\(\int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [404]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 30, antiderivative size = 237 \[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {24 c^5 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {12 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {2 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a+a \sin (e+f x))^{5/2}} \] Output:

24*c^5*cos(f*x+e)*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f 
*x+e))^(1/2)+12*c^4*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f/(a+a*sin(f*x+e 
))^(1/2)+3*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f/(a+a*sin(f*x+e))^(1 
/2)+2*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f/(a+a*sin(f*x+e))^(3/2)-1/2 
*c*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 15.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.85 \[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (273+\cos (4 (e+f x))+\cos (2 (e+f x)) \left (106-384 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+1152 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+320 \sin (e+f x)+1536 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (e+f x)+24 \sin (3 (e+f x))\right )}{16 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \] Input:

Integrate[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^(5/2),x]
 

Output:

(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(273 + 
 Cos[4*(e + f*x)] + Cos[2*(e + f*x)]*(106 - 384*Log[Cos[(e + f*x)/2] + Sin 
[(e + f*x)/2]]) + 1152*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 320*Sin[ 
e + f*x] + 1536*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] + 24 
*Sin[3*(e + f*x)]))/(16*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Si 
n[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.25 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {3042, 3218, 3042, 3218, 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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{9/2}}{(a \sin (e+f x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{9/2}}{(a \sin (e+f x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3218

\(\displaystyle -\frac {2 c \int \frac {(c-c \sin (e+f x))^{7/2}}{(\sin (e+f x) a+a)^{3/2}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 c \int \frac {(c-c \sin (e+f x))^{7/2}}{(\sin (e+f x) a+a)^{3/2}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3218

\(\displaystyle -\frac {2 c \left (-\frac {3 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 c \left (-\frac {3 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3219

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3219

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3216

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3146

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (\frac {2 c^2 \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {2 c \left (-\frac {3 c \left (2 c \left (\frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{2 f (a \sin (e+f x)+a)^{5/2}}\)

Input:

Int[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^(5/2),x]
 

Output:

-1/2*(c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(f*(a + a*Sin[e + f*x])^( 
5/2)) - (2*c*(-((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*(a + a*Sin[ 
e + f*x])^(3/2))) - (3*c*((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*Sqrt[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]]))))/a))/a
 

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 3218
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 
2*n + 1)))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b 
^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && GtQ[2*m + 
n + 1, 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])
 
Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.94

method result size
default \(\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (8 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{8}+32 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+192 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-192 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-173 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+154 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-29\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4} \sqrt {4}}{8 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}\) \(222\)

Input:

int((c-c*sin(f*x+e))^(9/2)/(a+sin(f*x+e)*a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/8/f*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)^3*(8*cos(1/4*Pi+ 
1/2*f*x+1/2*e)^8+32*cos(1/4*Pi+1/2*f*x+1/2*e)^6+192*ln(-cot(1/4*Pi+1/2*f*x 
+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e))*sin(1/4*Pi+1/2*f*x+1/2*e)^4-192*ln(2/(c 
os(1/4*Pi+1/2*f*x+1/2*e)+1))*sin(1/4*Pi+1/2*f*x+1/2*e)^4-173*cos(1/4*Pi+1/ 
2*f*x+1/2*e)^4+154*cos(1/4*Pi+1/2*f*x+1/2*e)^2-29)*(c*cos(1/4*Pi+1/2*f*x+1 
/2*e)^2)^(1/2)*c^4/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a^2*4^(1/2)
 

Fricas [F]

\[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

integral(-(c^4*cos(f*x + e)^4 - 8*c^4*cos(f*x + e)^2 + 8*c^4 + 4*(c^4*cos( 
f*x + e)^2 - 2*c^4)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x 
 + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*si 
n(f*x + e)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

integrate((-c*sin(f*x + e) + c)^(9/2)/(a*sin(f*x + e) + a)^(5/2), x)
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.79 \[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, \sqrt {a} c^{\frac {9}{2}} {\left (\frac {12 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 6 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{6}} - \frac {8 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

integrate((c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac 
")
 

Output:

-2*sqrt(a)*c^(9/2)*(12*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^3*sgn 
(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e 
))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 6*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 
2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^6 - (8*sin(-1/4*pi + 1/2*f*x + 1 
/2*e)^2 - 7)/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((c - c*sin(e + f*x))^(9/2)/(a + a*sin(e + f*x))^(5/2),x)
 

Output:

int((c - c*sin(e + f*x))^(9/2)/(a + a*sin(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{4} \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x -4 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )+6 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )-4 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}+3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+1}d x \right )}{a^{3}} \] Input:

int((c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x)
 

Output:

(sqrt(c)*sqrt(a)*c**4*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 
1)*sin(e + f*x)**4)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) 
+ 1),x) - 4*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + 
f*x)**3)/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x) + 6 
*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(s 
in(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x) - 4*int((sqrt( 
sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 
 + 3*sin(e + f*x)**2 + 3*sin(e + f*x) + 1),x) + int((sqrt(sin(e + f*x) + 1 
)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**3 + 3*sin(e + f*x)**2 + 3*sin( 
e + f*x) + 1),x)))/a**3