\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx\) [32]

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

Optimal result

Integrand size = 38, antiderivative size = 188 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^3 \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{35 a f \sqrt {c-c \sin (e+f x)}}+\frac {c^2 \cos (e+f x) (a+a \sin (e+f x))^{9/2} \sqrt {c-c \sin (e+f x)}}{14 a f}+\frac {3 c \cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{3/2}}{28 a f}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{9/2} (c-c \sin (e+f x))^{5/2}}{8 a f} \] Output:

1/35*c^3*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)/a/f/(c-c*sin(f*x+e))^(1/2)+1/14 
*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)*(c-c*sin(f*x+e))^(1/2)/a/f+3/28*c*c 
os(f*x+e)*(a+a*sin(f*x+e))^(9/2)*(c-c*sin(f*x+e))^(3/2)/a/f+1/8*cos(f*x+e) 
*(a+a*sin(f*x+e))^(9/2)*(c-c*sin(f*x+e))^(5/2)/a/f
 

Mathematica [A] (verified)

Time = 8.88 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.68 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a^3 c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-1960 \cos (2 (e+f x))-980 \cos (4 (e+f x))-280 \cos (6 (e+f x))-35 \cos (8 (e+f x))+19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x)))}{35840 f} \] Input:

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^( 
5/2),x]
 

Output:

(a^3*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]* 
(-1960*Cos[2*(e + f*x)] - 980*Cos[4*(e + f*x)] - 280*Cos[6*(e + f*x)] - 35 
*Cos[8*(e + f*x)] + 19600*Sin[e + f*x] + 3920*Sin[3*(e + f*x)] + 784*Sin[5 
*(e + f*x)] + 80*Sin[7*(e + f*x)]))/(35840*f)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3320, 3042, 3219, 3042, 3219, 3042, 3219, 3042, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {3}{4} c \int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{5/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} c \int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{5/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {3}{4} c \left (\frac {4}{7} c \int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{3/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} c \left (\frac {4}{7} c \int (\sin (e+f x) a+a)^{9/2} (c-c \sin (e+f x))^{3/2}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {3}{4} c \left (\frac {4}{7} c \left (\frac {1}{3} c \int (\sin (e+f x) a+a)^{9/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt {c-c \sin (e+f x)}}{6 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{4} c \left (\frac {4}{7} c \left (\frac {1}{3} c \int (\sin (e+f x) a+a)^{9/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt {c-c \sin (e+f x)}}{6 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {\frac {3}{4} c \left (\frac {4}{7} c \left (\frac {c^2 \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{15 f \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} \sqrt {c-c \sin (e+f x)}}{6 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{3/2}}{7 f}\right )+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{9/2} (c-c \sin (e+f x))^{5/2}}{8 f}}{a c}\)

Input:

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(5/2),x 
]
 

Output:

((c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*(c - c*Sin[e + f*x])^(5/2))/(8 
*f) + (3*c*((c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*(c - c*Sin[e + f*x] 
)^(3/2))/(7*f) + (4*c*((c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(15*f 
*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2)*Sq 
rt[c - c*Sin[e + f*x]])/(6*f)))/7))/4)/(a*c)
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3217
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f 
_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ 
n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
 

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]

\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{\frac {7}{2}} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]

Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x)
 

Output:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {{\left (35 \, a^{3} c^{2} \cos \left (f x + e\right )^{8} - 35 \, a^{3} c^{2} - 8 \, {\left (5 \, a^{3} c^{2} \cos \left (f x + e\right )^{6} + 6 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} + 8 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} + 16 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x, al 
gorithm="fricas")
 

Output:

-1/280*(35*a^3*c^2*cos(f*x + e)^8 - 35*a^3*c^2 - 8*(5*a^3*c^2*cos(f*x + e) 
^6 + 6*a^3*c^2*cos(f*x + e)^4 + 8*a^3*c^2*cos(f*x + e)^2 + 16*a^3*c^2)*sin 
(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + 
 e))
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(7/2)*(c-c*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x, al 
gorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^(7/2)*(-c*sin(f*x + e) + c)^(5/2)*cos(f*x + 
 e)^2, x)
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.34 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {32 \, {\left (35 \, a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{16} - 160 \, a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} + 280 \, a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 224 \, a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 70 \, a^{3} c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x, al 
gorithm="giac")
 

Output:

32/35*(35*a^3*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/ 
2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^16 - 160*a^3*c^2*sgn(cos(-1 
/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi 
+ 1/2*f*x + 1/2*e)^14 + 280*a^3*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sg 
n(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 - 224* 
a^3*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/ 
2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 + 70*a^3*c^2*sgn(cos(-1/4*pi + 1/2 
*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 
 1/2*e)^8)*sqrt(a)*sqrt(c)/f
 

Mupad [B] (verification not implemented)

Time = 20.82 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.00 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {7\,a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {7\,a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}-\frac {a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^3\,c^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \] Input:

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(7/2)*(c - c*sin(e + f*x))^(5/2),x 
)
 

Output:

(exp(- e*8i - f*x*8i)*(c - c*sin(e + f*x))^(1/2)*((35*a^3*c^2*exp(e*8i + f 
*x*8i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (7*a^3*c^2*exp(e* 
8i + f*x*8i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) - (7*a^3* 
c^2*exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/(128*f 
) - (a^3*c^2*exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2 
))/(64*f) - (a^3*c^2*exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*(a + a*sin(e + f* 
x))^(1/2))/(512*f) + (7*a^3*c^2*exp(e*8i + f*x*8i)*sin(3*e + 3*f*x)*(a + a 
*sin(e + f*x))^(1/2))/(32*f) + (7*a^3*c^2*exp(e*8i + f*x*8i)*sin(5*e + 5*f 
*x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (a^3*c^2*exp(e*8i + f*x*8i)*sin( 
7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos(e + f*x))
 

Reduce [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{3} c^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{5}d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}d x -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{3}d x \right )-2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )d x +\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}d x \right ) \] Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(5/2),x)
 

Output:

sqrt(c)*sqrt(a)*a**3*c**2*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) 
 + 1)*cos(e + f*x)**2*sin(e + f*x)**5,x) + int(sqrt(sin(e + f*x) + 1)*sqrt 
( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x)**4,x) - 2*int(sqrt(sin( 
e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x)**3,x) 
 - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2* 
sin(e + f*x)**2,x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)* 
cos(e + f*x)**2*sin(e + f*x),x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e 
 + f*x) + 1)*cos(e + f*x)**2,x))