\(\int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [94]

Optimal result

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

4/5*a*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/ 
2)-6/5*a*(g*cos(f*x+e))^(5/2)/c/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e) 
)^(3/2)+6/5*a*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))/c^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 = 4.39 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.26 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {4 i g \sqrt {e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) g} \left (\left (5+4 i e^{i (e+f x)}-3 e^{2 i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}}+e^{i (e+f x)} \left (-i+e^{i (e+f x)}\right )^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (e+f x)}\right )\right ) \sqrt {a (1+\sin (e+f x))}}{5 c \left (i c e^{-i (e+f x)} \left (-i+e^{i (e+f x)}\right )^2\right )^{3/2} \left (i+e^{i (e+f x)}\right ) \sqrt {1+e^{2 i (e+f x)}} f} \] Input:

Integrate[((g*Cos[e + f*x])^(3/2)*Sqrt[a + a*Sin[e + f*x]])/(c - c*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[a*(1 + Sin[e + f*x])])/(c*((I*c*(-I + E^(I*( 
e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*(I + E^(I*(e + f*x)))*Sqrt[1 + E^((2* 
I)*(e + f*x))]*f)
 

Rubi [A] (verified)

Time = 1.42 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3329

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3331

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3321

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

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

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. \(1462\) vs. \(2(158)=316\).

Time = 29.06 (sec) , antiderivative size = 1463, normalized size of antiderivative = 8.04

method	result	size
default	\(\text {Expression too large to display}\)	\(1463\)

Input:

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

Output:

g/c^2*(-2/21/f/(1+2^(1/2))*((cos(1/2*f*x+1/2*e)+1)*((2*cos(1/2*f*x+1/2*e)^ 
2+1)*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*(2*cos(1/2*f*x+1/2*e)^2-3))*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)*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*cos(1/2*f*x+1/2*e)^2+5)*si 
n(1/2*f*x+1/2*e)-2*cos(1/2*f*x+1/2*e)^3+7*cos(1/2*f*x+1/2*e))*2^(1/2)+(2*c 
os(1/2*f*x+1/2*e)^2+5)*sin(1/2*f*x+1/2*e)-2*cos(1/2*f*x+1/2*e)^3+7*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)*a)^(1/2)/(-(2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)- 
1)*c)^(1/2)/(4*cos(1/2*f*x+1/2*e)^5-4*cos(1/2*f*x+1/2*e)^4*sin(1/2*f*x+1/2 
*e)-4*cos(1/2*f*x+1/2*e)^3+4*cos(1/2*f*x+1/2*e)^2*sin(1/2*f*x+1/2*e)+cos(1 
/2*f*x+1/2*e)-sin(1/2*f*x+1/2*e))+1/105/f/(1+2^(1/2))/(2*2^(1/2)-3)*(126+6 
3*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e)+1)*((2*cos(1/2*f*x+1/2*e)^2+1 
)*sin(1/2*f*x+1/2*e)+cos(1/2*f*x+1/2*e)*(2*cos(1/2*f*x+1/2*e)^2-3))*2^(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)*(-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^(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)+4*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2 
*f*x+1/2*e)+1)*(24*((2*cos(1/2*f*x+1/2*e)^2+1)*sin(1/2*f*x+1/2*e)+cos(1...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.09 \[ \int \frac {(g \cos (e+f x))^{3/2} \sqrt {a+a \sin (e+f x)}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (i \, g \cos \left (f x + e\right )^{2} + 2 i \, g \sin \left (f x + e\right ) - 2 i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} \sqrt {a c g} {\left (-i \, g \cos \left (f x + e\right )^{2} - 2 i \, g \sin \left (f x + e\right ) + 2 i \, g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (3 \, g \sin \left (f x + e\right ) - g\right )}\right )}}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} + 2 \, c^{3} f \sin \left (f x + e\right ) - 2 \, c^{3} f\right )}} \] Input:

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/(c-c*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)*s 
qrt(-c*sin(f*x + e) + c)*(3*g*sin(f*x + e) - g))/(c^3*f*cos(f*x + e)^2 + 2 
*c^3*f*sin(f*x + e) - 2*c^3*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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