\(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [340]

Optimal result

Integrand size = 35, antiderivative size = 295 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=-\frac {2^{\frac {1}{2}+m} a \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2}-m,1,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m}}{(c-d) d (c+d)^2 f}+\frac {2^{\frac {1}{2}+m} (B c-A d) m \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \] Output:

-2^(1/2+m)*a*(A*d*(c*(1-m)-d*m)-B*(-c^2*m-c*d*m+d^2))*AppellF1(1/2,1,1/2-m 
,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(1+sin(f*x+e))^ 
(1/2-m)*(a+a*sin(f*x+e))^(-1+m)/(c-d)/d/(c+d)^2/f+2^(1/2+m)*(-A*d+B*c)*m*c 
os(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/d/(c^2-d^2)/f-(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f 
*x+e))^m/(c^2-d^2)/f/(c+d*sin(f*x+e))
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3463, 25, 3042, 3465, 3042, 3131, 3042, 3130, 3267, 154, 27, 153}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3465

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3131

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3130

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

\(\Big \downarrow \) 3267

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

\(\Big \downarrow \) 154

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 153

\(\displaystyle \frac {\frac {\sqrt {2} a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (A d (c (1-m)-d m)-B \left (c^2 (-m)-c d m+d^2\right )\right ) (a \sin (e+f x)+a)^m \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},1,m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (c-d) (a-a \sin (e+f x))}+\frac {a 2^{m+\frac {1}{2}} m (B c-A d) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{d f}}{a \left (c^2-d^2\right )}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

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*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/((c^2 - d^2)*f*(c + d* 
Sin[e + f*x]))) + ((2^(1/2 + m)*a*(B*c - A*d)*m*Cos[e + f*x]*Hypergeometri 
c2F1[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)/(d*f) + (Sqrt[2]*a^2*(A*d*(c*(1 - m) - d*m) - B* 
(d^2 - c^2*m - c*d*m))*AppellF1[1/2 + m, 1/2, 1, 3/2 + m, (1 + Sin[e + f*x 
])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*Sqrt[1 - Sin[e + f*x 
]]*(a + a*Sin[e + f*x])^m)/((c - d)*d*f*(1 + 2*m)*(a - a*Sin[e + f*x])))/( 
a*(c^2 - d^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

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 3267

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

rule 3463

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

rule 3465

Int[(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B 
/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b*S 
in[e + f*x])^m/(c + d*Sin[e + f*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] && 
 NeQ[m + 1/2, 0]
 

Maple [F]

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

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]

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] 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:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,0]%%%} / %%%{1,[0,0,2]%%%} Error: Bad Argument Valu 
e
 

Mupad [F(-1)]

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

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