\(\int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [145]

Optimal result

Integrand size = 42, antiderivative size = 186 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {28 c^2 (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {42 c^2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {4 c (g \cos (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{5 f g (a+a \sin (e+f x))^{5/2}} \] Output:

28/5*c^2*(g*cos(f*x+e))^(5/2)/a/f/g/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e) 
)^(1/2)+42/5*c^2*g*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2 
*f*x+1/2*e),2^(1/2))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-4 
/5*c*(g*cos(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2)/f/g/(a+a*sin(f*x+e))^(5/2 
)
 

Mathematica [A] (verified)

Time = 4.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.97 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c \sqrt {\cos (e+f x)} (g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \left (42 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+8 \sqrt {\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )+2 \cos \left (\frac {3}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )+2 \sin \left (\frac {3}{2} (e+f x)\right )\right )\right )}{5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (1+\sin (e+f x)))^{5/2}} \] Input:

Integrate[((g*Cos[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e 
 + f*x])^(5/2),x]
 

Output:

(c*Sqrt[Cos[e + f*x]]*(g*Cos[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(42* 
EllipticE[(e + f*x)/2, 2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 8*Sqrt 
[Cos[e + f*x]]*(Cos[(e + f*x)/2] + 2*Cos[(3*(e + f*x))/2] - Sin[(e + f*x)/ 
2] + 2*Sin[(3*(e + f*x))/2])))/(5*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^ 
3*(a*(1 + Sin[e + f*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3329, 3042, 3329, 3042, 3321, 3042, 3121, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {4 c \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2}}-\frac {7 c \left (-\frac {4 c (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}-\frac {6 c g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{5 a}\)

Input:

Int[((g*Cos[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + f*x 
])^(5/2),x]
 

Output:

(-4*c*(g*Cos[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(5*f*g*(a + a*Sin[e 
 + f*x])^(5/2)) - (7*c*((-4*c*(g*Cos[e + f*x])^(5/2))/(f*g*(a + a*Sin[e + 
f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]) - (6*c*g*Sqrt[Cos[e + f*x]]*Sqrt[g*C 
os[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt 
[c - c*Sin[e + f*x]])))/(5*a)
 

Defintions of rubi rules used

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3321

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

rule 3329

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(848\) vs. \(2(162)=324\).

Time = 18.36 (sec) , antiderivative size = 849, normalized size of antiderivative = 4.56

method	result	size
default	\(\text {Expression too large to display}\)	\(849\)

Input:

int((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-1/5/f*g/(1+2^(1/2))/(2*2^(1/2)-3)*c/a^2*(-24+21*(1+2*cos(1/2*f*x+1/2*e)*( 
cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e)+1)*sin(1/2*f*x+1/2*e)+cos(1/2*f* 
x+1/2*e)^2+2*cos(1/2*f*x+1/2*e))*2^(1/2)*((2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1 
/2)+2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))^(1/2)*(-2*(2^(1/2)*cos 
(1/2*f*x+1/2*e)-2^(1/2)-2*cos(1/2*f*x+1/2*e)+1)/(cos(1/2*f*x+1/2*e)+1))^(1 
/2)*EllipticE((1+2^(1/2))*(csc(1/2*f*x+1/2*e)-cot(1/2*f*x+1/2*e)),-2*2^(1/ 
2)+3)+42*(-2+(1+2*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1 
/2*e)+1)*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e))*2^( 
1/2)-4*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e)+1)*si 
n(1/2*f*x+1/2*e)-2*cos(1/2*f*x+1/2*e)^2-4*cos(1/2*f*x+1/2*e))*((2^(1/2)*co 
s(1/2*f*x+1/2*e)-2^(1/2)+2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))^( 
1/2)*(-2*(2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2)-2*cos(1/2*f*x+1/2*e)+1)/(cos( 
1/2*f*x+1/2*e)+1))^(1/2)*EllipticF((1+2^(1/2))*(csc(1/2*f*x+1/2*e)-cot(1/2 
*f*x+1/2*e)),-2*2^(1/2)+3)+2*(12+(-21+50*cos(1/2*f*x+1/2*e)^2+8*cos(1/2*f* 
x+1/2*e))*sin(1/2*f*x+1/2*e)-20*cos(1/2*f*x+1/2*e)^5+64*cos(1/2*f*x+1/2*e) 
^4+62*cos(1/2*f*x+1/2*e)^3-64*cos(1/2*f*x+1/2*e)^2-30*cos(1/2*f*x+1/2*e))* 
2^(1/2)+2*(21-50*cos(1/2*f*x+1/2*e)^2-8*cos(1/2*f*x+1/2*e))*sin(1/2*f*x+1/ 
2*e)+40*cos(1/2*f*x+1/2*e)^5-128*cos(1/2*f*x+1/2*e)^4-124*cos(1/2*f*x+1/2* 
e)^3+128*cos(1/2*f*x+1/2*e)^2+60*cos(1/2*f*x+1/2*e))*(g*(-1+2*cos(1/2*f*x+ 
1/2*e)^2))^(1/2)*(-(2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)-1)*c)^(1/2)...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.11 \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (21 \, \sqrt {\frac {1}{2}} {\left (-i \, c g \cos \left (f x + e\right )^{2} + 2 i \, c g \sin \left (f x + e\right ) + 2 i \, c g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 21 \, \sqrt {\frac {1}{2}} {\left (i \, c g \cos \left (f x + e\right )^{2} - 2 i \, c g \sin \left (f x + e\right ) - 2 i \, c g\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 4 \, {\left (4 \, c g \sin \left (f x + e\right ) + 3 \, c g\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \sin \left (f x + e\right ) - 2 \, a^{3} f\right )}} \] Input:

