Integrand size = 50, antiderivative size = 174 \[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(A+B+C) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {(A-B-3 C) \cos (e+f x) \log (1-\sin (e+f x))}{4 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {(A-B+C) \cos (e+f x) \log (1+\sin (e+f x))}{4 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
1/4*(A+B+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/a/f/(c-c*sin(f*x+e))^(3/2)-1 /4*(A-B-3*C)*cos(f*x+e)*ln(1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*s in(f*x+e))^(1/2)+1/4*(A-B+C)*cos(f*x+e)*ln(1+sin(f*x+e))/c/f/(a+a*sin(f*x+ e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 3.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13 \[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (A+B+C+(-A+B+3 C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(A-B+C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f \sqrt {a (1+\sin (e+f x))} (c-c \sin (e+f x))^{3/2}} \]
Integrate[(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2)/(Sqrt[a + a*Sin[e + f*x] ]*(c - c*Sin[e + f*x])^(3/2)),x]
((A + B + C + (-A + B + 3*C)*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*(Cos [(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (A - B + C)*Log[Cos[(e + f*x)/2] + S in[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2)*(Cos[(e + f*x)/2 ] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(2*f*Sqrt[a*( 1 + Sin[e + f*x])]*(c - c*Sin[e + f*x])^(3/2))
Time = 1.14 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3514, 27, 3042, 3448, 3042, 3216, 3042, 3146, 16}
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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)+C \sin (e+f x)^2}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3514 |
| \(\displaystyle \frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {\int -\frac {2 \left (a^2 (A-B-C)-2 a^2 C \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{4 a^2 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^2 (A-B-C)-2 a^2 C \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2 (A-B-C)-2 a^2 C \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3448 |
| \(\displaystyle \frac {\frac {a^2 (A-B+C) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{2 c}+\frac {1}{2} a (A-B-3 C) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A-B+C) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{2 c}+\frac {1}{2} a (A-B-3 C) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {\frac {a^3 (A-B+C) \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 c (A-B-3 C) \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^3 (A-B+C) \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 c (A-B-3 C) \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {\frac {a^2 (A-B+C) \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A-B-3 C) \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {a^2 (A-B+C) \cos (e+f x) \log (a \sin (e+f x)+a)}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A-B-3 C) \cos (e+f x) \log (c-c \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+B+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\) |
Int[(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2)/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
((A + B + C)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(4*a*f*(c - c*Sin[e + f*x])^(3/2)) + ((a^2*(A - B + C)*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(2* f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a^2*(A - B - 3*C)* Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]))/(2*a^2*c)
3.1.16.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp [(A*b + a*B)/(2*a*b) Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x] ], x], x] + Simp[(B*c + A*d)/(2*c*d) Int[Sqrt[c + d*Sin[e + f*x]]/Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(2*b*c*f*(2*m + 1))), x] - Si mp[1/(2*b*c*d*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - B*c*d*(m - n - 1) - C*(c ^2*m - d^2*(n + 1)) + d*((A*c + B*d)*(m + n + 2) - c*C*(3*m - n))*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)] || (EqQ[m + n + 2, 0] && NeQ[2*m + 1, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(746\) vs. \(2(156)=312\).
