\(\int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1415]

Optimal result

Integrand size = 37, antiderivative size = 321 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}-\frac {b g^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Output:

-2*2^(1/2)*(-a^2+b^2)^(1/2)*g^2*cos(f*x+e)^(1/2)*EllipticPi((d*sin(f*x+e)) 
^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b-(-a^2+b^2)^(1/2)),I)/a^2/d^(3/2) 
/f/(g*cos(f*x+e))^(1/2)+2*2^(1/2)*(-a^2+b^2)^(1/2)*g^2*cos(f*x+e)^(1/2)*El 
lipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b+(-a^2+b^2 
)^(1/2)),I)/a^2/d^(3/2)/f/(g*cos(f*x+e))^(1/2)-2*g*(g*cos(f*x+e))^(1/2)/a/ 
d/f/(d*sin(f*x+e))^(1/2)-b*g^2*InverseJacobiAM(e-1/4*Pi+f*x,2^(1/2))*sin(2 
*f*x+2*e)^(1/2)/a^2/d/f/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 21.77 (sec) , antiderivative size = 1095, normalized size of antiderivative = 3.41 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[(g*Cos[e + f*x])^(3/2)/((d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f* 
x])),x]
 

Output:

(-2*(g*Cos[e + f*x])^(3/2)*Tan[e + f*x])/(a*f*(d*Sin[e + f*x])^(3/2)) - (( 
g*Cos[e + f*x])^(3/2)*Sin[e + f*x]^(3/2)*((-2*b*(a + b*Sqrt[1 - Cos[e + f* 
x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*C 
os[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/((1 - Cos[e + f*x]^2)^(3/ 
4)*(5*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + 
f*x]^2)/(-a^2 + b^2)] + (-4*b^2*AppellF1[5/4, 3/4, 2, 9/4, Cos[e + f*x]^2, 
 (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + 3*(a^2 - b^2)*AppellF1[5/4, 7/4, 1, 
9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*( 
a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*b*(2*ArcTan[1 - ((1 + I)* 
Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4 
))] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4 
)*(-1 + Cos[e + f*x]^2)^(1/4))] + Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x] 
)/Sqrt[-1 + Cos[e + f*x]^2] - ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos 
[e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] + (I*a*Cos 
[e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] + ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4) 
*Sqrt[Cos[e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)]))/(Sqrt[a]*(-a^2 + b^2)^ 
(3/4)))*Sqrt[Sin[e + f*x]])/((1 - Cos[e + f*x]^2)^(1/4)*(a + b*Sin[e + f*x 
])) + (2*a*Sqrt[Sin[e + f*x]]*((Sqrt[a]*(-2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2 
)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + 2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^( 
1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + Log[-a + Sqrt[2]*Sqrt[a]*(a^2 - b^2...
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3378, 3042, 3043, 3053, 3042, 3120, 3387, 3042, 3386, 1542}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3378

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}dx}{a}\)

\(\Big \downarrow \) 3043

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{a^2 d \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{a^2 d \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3387

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3386

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} d \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \int \frac {1}{\left (\left (b-\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}+\frac {2 \sqrt {2} d \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \int \frac {1}{\left (\left (b+\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}\right )}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 1542

\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} \sqrt {d} \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (b-\sqrt {b^2-a^2}\right )}+\frac {2 \sqrt {2} \sqrt {d} \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (\sqrt {b^2-a^2}+b\right )}\right )}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\)

Input:

Int[(g*Cos[e + f*x])^(3/2)/((d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x 
]
 

Output:

-(((a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*((2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2]) 
*Sqrt[d]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x 
]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/((b - Sqrt[-a^2 + b^2])*f) + (2 
*Sqrt[2]*(1 + b/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(a/(b + Sqrt[-a^2 + 
b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1] 
)/((b + Sqrt[-a^2 + b^2])*f)))/(a^2*d^2*Sqrt[g*Cos[e + f*x]])) - (2*g*Sqrt 
[g*Cos[e + f*x]])/(a*d*f*Sqrt[d*Sin[e + f*x]]) - (b*g^2*EllipticF[e - Pi/4 
 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^2*d*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Si 
n[e + f*x]])
 

