\(\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx\) [390]

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

Optimal result

Integrand size = 30, antiderivative size = 188 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{56 c f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{56 c^2 f (c-c \sin (e+f x))^{13/2}}-\frac {a^4 \cos (e+f x)}{280 c^3 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{11/2}} \] Output:

1/8*a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(17/2)-3/56*a^2 
*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(15/2)+1/56*a^3*co 
s(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c^2/f/(c-c*sin(f*x+e))^(13/2)-1/280*a^4*co 
s(f*x+e)/c^3/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(11/2)
 

Mathematica [A] (verified)

Time = 13.20 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (40-28 \cos (2 (e+f x))+65 \sin (e+f x)-7 \sin (3 (e+f x)))}{140 c^8 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^8 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(17/2),x]
 

Output:

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(40 
- 28*Cos[2*(e + f*x)] + 65*Sin[e + f*x] - 7*Sin[3*(e + f*x)]))/(140*c^8*f* 
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^8*Sqrt[c - c*Sin 
[e + f*x]])
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3218, 3042, 3218, 3042, 3218, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{17/2}}dx\)

\(\Big \downarrow \) 3218

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{15/2}}dx}{8 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{15/2}}dx}{8 c}\)

\(\Big \downarrow \) 3218

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \left (\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{7 f (c-c \sin (e+f x))^{15/2}}-\frac {2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{13/2}}dx}{7 c}\right )}{8 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \left (\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{7 f (c-c \sin (e+f x))^{15/2}}-\frac {2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{13/2}}dx}{7 c}\right )}{8 c}\)

\(\Big \downarrow \) 3218

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \left (\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{7 f (c-c \sin (e+f x))^{15/2}}-\frac {2 a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{11/2}}dx}{6 c}\right )}{7 c}\right )}{8 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \left (\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{7 f (c-c \sin (e+f x))^{15/2}}-\frac {2 a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{11/2}}dx}{6 c}\right )}{7 c}\right )}{8 c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{17/2}}-\frac {3 a \left (\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{7 f (c-c \sin (e+f x))^{15/2}}-\frac {2 a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{6 f (c-c \sin (e+f x))^{13/2}}-\frac {a^2 \cos (e+f x)}{30 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{11/2}}\right )}{7 c}\right )}{8 c}\)

Input:

Int[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(17/2),x]
 

Output:

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(17/ 
2)) - (3*a*((a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(7*f*(c - c*Sin[e 
+ f*x])^(15/2)) - (2*a*((a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(6*f*(c 
- c*Sin[e + f*x])^(13/2)) - (a^2*Cos[e + f*x])/(30*c*f*Sqrt[a + a*Sin[e + 
f*x]]*(c - c*Sin[e + f*x])^(11/2))))/(7*c)))/(8*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 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])
 
Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.76

method result size
default \(\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{7} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{8} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{8}+4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+10 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+20 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+35\right ) a^{3} \sqrt {4}}{17920 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{8}}\) \(143\)

Input:

int((a+sin(f*x+e)*a)^(7/2)/(c-c*sin(f*x+e))^(17/2),x,method=_RETURNVERBOSE 
)
 

Output:

1/17920/f*tan(1/4*Pi+1/2*f*x+1/2*e)^7*sec(1/4*Pi+1/2*f*x+1/2*e)^8*(a*sin(1 
/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(cos(1/4*Pi+1/2*f*x+1/2*e)^8+4*cos(1/4*Pi+1/ 
2*f*x+1/2*e)^6+10*cos(1/4*Pi+1/2*f*x+1/2*e)^4+20*cos(1/4*Pi+1/2*f*x+1/2*e) 
^2+35)*a^3/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c^8*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.09 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=-\frac {{\left (14 \, a^{3} \cos \left (f x + e\right )^{2} - 17 \, a^{3} + {\left (7 \, a^{3} \cos \left (f x + e\right )^{2} - 18 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{35 \, {\left (c^{9} f \cos \left (f x + e\right )^{9} - 32 \, c^{9} f \cos \left (f x + e\right )^{7} + 160 \, c^{9} f \cos \left (f x + e\right )^{5} - 256 \, c^{9} f \cos \left (f x + e\right )^{3} + 128 \, c^{9} f \cos \left (f x + e\right ) + 8 \, {\left (c^{9} f \cos \left (f x + e\right )^{7} - 10 \, c^{9} f \cos \left (f x + e\right )^{5} + 24 \, c^{9} f \cos \left (f x + e\right )^{3} - 16 \, c^{9} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(17/2),x, algorithm="fri 
cas")
 

Output:

-1/35*(14*a^3*cos(f*x + e)^2 - 17*a^3 + (7*a^3*cos(f*x + e)^2 - 18*a^3)*si 
n(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^9*f*cos( 
f*x + e)^9 - 32*c^9*f*cos(f*x + e)^7 + 160*c^9*f*cos(f*x + e)^5 - 256*c^9* 
f*cos(f*x + e)^3 + 128*c^9*f*cos(f*x + e) + 8*(c^9*f*cos(f*x + e)^7 - 10*c 
^9*f*cos(f*x + e)^5 + 24*c^9*f*cos(f*x + e)^3 - 16*c^9*f*cos(f*x + e))*sin 
(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(17/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(17/2),x, algorithm="max 
ima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.82 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\frac {{\left (56 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 140 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 120 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 35 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{8960 \, c^{\frac {17}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{16}} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(17/2),x, algorithm="gia 
c")
 

Output:

1/8960*(56*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)^6 - 140*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 + 120*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 - 35*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/ 
(c^(17/2)*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/ 
2*e)^16)
 

Mupad [B] (verification not implemented)

Time = 23.15 (sec) , antiderivative size = 673, normalized size of antiderivative = 3.58 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\text {Too large to display} \] Input:

int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(17/2),x)
 

Output:

-((c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)* 
((a^3*exp(e*6i + f*x*6i)*(a + a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + 
 f*x*1i)*1i)/2))^(1/2)*64i)/(5*c^9*f) + (256*a^3*exp(e*7i + f*x*7i)*(a + a 
*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/(5*c^9* 
f) - (a^3*exp(e*8i + f*x*8i)*(a + a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e* 
1i + f*x*1i)*1i)/2))^(1/2)*832i)/(7*c^9*f) - (1024*a^3*exp(e*9i + f*x*9i)* 
(a + a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2))/( 
7*c^9*f) + (a^3*exp(e*10i + f*x*10i)*(a + a*((exp(- e*1i - f*x*1i)*1i)/2 - 
 (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*832i)/(7*c^9*f) + (256*a^3*exp(e*11i + 
f*x*11i)*(a + a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)) 
^(1/2))/(5*c^9*f) - (a^3*exp(e*12i + f*x*12i)*(a + a*((exp(- e*1i - f*x*1i 
)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*64i)/(5*c^9*f)))/(exp(e*1i + f 
*x*1i)*16i - 119*exp(e*2i + f*x*2i) - exp(e*3i + f*x*3i)*544i + 1700*exp(e 
*4i + f*x*4i) + exp(e*5i + f*x*5i)*3808i - 6188*exp(e*6i + f*x*6i) - exp(e 
*7i + f*x*7i)*7072i + 4862*exp(e*8i + f*x*8i) + 4862*exp(e*10i + f*x*10i) 
+ exp(e*11i + f*x*11i)*7072i - 6188*exp(e*12i + f*x*12i) - exp(e*13i + f*x 
*13i)*3808i + 1700*exp(e*14i + f*x*14i) + exp(e*15i + f*x*15i)*544i - 119* 
exp(e*16i + f*x*16i) - exp(e*17i + f*x*17i)*16i + exp(e*18i + f*x*18i) + 1 
)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{17/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{3} \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{9}-9 \sin \left (f x +e \right )^{8}+36 \sin \left (f x +e \right )^{7}-84 \sin \left (f x +e \right )^{6}+126 \sin \left (f x +e \right )^{5}-126 \sin \left (f x +e \right )^{4}+84 \sin \left (f x +e \right )^{3}-36 \sin \left (f x +e \right )^{2}+9 \sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{9}-9 \sin \left (f x +e \right )^{8}+36 \sin \left (f x +e \right )^{7}-84 \sin \left (f x +e \right )^{6}+126 \sin \left (f x +e \right )^{5}-126 \sin \left (f x +e \right )^{4}+84 \sin \left (f x +e \right )^{3}-36 \sin \left (f x +e \right )^{2}+9 \sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{9}-9 \sin \left (f x +e \right )^{8}+36 \sin \left (f x +e \right )^{7}-84 \sin \left (f x +e \right )^{6}+126 \sin \left (f x +e \right )^{5}-126 \sin \left (f x +e \right )^{4}+84 \sin \left (f x +e \right )^{3}-36 \sin \left (f x +e \right )^{2}+9 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{9}-9 \sin \left (f x +e \right )^{8}+36 \sin \left (f x +e \right )^{7}-84 \sin \left (f x +e \right )^{6}+126 \sin \left (f x +e \right )^{5}-126 \sin \left (f x +e \right )^{4}+84 \sin \left (f x +e \right )^{3}-36 \sin \left (f x +e \right )^{2}+9 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{9}} \] Input:

int((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(17/2),x)
 

Output:

(sqrt(c)*sqrt(a)*a**3*( - int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) 
 + 1)*sin(e + f*x)**3)/(sin(e + f*x)**9 - 9*sin(e + f*x)**8 + 36*sin(e + f 
*x)**7 - 84*sin(e + f*x)**6 + 126*sin(e + f*x)**5 - 126*sin(e + f*x)**4 + 
84*sin(e + f*x)**3 - 36*sin(e + f*x)**2 + 9*sin(e + f*x) - 1),x) - 3*int(( 
sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + 
 f*x)**9 - 9*sin(e + f*x)**8 + 36*sin(e + f*x)**7 - 84*sin(e + f*x)**6 + 1 
26*sin(e + f*x)**5 - 126*sin(e + f*x)**4 + 84*sin(e + f*x)**3 - 36*sin(e + 
 f*x)**2 + 9*sin(e + f*x) - 1),x) - 3*int((sqrt(sin(e + f*x) + 1)*sqrt( - 
sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**9 - 9*sin(e + f*x)**8 + 36* 
sin(e + f*x)**7 - 84*sin(e + f*x)**6 + 126*sin(e + f*x)**5 - 126*sin(e + f 
*x)**4 + 84*sin(e + f*x)**3 - 36*sin(e + f*x)**2 + 9*sin(e + f*x) - 1),x) 
- int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**9 
- 9*sin(e + f*x)**8 + 36*sin(e + f*x)**7 - 84*sin(e + f*x)**6 + 126*sin(e 
+ f*x)**5 - 126*sin(e + f*x)**4 + 84*sin(e + f*x)**3 - 36*sin(e + f*x)**2 
+ 9*sin(e + f*x) - 1),x)))/c**9