Integrand size = 39, antiderivative size = 413 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (c (A+C) d-d^2 (A-C+4 A m)-2 c^2 (C+2 C m)\right ) \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},\frac {1}{2},\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{d \left (c^2-d^2\right ) f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \operatorname {AppellF1}\left (\frac {3}{2}+m,\frac {1}{2},\frac {1}{2},\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{a d \left (c^2-d^2\right ) f (3+2 m) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
2*(A*d^2+C*c^2)*cos(f*x+e)*(a+a*sin(f*x+e))^m/d/(c^2-d^2)/f/(c+d*sin(f*x+e ))^(1/2)+(c*(A+C)*d-d^2*(4*A*m+A-C)-2*c^2*(2*C*m+C))*AppellF1(1/2+m,1/2,1/ 2,3/2+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))^m*2^(1/2)*((c+d*sin(f*x+e))/(c-d))^(1/2)/d/(c^2-d^2)/f/(1+2*m)/(1-si n(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)+(2*c^2*C*(1+m)+d^2*(2*A*m+A-C))*App ellF1(3/2+m,1/2,1/2,5/2+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)*2^(1/2)*((c+d*sin(f*x+e))/(c-d))^(1/2)/a/d/( c^2-d^2)/f/(3+2*m)/(1-sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx \]
Time = 1.27 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3042, 3523, 27, 3042, 3466, 3042, 3267, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin (e+f x)^2\right )}{(c+d \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}-\frac {2 \int -\frac {(\sin (e+f x) a+a)^m \left (a (c C (d-2 c m)+A d (c-2 d m))+a \left (2 C (m+1) c^2+d^2 (2 m A+A-C)\right ) \sin (e+f x)\right )}{2 \sqrt {c+d \sin (e+f x)}}dx}{a d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m \left (a (c C (d-2 c m)+A d (c-2 d m))+a \left (2 C (m+1) c^2+d^2 (2 m A+A-C)\right ) \sin (e+f x)\right )}{\sqrt {c+d \sin (e+f x)}}dx}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m \left (a (c C (d-2 c m)+A d (c-2 d m))+a \left (2 C (m+1) c^2+d^2 (2 m A+A-C)\right ) \sin (e+f x)\right )}{\sqrt {c+d \sin (e+f x)}}dx}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3466 |
| \(\displaystyle \frac {a \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c+d \sin (e+f x)}}dx+\left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \int \frac {(\sin (e+f x) a+a)^{m+1}}{\sqrt {c+d \sin (e+f x)}}dx}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c+d \sin (e+f x)}}dx+\left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \int \frac {(\sin (e+f x) a+a)^{m+1}}{\sqrt {c+d \sin (e+f x)}}dx}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3267 |
| \(\displaystyle \frac {\frac {a^3 \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {a^2 \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 157 |
| \(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 156 |
| \(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}+\frac {a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 155 |
| \(\displaystyle \frac {\frac {\sqrt {2} a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) (a \sin (e+f x)+a)^m \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (a-a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} a \sqrt {1-\sin (e+f x)} \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) (a \sin (e+f x)+a)^{m+1} \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},\frac {1}{2},m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (a-a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}}{a d \left (c^2-d^2\right )}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}\) |
(2*(c^2*C + A*d^2)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(d*(c^2 - d^2)*f*S qrt[c + d*Sin[e + f*x]]) + ((Sqrt[2]*a^2*(c*(A + C)*d - d^2*(A - C + 4*A*m ) - 2*c^2*(C + 2*C*m))*AppellF1[1/2 + m, 1/2, 1/2, 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*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(f*(1 + 2* m)*(a - a*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) + (Sqrt[2]*a*(2*c^2*C*(1 + m) + d^2*(A - C + 2*A*m))*AppellF1[3/2 + m, 1/2, 1/2, 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)*Sqrt[(c + d*Sin[e + f*x])/(c - d)] )/(f*(3 + 2*m)*(a - a*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]))/(a*d*(c^2 - d^2))
3.1.14.3.1 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)/(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])
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral((C*cos(f*x + e)^2 - A - C)*sqrt(d*sin(f*x + e) + c)*(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)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]