Integrand size = 35, antiderivative size = 226 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{2 d f (a+a \sin (e+f x))^{3/2}}-\frac {(A-4 A n+B (3+4 n)) \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{4 a f \sqrt {a+a \sin (e+f x)}}-\frac {(A-B) (1+2 n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right ) \sin ^{-n}(e+f x) (d \sin (e+f x))^n}{2 a f \sqrt {a+a \sin (e+f x)}} \]
1/2*(A-B)*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(a+a*sin(f*x+e))^(3/2)-1/4*( A-4*A*n+B*(3+4*n))*AppellF1(1/2,-n,1,3/2,1-sin(f*x+e),1/2-1/2*sin(f*x+e))* cos(f*x+e)*(d*sin(f*x+e))^n/a/f/(sin(f*x+e)^n)/(a+a*sin(f*x+e))^(1/2)-1/2* (A-B)*(1+2*n)*cos(f*x+e)*hypergeom([1/2, -n],[3/2],1-sin(f*x+e))*(d*sin(f* x+e))^n/a/f/(sin(f*x+e)^n)/(a+a*sin(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(523\) vs. \(2(226)=452\).
Time = 27.19 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.31 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sec (e+f x) (d \sin (e+f x))^n \left (a B (1+\sin (e+f x)) \left (a \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-n} (1+\sin (e+f x))-\frac {4 \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-n} \left (-2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{-1+4 n^2}\right )+A \left (a^2 \operatorname {AppellF1}\left (1,\frac {1}{2},-n,2,\frac {1}{2} (1+\sin (e+f x)),1+\sin (e+f x)\right ) \sqrt {2-2 \sin (e+f x)} (-\sin (e+f x))^{-n} (1+\sin (e+f x))^2-\frac {4 a \sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}} (1+\sin (e+f x)) \left (1-\frac {1}{1+\sin (e+f x)}\right )^{-n} \left (2 a (1+2 n) \operatorname {AppellF1}\left (\frac {1}{2}-n,-\frac {1}{2},-n,\frac {3}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right )+a (-1+2 n) \operatorname {AppellF1}\left (-\frac {1}{2}-n,-\frac {1}{2},-n,\frac {1}{2}-n,\frac {2}{1+\sin (e+f x)},\frac {1}{1+\sin (e+f x)}\right ) (1+\sin (e+f x))\right )}{-1+4 n^2}\right )\right )}{8 a^3 f \sqrt {a (1+\sin (e+f x))}} \]
(Sec[e + f*x]*(d*Sin[e + f*x])^n*(a*B*(1 + Sin[e + f*x])*((a*AppellF1[1, 1 /2, -n, 2, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sqrt[2 - 2*Sin[e + f*x] ]*(1 + Sin[e + f*x]))/(-Sin[e + f*x])^n - (4*Sqrt[(-1 + Sin[e + f*x])/(1 + Sin[e + f*x])]*(-2*a*(1 + 2*n)*AppellF1[1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)] + a*(-1 + 2*n)*AppellF1[-1/2 - n , -1/2, -n, 1/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1)]*(1 + S in[e + f*x])))/((-1 + 4*n^2)*(1 - (1 + Sin[e + f*x])^(-1))^n)) + A*((a^2*A ppellF1[1, 1/2, -n, 2, (1 + Sin[e + f*x])/2, 1 + Sin[e + f*x]]*Sqrt[2 - 2* Sin[e + f*x]]*(1 + Sin[e + f*x])^2)/(-Sin[e + f*x])^n - (4*a*Sqrt[(-1 + Si n[e + f*x])/(1 + Sin[e + f*x])]*(1 + Sin[e + f*x])*(2*a*(1 + 2*n)*AppellF1 [1/2 - n, -1/2, -n, 3/2 - n, 2/(1 + Sin[e + f*x]), (1 + Sin[e + f*x])^(-1) ] + a*(-1 + 2*n)*AppellF1[-1/2 - n, -1/2, -n, 1/2 - n, 2/(1 + Sin[e + f*x] ), (1 + Sin[e + f*x])^(-1)]*(1 + Sin[e + f*x])))/((-1 + 4*n^2)*(1 - (1 + S in[e + f*x])^(-1))^n))))/(8*a^3*f*Sqrt[a*(1 + Sin[e + f*x])])
Time = 1.38 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3457, 27, 3042, 3466, 3042, 3255, 77, 75, 3266, 3042, 3265, 3042, 3264, 148, 333}
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.