Time = 3.90 (sec) , antiderivative size = 747, normalized size of antiderivative = 4.29
| method | result | size |
| default | \(\frac {-A \sin \left (f x +e \right )-C -A -B -C \sin \left (f x +e \right )+A \sin \left (f x +e \right ) \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-A \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-B \sin \left (f x +e \right ) \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+B \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-B \sin \left (f x +e \right )+A \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \sin \left (f x +e \right ) \cos \left (f x +e \right )-B \cos \left (f x +e \right ) \sin \left (f x +e \right )-B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+C \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+3 C \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 C \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+B \left (\cos ^{2}\left (f x +e \right )\right )-C \sin \left (f x +e \right ) \cos \left (f x +e \right )-C \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-3 C \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+2 C \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 C \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-C \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-3 C \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+C \left (\cos ^{2}\left (f x +e \right )\right )-B \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+B \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+A \left (\cos ^{2}\left (f x +e \right )\right )}{2 c f \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\) | \(747\) |
| parts | \(\text {Expression too large to display}\) | \(815\) |
int((A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e) )^(1/2),x,method=_RETURNVERBOSE)
1/2/c/f*(-A*sin(f*x+e)-C*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)-3*C*cos (f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)+2*C*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)) )-C-A-B-C*sin(f*x+e)+A*sin(f*x+e)*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)- A*ln(csc(f*x+e)-cot(f*x+e)-1)*sin(f*x+e)*cos(f*x+e)-B*sin(f*x+e)*cos(f*x+e )*ln(-cot(f*x+e)+csc(f*x+e)+1)+B*ln(csc(f*x+e)-cot(f*x+e)-1)*sin(f*x+e)*co s(f*x+e)-B*sin(f*x+e)+B*cos(f*x+e)^2+A*cos(f*x+e)^2+C*cos(f*x+e)^2+A*cos(f *x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)-A*sin(f*x+e)*cos(f*x+e)-B*cos(f*x+e)*sin (f*x+e)+C*ln(-cot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)*cos(f*x+e)+3*C*ln(csc(f* x+e)-cot(f*x+e)-1)*sin(f*x+e)*cos(f*x+e)-2*C*ln(2/(1+cos(f*x+e)))*sin(f*x+ e)*cos(f*x+e)-C*sin(f*x+e)*cos(f*x+e)-C*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+ e)+1)-3*C*cos(f*x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)+2*C*cos(f*x+e)*ln(2/(1+co s(f*x+e)))-B*cos(f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)+A*cos(f*x+e)^2*ln(cs c(f*x+e)-cot(f*x+e)-1)-B*cos(f*x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)-A*cos(f*x+ e)*ln(-cot(f*x+e)+csc(f*x+e)+1)+B*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)+ B*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)-A*cos(f*x+e)^2*ln(-cot(f*x+e)+ csc(f*x+e)+1))/(-cos(f*x+e)+sin(f*x+e)-1)/(-c*(sin(f*x+e)-1))^(1/2)/(a*(1+ sin(f*x+e)))^(1/2)
\[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin( f*x+e))^(1/2),x, algorithm="fricas")
integral((C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a*c^2*cos(f*x + e)^2*sin(f*x + e) - a*c^2*c os(f*x + e)^2), x)
\[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {A + B \sin {\left (e + f x \right )} + C \sin ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]
integrate((A+B*sin(f*x+e)+C*sin(f*x+e)**2)/(c-c*sin(f*x+e))**(3/2)/(a+a*si n(f*x+e))**(1/2),x)
Integral((A + B*sin(e + f*x) + C*sin(e + f*x)**2)/(sqrt(a*(sin(e + f*x) + 1))*(-c*(sin(e + f*x) - 1))**(3/2)), x)
\[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin( f*x+e))^(1/2),x, algorithm="maxima")
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)/(sqrt(a*sin(f*x + e) + a )*(-c*sin(f*x + e) + c)^(3/2)), x)
Time = 0.47 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.26 \[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {{\left (A \sqrt {a} - B \sqrt {a} - 3 \, C \sqrt {a}\right )} \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a c^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a c^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {A \sqrt {a} + B \sqrt {a} + C \sqrt {a}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a c^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{4 \, f} \]
integrate((A+B*sin(f*x+e)+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin( f*x+e))^(1/2),x, algorithm="giac")
1/4*((A*sqrt(a) - B*sqrt(a) - 3*C*sqrt(a))*log(-cos(-1/4*pi + 1/2*f*x + 1/ 2*e)^2 + 1)/(a*c^(3/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*(A*sqrt(a) - B*sqrt(a) + C*sqrt(a))*log(abs(cos( -1/4*pi + 1/2*f*x + 1/2*e)))/(a*c^(3/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) )*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (A*sqrt(a) + B*sqrt(a) + C*sqrt(a ))/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*a*c^(3/2)*sgn(cos(-1/4*pi + 1/2 *f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f
Timed out. \[ \int \frac {A+B \sin (e+f x)+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]
int((A + B*sin(e + f*x) + C*sin(e + f*x)^2)/((a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(3/2)),x)