Integrand size = 48, antiderivative size = 197 \[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {2 c (C-6 C m+A (5+2 m)-B (5+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 c (5 B+2 C+2 B m+4 C m) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) (5+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2}}{c f (5+2 m)} \]
2*C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3/2)/c/f/(5+2*m)+2*c*( C-6*C*m+A*(5+2*m)-B*(5+2*m))*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(4*m^2+12*m+5 )/(c-c*sin(f*x+e))^(1/2)+2*c*(2*B*m+4*C*m+5*B+2*C)*cos(f*x+e)*(a+a*sin(f*x +e))^(1+m)/a/f/(4*m^2+16*m+15)/(c-c*sin(f*x+e))^(1/2)
Time = 2.07 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.90 \[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} \left (30 A-20 B+19 C+32 A m-8 B m+8 C m+8 A m^2+4 C m^2-C \left (3+8 m+4 m^2\right ) \cos (2 (e+f x))+2 (1+2 m) (5 B-4 C+2 B m) \sin (e+f x)\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
Integrate[(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]]*(A + B*Sin[e + f *x] + C*Sin[e + f*x]^2),x]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c *Sin[e + f*x]]*(30*A - 20*B + 19*C + 32*A*m - 8*B*m + 8*C*m + 8*A*m^2 + 4* C*m^2 - C*(3 + 8*m + 4*m^2)*Cos[2*(e + f*x)] + 2*(1 + 2*m)*(5*B - 4*C + 2* B*m)*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - S in[(e + f*x)/2]))
Time = 1.00 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3042, 3518, 27, 3042, 3450, 3042, 3217}
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.
| \(\displaystyle \int \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m \left (A+B \sin (e+f x)+C \sin (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 3518 |
| \(\displaystyle \frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}-\frac {2 \int -\frac {1}{2} (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)} (a c (C (3-2 m)+A (2 m+5))+a c (2 m B+5 B+2 C+4 C m) \sin (e+f x))dx}{a c (2 m+5)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)} (a c (C (3-2 m)+A (2 m+5))+a c (2 m B+5 B+2 C+4 C m) \sin (e+f x))dx}{a c (2 m+5)}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)} (a c (C (3-2 m)+A (2 m+5))+a c (2 m B+5 B+2 C+4 C m) \sin (e+f x))dx}{a c (2 m+5)}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}\) |
| \(\Big \downarrow \) 3450 |
| \(\displaystyle \frac {a c (A (2 m+5)-B (2 m+5)-6 C m+C) \int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)}dx+c (2 B m+5 B+4 C m+2 C) \int (\sin (e+f x) a+a)^{m+1} \sqrt {c-c \sin (e+f x)}dx}{a c (2 m+5)}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a c (A (2 m+5)-B (2 m+5)-6 C m+C) \int (\sin (e+f x) a+a)^m \sqrt {c-c \sin (e+f x)}dx+c (2 B m+5 B+4 C m+2 C) \int (\sin (e+f x) a+a)^{m+1} \sqrt {c-c \sin (e+f x)}dx}{a c (2 m+5)}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {\frac {2 a c^2 (A (2 m+5)-B (2 m+5)-6 C m+C) \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {2 c^2 (2 B m+5 B+4 C m+2 C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{f (2 m+3) \sqrt {c-c \sin (e+f x)}}}{a c (2 m+5)}+\frac {2 C \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{c f (2 m+5)}\) |
Int[(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]]*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]
(2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(c*f* (5 + 2*m)) + ((2*a*c^2*(C - 6*C*m + A*(5 + 2*m) - B*(5 + 2*m))*Cos[e + f*x ]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + (2*c^2* (5*B + 2*C + 2*B*m + 4*C*m)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(f* (3 + 2*m)*Sqrt[c - c*Sin[e + f*x]]))/(a*c*(5 + 2*m))
3.1.20.3.1 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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f _.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [B/d Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] - Simp[(B*c - A*d)/d Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x ], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[ a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (b*B*d*(m + n + 2) - b*c*C*(2*m + 1))* Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c -c \sin \left (f x +e \right )}\, \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )d x\]
Time = 0.29 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.57 \[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {2 \, {\left ({\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{3} - 4 \, {\left (A + B + C\right )} m^{2} + {\left (4 \, {\left (B + C\right )} m^{2} + 12 \, B m + 5 \, B - C\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (2 \, A + B\right )} m - {\left (4 \, {\left (A + C\right )} m^{2} + 4 \, {\left (4 \, A - B + 2 \, C\right )} m + 15 \, A - 10 \, B + 11 \, C\right )} \cos \left (f x + e\right ) - {\left (4 \, {\left (A + B + C\right )} m^{2} - {\left (4 \, C m^{2} + 8 \, C m + 3 \, C\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (2 \, A + B\right )} m + {\left (4 \, B m^{2} + 4 \, {\left (3 \, B - 2 \, C\right )} m + 5 \, B - 4 \, C\right )} \cos \left (f x + e\right ) + 15 \, A - 5 \, B + 7 \, C\right )} \sin \left (f x + e\right ) - 15 \, A + 5 \, B - 7 \, C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)+C*sin( f*x+e)^2),x, algorithm="fricas")
-2*((4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^3 - 4*(A + B + C)*m^2 + (4*(B + C )*m^2 + 12*B*m + 5*B - C)*cos(f*x + e)^2 - 8*(2*A + B)*m - (4*(A + C)*m^2 + 4*(4*A - B + 2*C)*m + 15*A - 10*B + 11*C)*cos(f*x + e) - (4*(A + B + C)* m^2 - (4*C*m^2 + 8*C*m + 3*C)*cos(f*x + e)^2 + 8*(2*A + B)*m + (4*B*m^2 + 4*(3*B - 2*C)*m + 5*B - 4*C)*cos(f*x + e) + 15*A - 5*B + 7*C)*sin(f*x + e) - 15*A + 5*B - 7*C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f *m^3 + 36*f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*sin(f*x + e) + 15*f)
\[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}\right )\, dx \]
Integral((a*(sin(e + f*x) + 1))**m*sqrt(-c*(sin(e + f*x) - 1))*(A + B*sin( e + f*x) + C*sin(e + f*x)**2), x)
Leaf count of result is larger than twice the leaf count of optimal. 644 vs. \(2 (187) = 374\).
