Integrand size = 41, antiderivative size = 392 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {2^{\frac {1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac {\sqrt {2} C \operatorname {AppellF1}\left (\frac {3}{2}+m,\frac {1}{2},1+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{a (c-d) d f (3+2 m) \sqrt {1-\sin (e+f x)}} \]
(A*d^2+C*c^2)*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-1-m)/d/(c^2 -d^2)/f/(1+m)-2^(1/2+m)*a*(c*(A+C)*d*(1+m)+d^2*(-A*m+C*m+C)-c^2*(2*C*m+C)) *cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2*(c-d)*(1-sin(f*x+e))/(c+d*sin (f*x+e)))*(a+a*sin(f*x+e))^(-1+m)*((c+d)*(1+sin(f*x+e))/(c+d*sin(f*x+e)))^ (1/2-m)/(c-d)/d/(c+d)^2/f/(1+m)/((c+d*sin(f*x+e))^m)+C*AppellF1(3/2+m,1+m, 1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin( f*x+e))^(1+m)*((c+d*sin(f*x+e))/(c-d))^m*2^(1/2)/a/(c-d)/d/f/(3+2*m)/((c+d *sin(f*x+e))^m)/(1-sin(f*x+e))^(1/2)
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx \]
Time = 1.29 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3523, 25, 3042, 3466, 3042, 3267, 142, 157, 27, 156, 155}
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.
| \(\displaystyle \int (a \sin (e+f x)+a)^m \left (A+C \sin ^2(e+f x)\right ) (c+d \sin (e+f x))^{-m-2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^m \left (A+C \sin (e+f x)^2\right ) (c+d \sin (e+f x))^{-m-2}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}-\frac {\int -(\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+c C (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+c C (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1} \left (a (A d (m c+c-d m)+c C (m d+d-c m))+a C \left (c^2-d^2\right ) (m+1) \sin (e+f x)\right )dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3466 |
| \(\displaystyle \frac {a \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1}dx+C (m+1) \left (c^2-d^2\right ) \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{-m-1}dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) \int (\sin (e+f x) a+a)^m (c+d \sin (e+f x))^{-m-1}dx+C (m+1) \left (c^2-d^2\right ) \int (\sin (e+f x) a+a)^{m+1} (c+d \sin (e+f x))^{-m-1}dx}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3267 |
| \(\displaystyle \frac {\frac {a^3 \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {a^2 C (m+1) \left (c^2-d^2\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 142 |
| \(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 157 |
| \(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} (c+d \sin (e+f x))^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {\frac {a^2 C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}} \left (\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}\right )^{-m-1}}{\sqrt {1-\sin (e+f x)}}d\sin (e+f x)}{f (c-d) (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle \frac {\frac {\sqrt {2} a C (m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},m+1,m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (c-d) (a-a \sin (e+f x))}-\frac {a^2 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{f (c+d)}}{a d (m+1) \left (c^2-d^2\right )}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}\) |
((c^2*C + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^ (-1 - m))/(d*(c^2 - d^2)*f*(1 + m)) + (-((2^(1/2 + m)*a^2*(c*(A + C)*d*(1 + m) + d^2*(C - A*m + C*m) - c^2*(C + 2*C*m))*Cos[e + f*x]*Hypergeometric2 F1[1/2, 1/2 - m, 3/2, ((c - d)*(1 - Sin[e + f*x]))/(2*(c + d*Sin[e + f*x]) )]*(a + a*Sin[e + f*x])^(-1 + m)*(((c + d)*(1 + Sin[e + f*x]))/(c + d*Sin[ e + f*x]))^(1/2 - m))/((c + d)*f*(c + d*Sin[e + f*x])^m)) + (Sqrt[2]*a*C*( c^2 - d^2)*(1 + m)*AppellF1[3/2 + m, 1/2, 1 + m, 5/2 + 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])^(1 + m)*((c + d*Sin[e + f*x])/(c - d))^m)/((c - d) *f*(3 + 2*m)*(a - a*Sin[e + f*x])*(c + d*Sin[e + f*x])^m))/(a*d*(c^2 - d^2 )*(1 + m))
3.1.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f *x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] && !IntegerQ[n]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*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, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
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, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & !GtQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A*b - a*B)/b Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x ] + Simp[B/b 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] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-2-m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x, algorithm="fricas")
integral(-(C*cos(f*x + e)^2 - A - C)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e ) + c)^(-m - 2), x)
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(-2-m)*(A+C*sin(f*x+e)^2),x, algorithm="maxima")
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx=\int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \]