\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\) [109]

Optimal result

Integrand size = 38, antiderivative size = 159 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {32 (5 A-7 B) c^3 \cos (e+f x)}{15 a f \sqrt {c-c \sin (e+f x)}}-\frac {8 (5 A-7 B) c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{15 a f}-\frac {(5 A-7 B) c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{a c f} \] Output:

-32/15*(5*A-7*B)*c^3*cos(f*x+e)/a/f/(c-c*sin(f*x+e))^(1/2)-8/15*(5*A-7*B)* 
c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f-1/5*(5*A-7*B)*c*cos(f*x+e)*(c-c* 
sin(f*x+e))^(3/2)/a/f-(A-B)*sec(f*x+e)*(c-c*sin(f*x+e))^(7/2)/a/c/f
 

Mathematica [A] (verified)

Time = 12.56 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (450 A-600 B+2 (5 A-16 B) \cos (2 (e+f x))+25 (8 A-13 B) \sin (e+f x)+3 B \sin (3 (e+f x)))}{30 a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \] Input:

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

Output:

-1/30*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]* 
(450*A - 600*B + 2*(5*A - 16*B)*Cos[2*(e + f*x)] + 25*(8*A - 13*B)*Sin[e + 
 f*x] + 3*B*Sin[3*(e + f*x)]))/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])* 
(1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3446, 3042, 3334, 3042, 3126, 3042, 3126, 3042, 3125}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{2} c (5 A-7 B) \int (c-c \sin (e+f x))^{5/2}dx-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (5 A-7 B) \int (c-c \sin (e+f x))^{5/2}dx-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{f}}{a c}\)

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3126

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3125

\(\displaystyle \frac {-\frac {1}{2} c (5 A-7 B) \left (\frac {8}{5} c \left (\frac {8 c^2 \cos (e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )+\frac {2 c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}\right )-\frac {(A-B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{f}}{a c}\)

Input:

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

Output:

(-(((A - B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/f) - ((5*A - 7*B)*c*( 
(2*c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5*f) + (8*c*((8*c^2*Cos[e + 
 f*x])/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]*Sqrt[c - c*Sin[e 
 + f*x]])/(3*f)))/5))/2)/(a*c)
 

Defintions of rubi rules used

rule 3042

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

rule 3125

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos 
[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[a^2 - b^2, 0]
 

rule 3126

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

rule 3334

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

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin 
[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* 
d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-3 B \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-5 A +16 B \right ) \cos \left (f x +e \right )^{2}+\left (-50 A +82 B \right ) \sin \left (f x +e \right )-110 A +142 B \right )}{15 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(95\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x,method=_RET 
URNVERBOSE)
 

Output:

-2/15*c^3/a*(sin(f*x+e)-1)*(-3*B*cos(f*x+e)^2*sin(f*x+e)+(-5*A+16*B)*cos(f 
*x+e)^2+(-50*A+82*B)*sin(f*x+e)-110*A+142*B)/cos(f*x+e)/(c-c*sin(f*x+e))^( 
1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left ({\left (5 \, A - 16 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (55 \, A - 71 \, B\right )} c^{2} + {\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (25 \, A - 41 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, a f \cos \left (f x + e\right )} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algo 
rithm="fricas")
 

Output:

-2/15*((5*A - 16*B)*c^2*cos(f*x + e)^2 + 2*(55*A - 71*B)*c^2 + (3*B*c^2*co 
s(f*x + e)^2 + 2*(25*A - 41*B)*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c 
)/(a*f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (145) = 290\).

Time = 0.14 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.43 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (\frac {5 \, {\left (23 \, c^{\frac {5}{2}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {40 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {65 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {20 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {23 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} A}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} - \frac {2 \, {\left (79 \, c^{\frac {5}{2}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {205 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {170 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {205 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {79 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} B}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}}\right )}}{15 \, f} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algo 
rithm="maxima")
 

Output:

2/15*(5*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5 
/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 40*c^(5/2)*sin(f*x + e)^3/(cos(f 
*x + e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(5/ 
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f* 
x + e) + 1)^6)*A/((a + a*sin(f*x + e)/(cos(f*x + e) + 1))*(sin(f*x + e)^2/ 
(cos(f*x + e) + 1)^2 + 1)^(5/2)) - 2*(79*c^(5/2) + 79*c^(5/2)*sin(f*x + e) 
/(cos(f*x + e) + 1) + 205*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 17 
0*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 205*c^(5/2)*sin(f*x + e)^4 
/(cos(f*x + e) + 1)^4 + 79*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7 
9*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*B/((a + a*sin(f*x + e)/(cos 
(f*x + e) + 1))*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (145) = 290\).

Time = 0.36 (sec) , antiderivative size = 573, normalized size of antiderivative = 3.60 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algo 
rithm="giac")
 

Output:

8/15*sqrt(2)*sqrt(c)*(15*(A*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*c^ 
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) 
 - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)) - (25*A*c^2*sgn(sin(-1/4* 
pi + 1/2*f*x + 1/2*e)) - 41*B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 11 
0*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1 
/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 190*B*c^2*(cos(-1/4*pi + 1/2 
*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2* 
f*x + 1/2*e) + 1) + 160*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(s 
in(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 32 
0*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 
 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 90*A*c^2*(cos(-1/4*pi + 
1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 
 1/2*f*x + 1/2*e) + 1)^3 + 90*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3 
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^ 
3 + 15*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2* 
f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 15*B*c^2*(cos(-1/4* 
pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4 
*pi + 1/2*f*x + 1/2*e) + 1)^4)/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(c 
os(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^5))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*c**2*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) + 1),x)*a + int 
((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x) + 1),x)*b + int 
((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) + 1),x)*a - 2*i 
nt((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) + 1),x)*b - 2 
*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x)*a + in 
t((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x)*b))/a