\(\int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [146]

Optimal result

Integrand size = 42, antiderivative size = 182 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {4 c (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}+\frac {6 c (g \cos (e+f x))^{5/2}}{5 a f g (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {6 c g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

-4/5*c*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(1 
/2)+6/5*c*(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)+6/5*c*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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.69 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {4 i g \sqrt {e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) g} \left (\left (5-4 i e^{i (e+f x)}-3 e^{2 i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}}+e^{i (e+f x)} \left (i+e^{i (e+f x)}\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {c-c \sin (e+f x)}}{5 a \left (-i+e^{i (e+f x)}\right ) \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^{3/2} \sqrt {1+e^{2 i (e+f x)}} f} \] Input:

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

Output:

(((4*I)/5)*g*Sqrt[((1 + E^((2*I)*(e + f*x)))*g)/E^(I*(e + f*x))]*((5 - (4* 
I)*E^(I*(e + f*x)) - 3*E^((2*I)*(e + f*x)))*Sqrt[1 + E^((2*I)*(e + f*x))] 
+ E^(I*(e + f*x))*(I + E^(I*(e + f*x)))^3*Hypergeometric2F1[1/2, 3/4, 7/4, 
 -E^((2*I)*(e + f*x))])*Sqrt[c - c*Sin[e + f*x]])/(a*(-I + E^(I*(e + f*x)) 
)*(((-I)*a*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*Sqrt[1 + E^((2* 
I)*(e + f*x))]*f)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, 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, 3331, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3331

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(-4*c*(g*Cos[e + f*x])^(5/2))/(5*f*g*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c 
*Sin[e + f*x]]) - (3*c*((-2*(g*Cos[e + f*x])^(5/2))/(f*g*(a + a*Sin[e + f* 
x])^(3/2)*Sqrt[c - c*Sin[e + f*x]]) - (2*g*Sqrt[Cos[e + f*x]]*Sqrt[g*Cos[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]
 

rule 3331

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(821\) vs. \(2(158)=316\).

Time = 13.50 (sec) , antiderivative size = 822, normalized size of antiderivative = 4.52

method	result	size
default	\(\text {Expression too large to display}\)	\(822\)

Input:

int((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/2),x,m 
ethod=_RETURNVERBOSE)
 

Output:

-1/5/f*g/(2*2^(1/2)-3)/(1+2^(1/2))/a^2*(-2+3*(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 
)+6*(-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)*sin(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)*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)*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*(1+(-3+10*cos(1/2*f*x+1/2*e)^2+4*cos(1/2*f*x+1/2*e 
))*sin(1/2*f*x+1/2*e)+12*cos(1/2*f*x+1/2*e)^4+6*cos(1/2*f*x+1/2*e)^3-12*co 
s(1/2*f*x+1/2*e)^2-5*cos(1/2*f*x+1/2*e))*2^(1/2)+2*(3-10*cos(1/2*f*x+1/2*e 
)^2-4*cos(1/2*f*x+1/2*e))*sin(1/2*f*x+1/2*e)-24*cos(1/2*f*x+1/2*e)^4-12*co 
s(1/2*f*x+1/2*e)^3+24*cos(1/2*f*x+1/2*e)^2+10*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)/((2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)+1)*a)^(1/2)/(4*cos...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (-i \, g \cos \left (f x + e\right )^{2} + 2 i \, g \sin \left (f x + e\right ) + 2 i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (i \, g \cos \left (f x + e\right )^{2} - 2 i \, g \sin \left (f x + e\right ) - 2 i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (3 \, g \sin \left (f x + e\right ) + g\right )}\right )}}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \sin \left (f x + e\right ) - 2 \, a^{3} f\right )}} \] Input:

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="fricas")
 

Output:

2/5*(3*sqrt(1/2)*sqrt(a*c*g)*(-I*g*cos(f*x + e)^2 + 2*I*g*sin(f*x + e) + 2 
*I*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*s 
in(f*x + e))) + 3*sqrt(1/2)*sqrt(a*c*g)*(I*g*cos(f*x + e)^2 - 2*I*g*sin(f* 
x + e) - 2*I*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x 
+ e) - I*sin(f*x + e))) - sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq 
rt(-c*sin(f*x + e) + c)*(3*g*sin(f*x + e) + g))/(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. \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))** 
(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="maxima")
 

Output:

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

Giac [F(-1)]

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

integrate((g*cos(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(5/ 
2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(g)*sqrt(c)*sqrt(a)*g*(2*sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) 
+ 1)*sqrt(cos(e + f*x))*sin(e + f*x) - 2*sqrt(sin(e + f*x) + 1)*sqrt( - si 
n(e + f*x) + 1)*sqrt(cos(e + f*x)) + int((sqrt(sin(e + f*x) + 1)*sqrt( - s 
in(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2)/(cos(e + f*x)*sin(e + 
 f*x)**2 + 2*cos(e + f*x)*sin(e + f*x) + cos(e + f*x)),x)*sin(e + f*x)**2* 
f + 2*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)**2 + 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)*sq 
rt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2)/(cos(e + f*x)* 
sin(e + f*x)**2 + 2*cos(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))/(cos(e + f*x)*sin(e + f*x)**2 + 2*cos(e + f*x)*sin(e + f*x) + cos( 
e + f*x)),x)*sin(e + f*x)**2*f - 2*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin 
(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x))/(cos(e + f*x)*sin(e + f*x) 
**2 + 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))/(cos(e + f*x)*sin(e + f*x)**2 + 2*cos(e + f*x)*sin(e + f*x) + cos 
(e + f*x)),x)*f))/(6*a**3*f*(sin(e + f*x)**2 + 2*sin(e + f*x) + 1))