Defintions of rubi rules used

rule 1542

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ 
{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x 
], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
 

rule 3042

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

rule 3043

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( 
m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ 
(a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & 
& NeQ[m, -1]
 

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3120

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

rule 3378

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int[ 
(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] + (-Simp[b*(g^2/(a^2*d) 
)   Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] - Simp[g^ 
2*((a^2 - b^2)/(a^2*d^2))   Int[(g*Cos[e + f*x])^(p - 2)*((d*Sin[e + f*x])^ 
(n + 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g}, x] && N 
eQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && GtQ[p, 1] && (LeQ[n, -2] || (EqQ 
[n, -3/2] && EqQ[p, 3/2]))
 

rule 3386

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 
2]}, Simp[2*Sqrt[2]*d*((b + q)/(f*q))   Subst[Int[1/((d*(b + q) + a*x^2)*Sq 
rt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - 
 Simp[2*Sqrt[2]*d*((b - q)/(f*q))   Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 
 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x]] /; F 
reeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 3387

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1714\) vs. \(2(275)=550\).

Time = 3.31 (sec) , antiderivative size = 1715, normalized size of antiderivative = 5.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1715\)

Input:

int((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x,method=_R 
ETURNVERBOSE)
 

Output:

4/f*cot(f*x+e)*csc(f*x+e)/d/(d*sin(f*x+e))^(1/2)*(1-cos(f*x+e))*g*(g*cos(f 
*x+e))^(1/2)*(2*(-a^2+b^2)^(1/2)*EllipticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2) 
,1/2*2^(1/2))*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e) 
)^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*a*b-2*EllipticF((csc(f*x+e)-cot(f* 
x+e)+1)^(1/2),1/2*2^(1/2))*b^2*(-a^2+b^2)^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^ 
(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)- 
(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(csc(f 
*x+e)-cot(f*x+e)+1)^(1/2)*EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b 
+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^3+(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2) 
*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*EllipticPi 
((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^ 
2*b-(-a^2+b^2)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot 
(f*x+e))^(1/2)*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*EllipticPi((csc(f*x+e)-cot( 
f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^2+(csc(f*x+e)-cot 
(f*x+e)+1)^(1/2)*(-2*csc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x 
+e))^(1/2)*EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/ 
2)+a),1/2*2^(1/2))*a*b^2-EllipticPi((csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+ 
(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*b^3*(csc(f*x+e)-cot(f*x+e)+1)^(1/2)*(-2*c 
sc(f*x+e)+2*cot(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)+EllipticPi( 
(csc(f*x+e)-cot(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*...
 

Fricas [F(-1)]

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

integrate((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((g*cos(f*x+e))**(3/2)/(d*sin(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

Integral((g*cos(e + f*x))**(3/2)/((d*sin(e + f*x))**(3/2)*(a + b*sin(e + f 
*x))), x)
 

Maxima [F]

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

integrate((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="maxima")
 

Output:

integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(3 
/2)), x)
 

Giac [F]

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

integrate((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, al 
gorithm="giac")
 

Output:

integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)*(d*sin(f*x + e))^(3 
/2)), x)
 

Mupad [F(-1)]

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

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x 
)
                                                                                    
                                                                                    
 

Output:

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(3/2)*(a + b*sin(e + f*x))), 
x)
 

Reduce [F]

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

int((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)
 

Output:

(sqrt(g)*sqrt(d)*int((sqrt(sin(e + f*x))*sqrt(cos(e + f*x))*cos(e + f*x))/ 
(sin(e + f*x)**3*b + sin(e + f*x)**2*a),x)*g)/d**2