\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx\) [174]

Optimal result

Integrand size = 40, antiderivative size = 197 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {4 (A-B) c^3 \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 (A-B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {(A-B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a+a \sin (e+f x)}} \] Output:

4*(A-B)*c^3*cos(f*x+e)*ln(1+sin(f*x+e))/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin( 
f*x+e))^(1/2)+2*(A-B)*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x 
+e))^(1/2)+1/2*(A-B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e) 
)^(1/2)-1/3*B*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 11.97 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)} \left (3 (A-3 B) \cos (2 (e+f x))-96 A \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+96 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(36 A-51 B) \sin (e+f x)+B \sin (3 (e+f x))\right )}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \sqrt {a (1+\sin (e+f x))}} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/Sqrt[a + a*Sin 
[e + f*x]],x]
 

Output:

-1/12*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^2*Sqr 
t[c - c*Sin[e + f*x]]*(3*(A - 3*B)*Cos[2*(e + f*x)] - 96*A*Log[Cos[(e + f* 
x)/2] + Sin[(e + f*x)/2]] + 96*B*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] 
+ (36*A - 51*B)*Sin[e + f*x] + B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - 
 Sin[(e + f*x)/2])^5*Sqrt[a*(1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {3042, 3452, 3042, 3219, 3042, 3219, 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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{\sqrt {a \sin (e+f x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{\sqrt {a \sin (e+f x)+a}}dx\)

\(\Big \downarrow \) 3452

\(\displaystyle (A-B) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3219

\(\displaystyle (A-B) \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3219

\(\displaystyle (A-B) \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3216

\(\displaystyle (A-B) \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-B) \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3146

\(\displaystyle (A-B) \left (2 c \left (\frac {2 c^2 \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 16

\(\displaystyle (A-B) \left (2 c \left (\frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )-\frac {B \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f \sqrt {a \sin (e+f x)+a}}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/Sqrt[a + a*Sin[e + f 
*x]],x]
 

Output:

-1/3*(B*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*Sqrt[a + a*Sin[e + f*x 
]]) + (A - B)*((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*f*Sqrt[a + a 
*Sin[e + f*x]]) + 2*c*((2*c^2*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(f*Sqr 
t[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*Sqrt[c - 
 c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])))
 

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 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 3219

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m]) &&  !( 
ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
 

rule 3452

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1))   Int[( 
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, 
e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, 
 -2^(-1)] && NeQ[m + n + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(441\) vs. \(2(177)=354\).

Time = 7.30 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.24

method	result	size
parts	\(\frac {A \sqrt {4}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-4 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+4 \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 )-3\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}+\frac {2 B \,c^{2} \left (\left (12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{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 (-12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{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 (\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+9 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-12\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\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 ) c}}{3 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}}\)	\(442\)
default	\(-\frac {A \sqrt {4}\, \left (-\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+4 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-4 \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 )+3\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}+\frac {2 B \,c^{2} \left (\left (-12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{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 (12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{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 (\left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-9 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+12\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+12 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\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 ) c}}{3 f \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sin \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}}\)	\(445\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

A/f*4^(1/2)*(cos(1/4*Pi+1/2*f*x+1/2*e)^4+2*cos(1/4*Pi+1/2*f*x+1/2*e)^2-4*l 
n(2/(cos(1/4*Pi+1/2*f*x+1/2*e)+1))+4*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4 
*Pi+1/2*f*x+1/2*e))-3)*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*c^2/(a*sin(1/ 
4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*tan(1/4*Pi+1/2*f*x+1/2*e)+2/3*B*c^2/f*((12*si 
n(1/2*f*x+1/2*e)+12*cos(1/2*f*x+1/2*e))*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))+(-12*sin(1/2*f*x+1/2*e)-12*cos(1/2*f*x 
+1/2*e))*ln(2/(cos(1/2*f*x+1/2*e)+1))+cos(1/2*f*x+1/2*e)*((4*cos(1/2*f*x+1 
/2*e)^4-4*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)+5*cos(1/2*f*x+1/2*e)^2+9 
*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-12)*sin(1/2*f*x+1/2*e)^2-12*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)*cos(1/2*f*x+1/2*e 
)-1)*c)^(1/2)/(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)*cos(1/2*f*x+1/2*e)+1)*a)^(1/2)
 

Fricas [F]

\[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x 
, algorithm="fricas")
 

Output:

integral(-((A - 2*B)*c^2*cos(f*x + e)^2 - 2*(A - B)*c^2 + (B*c^2*cos(f*x + 
 e)^2 + 2*(A - B)*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/sqrt(a*sin( 
f*x + e) + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**(1/2) 
,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x 
, algorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*(-c*sin(f*x + e) + c)^(5/2)/sqrt(a*sin(f*x 
+ e) + a), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),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. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x)) 
^(1/2),x)
 

Output:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x)) 
^(1/2), x)
 

Reduce [F]

\[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{2} \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )+1}d x \right ) a -2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )+1}d x \right ) b -2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x \right ) a \right )}{a} \] Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x)
 

Output:

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