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(A+B \sin (e+f x)) (d \sin (e+f x))^n}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(d \sin (e+f x))^n (2 a d (-n A+A+B+B n)+a (A-B) d (2 n+1) \sin (e+f x))}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d \sin (e+f x))^n (2 a d (-n A+A+B+B n)+a (A-B) d (2 n+1) \sin (e+f x))}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(d \sin (e+f x))^n (2 a d (-n A+A+B+B n)+a (A-B) d (2 n+1) \sin (e+f x))}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3466 |
| \(\displaystyle \frac {a d (-4 A n+A+4 B n+3 B) \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx+d (2 n+1) (A-B) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a d (-4 A n+A+4 B n+3 B) \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx+d (2 n+1) (A-B) \int (d \sin (e+f x))^n \sqrt {\sin (e+f x) a+a}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3255 |
| \(\displaystyle \frac {\frac {a^2 d (2 n+1) (A-B) \cos (e+f x) \int \frac {(d \sin (e+f x))^n}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+a d (-4 A n+A+4 B n+3 B) \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 77 |
| \(\displaystyle \frac {\frac {a^2 d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {a-a \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+a d (-4 A n+A+4 B n+3 B) \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle \frac {a d (-4 A n+A+4 B n+3 B) \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x) a+a}}dx-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3266 |
| \(\displaystyle \frac {\frac {a d (-4 A n+A+4 B n+3 B) \sqrt {\sin (e+f x)+1} \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x)+1}}dx}{\sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a d (-4 A n+A+4 B n+3 B) \sqrt {\sin (e+f x)+1} \int \frac {(d \sin (e+f x))^n}{\sqrt {\sin (e+f x)+1}}dx}{\sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3265 |
| \(\displaystyle \frac {\frac {a d (-4 A n+A+4 B n+3 B) \sqrt {\sin (e+f x)+1} \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {\sin (e+f x)+1}}dx}{\sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a d (-4 A n+A+4 B n+3 B) \sqrt {\sin (e+f x)+1} \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin (e+f x)^n}{\sqrt {\sin (e+f x)+1}}dx}{\sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3264 |
| \(\displaystyle \frac {-\frac {a d (-4 A n+A+4 B n+3 B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sqrt {1-\sin (e+f x)} (\sin (e+f x)+1)}d(1-\sin (e+f x))}{f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 148 |
| \(\displaystyle \frac {-\frac {2 a d (-4 A n+A+4 B n+3 B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \int \frac {\sin ^n(e+f x)}{\sin (e+f x)+1}d\sqrt {1-\sin (e+f x)}}{f \sqrt {1-\sin (e+f x)} \sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 333 |
| \(\displaystyle \frac {-\frac {a d (-4 A n+A+4 B n+3 B) \cos (e+f x) \sin ^{-n}(e+f x) \operatorname {AppellF1}\left (\frac {1}{2},-n,1,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) (d \sin (e+f x))^n}{f \sqrt {a \sin (e+f x)+a}}-\frac {2 a d (2 n+1) (A-B) \cos (e+f x) \sin ^{-n}(e+f x) (d \sin (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},1-\sin (e+f x)\right )}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2 d}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{2 d f (a \sin (e+f x)+a)^{3/2}}\) |
((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(2*d*f*(a + a*Sin[e + f*x] )^(3/2)) + (-((a*d*(A + 3*B - 4*A*n + 4*B*n)*AppellF1[1/2, -n, 1, 3/2, 1 - Sin[e + f*x], (1 - Sin[e + f*x])/2]*Cos[e + f*x]*(d*Sin[e + f*x])^n)/(f*S in[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]])) - (2*a*(A - B)*d*(1 + 2*n)*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, 1 - Sin[e + f*x]]*(d*Sin[e + f*x])^ n)/(f*Sin[e + f*x]^n*Sqrt[a + a*Sin[e + f*x]]))/(4*a^2*d)
3.1.11.3.1 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[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m]) Int[((-d)*(x/ c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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[(c + d*x)^n/Sqrt[a - b*x], x] , x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[2*n]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])) Subst[Int[(a - x)^n*((2*a - x)^(m - 1 /2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n} , x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_), x_Symbol] :> Simp[(d/b)^IntPart[n]*((d*Sin[e + f*x])^FracPart[n ]/(b*Sin[e + f*x])^FracPart[n]) Int[(a + b*Sin[e + f*x])^m*(b*Sin[e + f*x ])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !I ntegerQ[m] && GtQ[a, 0] && !GtQ[d/b, 0]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Sin[e + f*x])^FracPart[m ]/(1 + (b/a)*Sin[e + f*x])^FracPart[m]) Int[(1 + (b/a)*Sin[e + f*x])^m*(d *Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 2, 0] && !IntegerQ[m] && !GtQ[a, 0]
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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] && (IntegerQ[2*n] || EqQ[c, 0])
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]
\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )\right )}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral(-(B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e))^n /(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (d \sin {\left (e + f x \right )}\right )^{n} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]