\(\int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [59]

Optimal result

Integrand size = 38, antiderivative size = 191 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {12 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {3 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{5/2}}{a f (a+a \sin (e+f x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 8.80 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} \left (-44-18 \cos (2 (e+f x))-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{8 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \] Input:

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

Output:

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f*x]]*(-44 
 - 18*Cos[2*(e + f*x)] - 192*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (3 
9 - 192*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3*(e 
+ f*x)]))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]) 
)^(5/2))
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 197, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

(-((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*S 
in[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*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

integrate(cos(f*x+e)**2*(c-c*sin(f*x+e))**(5/2)/(a+a*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. \(2 (173) = 346\).

Time = 0.18 (sec) , antiderivative size = 1120, normalized size of antiderivative = 5.86 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(144*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) - 72*c^( 
5/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/a^(5/2) - (46*c^(5/2) + 
199*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 335*c^(5/2)*sin(f*x + e)^2/( 
cos(f*x + e) + 1)^2 + 509*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 49 
6*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 373*c^(5/2)*sin(f*x + e)^5 
/(cos(f*x + e) + 1)^5 + 219*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 
63*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(a^(5/2) + 4*a^(5/2)*sin(f 
*x + e)/(cos(f*x + e) + 1) + 8*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 
 + 12*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a^(5/2)*sin(f*x + e 
)^4/(cos(f*x + e) + 1)^4 + 12*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 
+ 8*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*a^(5/2)*sin(f*x + e)^7 
/(cos(f*x + e) + 1)^7 + a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) + (46 
*c^(5/2) + 121*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 149*c^(5/2)*sin(f 
*x + e)^2/(cos(f*x + e) + 1)^2 + 179*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) 
+ 1)^3 + 148*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 43*c^(5/2)*sin( 
f*x + e)^5/(cos(f*x + e) + 1)^5 + 33*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) 
+ 1)^6 - 15*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)/(a^(5/2) + 4*a^(5 
/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 8*a^(5/2)*sin(f*x + e)^2/(cos(f*x + 
e) + 1)^2 + 12*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 14*a^(5/2)*si 
n(f*x + e)^4/(cos(f*x + e) + 1)^4 + 12*a^(5/2)*sin(f*x + e)^5/(cos(f*x ...
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {2 \, \sqrt {a} c^{\frac {5}{2}} {\left (\frac {6 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 4 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{6}} - \frac {2}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*sqrt(a)*c**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)))/a**3