\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [25]

Optimal result

Integrand size = 38, antiderivative size = 192 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{c f (c-c \sin (e+f x))^{3/2}}+\frac {12 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 8.71 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} \left (44+18 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{8 c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(4 
4 + 18*Cos[2*(e + f*x)] + 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + ( 
39 - 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3*(e 
 + f*x)]))/(8*c^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f* 
x])^2*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3320, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

rule 3320

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 
2)*c^(p/2))   Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + 
p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && 
EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
 

Maple [F]

\[\int \frac {\cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, al 
gorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (174) = 348\).

Time = 0.47 (sec) , antiderivative size = 1120, normalized size of antiderivative = 5.83 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, al 
gorithm="maxima")
 

Output:

-1/6*(144*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) - 72*a^ 
(5/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/c^(5/2) + (46*a^(5/2) - 
 121*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 149*a^(5/2)*sin(f*x + e)^2/ 
(cos(f*x + e) + 1)^2 - 179*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1 
48*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 43*a^(5/2)*sin(f*x + e)^5 
/(cos(f*x + e) + 1)^5 + 33*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1 
5*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(5/2) - 4*c^(5/2)*sin(f* 
x + e)/(cos(f*x + e) + 1) + 8*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 
- 12*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*c^(5/2)*sin(f*x + e) 
^4/(cos(f*x + e) + 1)^4 - 12*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 
 8*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 4*c^(5/2)*sin(f*x + e)^7/ 
(cos(f*x + e) + 1)^7 + c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) - (46* 
a^(5/2) - 199*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 335*a^(5/2)*sin(f* 
x + e)^2/(cos(f*x + e) + 1)^2 - 509*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 
 1)^3 + 496*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 373*a^(5/2)*sin( 
f*x + e)^5/(cos(f*x + e) + 1)^5 + 219*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) 
 + 1)^6 - 63*a^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(c^(5/2) - 4*c^( 
5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*c^(5/2)*sin(f*x + e)^2/(cos(f*x + 
 e) + 1)^2 - 12*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*c^(5/2)*s 
in(f*x + e)^4/(cos(f*x + e) + 1)^4 - 12*c^(5/2)*sin(f*x + e)^5/(cos(f*x...
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, a^{\frac {5}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, c^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{6}} - \frac {2}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

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

Output:

-2*a^(5/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^3*sgn( 
sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (c^3*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*s 
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*c^3*cos(-1/4*pi + 1/2*f*x + 1/2*e)^ 
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c^6 - 2/((cos(-1/4*pi + 1/2*f*x + 1 
/2*e)^2 - 1)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1 
/2*f*x + 1/2*e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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