\(\int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [7]

Optimal result

Integrand size = 42, antiderivative size = 167 \[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(A+C) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {(A-3 C) \cos (e+f x) \log (1-\sin (e+f x))}{4 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {(A+C) \cos (e+f x) \log (1+\sin (e+f x))}{4 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

1/4*(A+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-3*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)+1/4*(A+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)
 

Mathematica [A] (verified)

Time = 4.78 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.14 \[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (A+C-(A-3 C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(A+C) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f \sqrt {a (1+\sin (e+f x))} (c-c \sin (e+f x))^{3/2}} \] Input:

Integrate[(A + C*Sin[e + f*x]^2)/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + 
f*x])^(3/2)),x]
 

Output:

((A + C - (A - 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 + C)*Log[Cos[(e + f*x)/2] + Sin[(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))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3515, 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.

\(\displaystyle \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin (e+f x)^2}{\sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3515

\(\displaystyle \frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {\int -\frac {2 \left (a^2 (A-C)-2 a^2 C \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{4 a^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a^2 (A-C)-2 a^2 C \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a^2 (A-C)-2 a^2 C \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3448

\(\displaystyle \frac {\frac {a^2 (A+C) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{2 c}+\frac {1}{2} a (A-3 C) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (A+C) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{2 c}+\frac {1}{2} a (A-3 C) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3216

\(\displaystyle \frac {\frac {a^3 (A+C) \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 c (A-3 C) \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^3 (A+C) \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 c (A-3 C) \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{2 \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3146

\(\displaystyle \frac {\frac {a^2 (A+C) \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A-3 C) \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\frac {a^2 (A+C) \cos (e+f x) \log (a \sin (e+f x)+a)}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A-3 C) \cos (e+f x) \log (c-c \sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{2 a^2 c}+\frac {(A+C) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 a f (c-c \sin (e+f x))^{3/2}}\)

Input:

Int[(A + C*Sin[e + f*x]^2)/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^ 
(3/2)),x]
 

Output:

((A + 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 + 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 - 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)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3146

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])
 

rule 3216

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 
]
 

rule 3448

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]
 

rule 3515

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[(a*A + 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] - Simp[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)) - C*(c^2*m - d^2*(n + 1)) + d*(A*c*(m + n + 2) - c*C*(3*m - n))* 
Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(587\) vs. \(2(149)=298\).

Time = 6.96 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.52

method	result	size
parts	\(\frac {A \sqrt {4}\, \left (2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{8 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c}+\frac {C \left (\left (\left (6 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )\right ) \ln \left (\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (\left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )\right ) \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (\left (-8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+4\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )\right ) \ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )\right )}{2 c f \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}}\)	\(588\)

Input:

int((A+C*sin(f*x+e)^2)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x,met 
hod=_RETURNVERBOSE)
 

Output:

1/8*A/f*4^(1/2)*(2*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e) 
)*cos(1/4*Pi+1/2*f*x+1/2*e)^2-2*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1 
/2*f*x+1/2*e)-1)*cos(1/4*Pi+1/2*f*x+1/2*e)^2-2*ln(-cot(1/4*Pi+1/2*f*x+1/2* 
e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)*cos(1/4*Pi+1/2*f*x+1/2*e)^2+sin(1/4*Pi+1/2 
*f*x+1/2*e)^2)/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*cos(1/4*Pi+1/2*f*x 
+1/2*e)^2)^(1/2)/c*tan(1/4*Pi+1/2*f*x+1/2*e)+1/2*C/c/f*(((6*cos(1/2*f*x+1/ 
2*e)^2-3)*sin(1/2*f*x+1/2*e)-3*cos(1/2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e) 
^2))*ln(2*(sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1))+ 
((-1+2*cos(1/2*f*x+1/2*e)^2)*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e)*(-1+2*c 
os(1/2*f*x+1/2*e)^2))*ln(-2*(cos(1/2*f*x+1/2*e)+sin(1/2*f*x+1/2*e))/(cos(1 
/2*f*x+1/2*e)+1))+((-8*cos(1/2*f*x+1/2*e)^2+4)*sin(1/2*f*x+1/2*e)+4*cos(1/ 
2*f*x+1/2*e)*(-1+2*cos(1/2*f*x+1/2*e)^2))*ln(2/(cos(1/2*f*x+1/2*e)+1))+cos 
(1/2*f*x+1/2*e)*(-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-2*sin(1/2*f*x+1/ 
2*e)^2))/(sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e))/((2*sin(1/2*f*x+1/2*e)*co 
s(1/2*f*x+1/2*e)+1)*a)^(1/2)/(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1) 
*c)^(1/2)
 

Fricas [F]

\[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {C \sin \left (f x + e\right )^{2} + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*sin(f*x+e)^2)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2) 
,x, algorithm="fricas")
 

Output:

integral((C*cos(f*x + e)^2 - 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*cos(f*x + e)^2), x 
)
 

Sympy [F]

\[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {A + C \sin ^{2}{\left (e + f x \right )}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((A+C*sin(f*x+e)**2)/(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(3 
/2),x)
 

Output:

Integral((A + C*sin(e + f*x)**2)/(sqrt(a*(sin(e + f*x) + 1))*(-c*(sin(e + 
f*x) - 1))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {C \sin \left (f x + e\right )^{2} + A}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+C*sin(f*x+e)^2)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2) 
,x, algorithm="maxima")
 

Output:

integrate((C*sin(f*x + e)^2 + A)/(sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e 
) + c)^(3/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*sin(f*x+e)^2)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2) 
,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {C\,{\sin \left (e+f\,x\right )}^2+A}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((A + C*sin(e + f*x)^2)/((a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x) 
)^(3/2)),x)
 

Output:

int((A + C*sin(e + f*x)^2)/((a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x) 
)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {A+C \sin ^2(e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-\sin \left (f x +e \right )^{2}-\sin \left (f x +e \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-\sin \left (f x +e \right )^{2}-\sin \left (f x +e \right )+1}d x \right ) a \right )}{a \,c^{2}} \] Input:

int((A+C*sin(f*x+e)^2)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x)
 

Output:

(sqrt(c)*sqrt(a)*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*si 
n(e + f*x)**2)/(sin(e + f*x)**3 - sin(e + f*x)**2 - sin(e + f*x) + 1),x)*c 
 + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**3 
 - sin(e + f*x)**2 - sin(e + f*x) + 1),x)*a))/(a*c**2)