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="fricas")
 

Output:

2/5*(21*sqrt(1/2)*(-I*c*g*cos(f*x + e)^2 + 2*I*c*g*sin(f*x + e) + 2*I*c*g) 
*sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e 
) + I*sin(f*x + e))) + 21*sqrt(1/2)*(I*c*g*cos(f*x + e)^2 - 2*I*c*g*sin(f* 
x + e) - 2*I*c*g)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(f*x + e) - I*sin(f*x + e))) - 4*(4*c*g*sin(f*x + e) + 3*c*g)*sqr 
t(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3 
*f*cos(f*x + e)^2 - 2*a^3*f*sin(f*x + e) - 2*a^3*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))** 
(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="maxima")
 

Output:

integrate((g*cos(f*x + e))^(3/2)*(-c*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + 
e) + a)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int(((g*cos(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x 
))^(5/2),x)
 

Output:

int(((g*cos(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(3/2))/(a + a*sin(e + f*x 
))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

int((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2),x)
 

Output:

(sqrt(g)*sqrt(c)*sqrt(a)*c*g*(2*sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x 
) + 1)*sqrt(cos(e + f*x))*sin(e + f*x) - 2*int((sqrt(sin(e + f*x) + 1)*sqr 
t( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x)**2)/(s 
in(e + f*x)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 1),x)*sin(e + f*x)** 
2*f - 4*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + 
 f*x))*cos(e + f*x)*sin(e + f*x)**2)/(sin(e + f*x)**4 + 2*sin(e + f*x)**3 
- 2*sin(e + f*x) - 1),x)*sin(e + f*x)*f - 2*int((sqrt(sin(e + f*x) + 1)*sq 
rt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x)**2)/( 
sin(e + f*x)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 1),x)*f + int((sqrt 
(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f* 
x)**4)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e + f*x)*sin(e + f*x)**3 - 2* 
cos(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*sin(e + f*x)**2*f + 2*int((sq 
rt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + 
f*x)**4)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e + f*x)*sin(e + f*x)**3 - 
2*cos(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*sin(e + f*x)*f + int((sqrt( 
sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x 
)**4)/(cos(e + f*x)*sin(e + f*x)**4 + 2*cos(e + f*x)*sin(e + f*x)**3 - 2*c 
os(e + f*x)*sin(e + f*x) - cos(e + f*x)),x)*f - int((sqrt(sin(e + f*x) + 1 
)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2)/(cos(e + f 
*x)*sin(e + f*x)**4 + 2*cos(e + f*x)*sin(e + f*x)**3 - 2*cos(e + f*x)*s...