Time = 0.38 (sec) , antiderivative size = 644, normalized size of antiderivative = 3.27 \[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {2 \, {\left (\frac {2 \, {\left (\frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - a^{m} \sqrt {c} - \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} B e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + \frac {{\left (4 \, m^{2} + 8 \, m + 3\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 3\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} - \frac {4 \, {\left (\frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {{\left (4 \, m^{2} + 4 \, m + 5\right )} a^{m} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {4 \, a^{m} \sqrt {c} m \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2 \, a^{m} \sqrt {c} - \frac {2 \, a^{m} \sqrt {c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} C e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + \frac {2 \, {\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac {{\left (a^{m} \sqrt {c} + \frac {a^{m} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} A e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (2 \, m + 1\right )} \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{f} \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)+C*sin( f*x+e)^2),x, algorithm="maxima")
-2*(2*(2*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a^m*sqrt(c)*m*s in(f*x + e)^2/(cos(f*x + e) + 1)^2 - a^m*sqrt(c) - a^m*sqrt(c)*sin(f*x + e )^3/(cos(f*x + e) + 1)^3)*B*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1 ) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((4*m^2 + 8*m + (4*m^2 + 8*m + 3)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3)*sqrt(sin(f*x + e)^2/( cos(f*x + e) + 1)^2 + 1)) - 4*(4*a^m*sqrt(c)*m*sin(f*x + e)/(cos(f*x + e) + 1) - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - (4*m^2 + 4*m + 5)*a^m*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 4*a^m *sqrt(c)*m*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2*a^m*sqrt(c) - 2*a^m*sqr t(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*C*e^(2*m*log(sin(f*x + e)/(cos(f *x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^ 3 + 36*m^2 + 46*m + 2*(8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)) + (a^m*sqrt(c) + a^m*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1))*A*e^(2*m*log(sin(f*x + e)/( cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/( (2*m + 1)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)))/f
\[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=\int { {\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e)+C*sin( f*x+e)^2),x, algorithm="giac")
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c )*(a*sin(f*x + e) + a)^m, x)
Time = 19.68 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.59 \[ \int (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx=-\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (30\,A-15\,B+15\,C+32\,A\,m+4\,B\,m+8\,A\,m^2+4\,B\,m^2+4\,C\,m^2\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (A\,30{}\mathrm {i}-B\,15{}\mathrm {i}+C\,15{}\mathrm {i}+A\,m\,32{}\mathrm {i}+B\,m\,4{}\mathrm {i}+A\,m^2\,8{}\mathrm {i}+B\,m^2\,4{}\mathrm {i}+C\,m^2\,4{}\mathrm {i}\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {C\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,4{}\mathrm {i}+m\,8{}\mathrm {i}+3{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {C\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (10\,B-5\,C+4\,B\,m+2\,C\,m\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (B\,10{}\mathrm {i}-C\,5{}\mathrm {i}+B\,m\,4{}\mathrm {i}+C\,m\,2{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}\right )}{{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (8\,m^3+36\,m^2+46\,m+15\right )}{m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}}} \]
int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(1/2)*(A + B*sin(e + f*x) + C*sin(e + f*x)^2),x)
-((c - c*sin(e + f*x))^(1/2)*((C*(a + a*sin(e + f*x))^m*(m*8i + m^2*4i + 3 i))/(2*f*(m*46i + m^2*36i + m^3*8i + 15i)) - (exp(e*2i + f*x*2i)*(a + a*si n(e + f*x))^m*(A*30i - B*15i + C*15i + A*m*32i + B*m*4i + A*m^2*8i + B*m^2 *4i + C*m^2*4i))/(f*(m*46i + m^2*36i + m^3*8i + 15i)) - (exp(e*3i + f*x*3i )*(a + a*sin(e + f*x))^m*(30*A - 15*B + 15*C + 32*A*m + 4*B*m + 8*A*m^2 + 4*B*m^2 + 4*C*m^2))/(f*(m*46i + m^2*36i + m^3*8i + 15i)) + (C*exp(e*5i + f *x*5i)*(a + a*sin(e + f*x))^m*(8*m + 4*m^2 + 3))/(2*f*(m*46i + m^2*36i + m ^3*8i + 15i)) + (exp(e*1i + f*x*1i)*(2*m + 1)*(a + a*sin(e + f*x))^m*(10*B - 5*C + 4*B*m + 2*C*m))/(2*f*(m*46i + m^2*36i + m^3*8i + 15i)) + (exp(e*4 i + f*x*4i)*(2*m + 1)*(a + a*sin(e + f*x))^m*(B*10i - C*5i + B*m*4i + C*m* 2i))/(2*f*(m*46i + m^2*36i + m^3*8i + 15i))))/(exp(e*3i + f*x*3i) + (exp(e *2i + f*x*2i)*(46*m + 36*m^2 + 8*m^3 + 15))/(m*46i + m^2*36i + m^3*8i + 15 i))