*m)*(3 + 2*m))))/(c 
*(5 + 2*m))
 

Defintions of rubi rules used

rule 3042

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

rule 3221

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[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] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne 
Q[m, -2^(-1)]
 

rule 3222

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[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[(m + n + 1)/(a*(2*m + 1) 
)   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free 
Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && 
 ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || 
!SumSimplerQ[n, 1])
 

rule 3451

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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[ 
{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2* 
m + 1, 0]
 

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{-3-m}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.72 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\frac {{\left ({\left (2 \, B m - 2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} + {\left (4 \, B m^{2} - 4 \, {\left (A - 3 \, B\right )} m - 6 \, A + 9 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (4 \, A m^{2} + 4 \, {\left (3 \, A - B\right )} m + 9 \, A - 6 \, B\right )} \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 3}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f} \] Input:

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-m),x, a 
lgorithm="fricas")
 

Output:

((2*B*m - 2*A + 3*B)*cos(f*x + e)^3 + (4*B*m^2 - 4*(A - 3*B)*m - 6*A + 9*B 
)*cos(f*x + e)*sin(f*x + e) + (4*A*m^2 + 4*(3*A - B)*m + 9*A - 6*B)*cos(f* 
x + e))*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 3)/(8*f*m^3 + 3 
6*f*m^2 + 46*f*m + 15*f)
 

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 3} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \] Input:

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(-3-m),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 3)*(A + 
B*sin(e + f*x)), x)
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-m),x, a 
lgorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c 
)^(-m - 3), x)
 

Giac [F]

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

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-3-m),x, a 
lgorithm="giac")
 

Output:

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c 
)^(-m - 3), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 39.01 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.25 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-3-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (30\,A\,\cos \left (e+f\,x\right )-15\,B\,\cos \left (e+f\,x\right )-2\,A\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (3\,e+3\,f\,x\right )-12\,A\,\sin \left (2\,e+2\,f\,x\right )+18\,B\,\sin \left (2\,e+2\,f\,x\right )+8\,B\,m^2\,\sin \left (2\,e+2\,f\,x\right )+48\,A\,m\,\cos \left (e+f\,x\right )-10\,B\,m\,\cos \left (e+f\,x\right )+16\,A\,m^2\,\cos \left (e+f\,x\right )+2\,B\,m\,\cos \left (3\,e+3\,f\,x\right )-8\,A\,m\,\sin \left (2\,e+2\,f\,x\right )+24\,B\,m\,\sin \left (2\,e+2\,f\,x\right )\right )}{c^3\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (8\,m^3+36\,m^2+46\,m+15\right )\,\left (15\,\sin \left (e+f\,x\right )+6\,\cos \left (2\,e+2\,f\,x\right )-\sin \left (3\,e+3\,f\,x\right )-10\right )} \] Input:

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

Output:

-((a*(sin(e + f*x) + 1))^m*(30*A*cos(e + f*x) - 15*B*cos(e + f*x) - 2*A*co 
s(3*e + 3*f*x) + 3*B*cos(3*e + 3*f*x) - 12*A*sin(2*e + 2*f*x) + 18*B*sin(2 
*e + 2*f*x) + 8*B*m^2*sin(2*e + 2*f*x) + 48*A*m*cos(e + f*x) - 10*B*m*cos( 
e + f*x) + 16*A*m^2*cos(e + f*x) + 2*B*m*cos(3*e + 3*f*x) - 8*A*m*sin(2*e 
+ 2*f*x) + 24*B*m*sin(2*e + 2*f*x)))/(c^3*f*(-c*(sin(e + f*x) - 1))^m*(46* 
m + 36*m^2 + 8*m^3 + 15)*(15*sin(e + f*x) + 6*cos(2*e + 2*f*x) - sin(3*e + 
 3*f*x) - 10))
 

Reduce [F]

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

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

Output:

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