\(\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^2} \, dx\) [1438]

Optimal result

Integrand size = 37, antiderivative size = 331 \[ \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^2} \, dx=\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}-\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}+\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{2 a^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Output:

2^(1/2)*b*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/(-a^2+b^2)^(1/2)/d^(1/2)/ 
f/(g*cos(f*x+e))^(1/2)-2^(1/2)*b*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/(- 
a^2+b^2)^(1/2)/d^(1/2)/f/(g*cos(f*x+e))^(1/2)+g*(g*cos(f*x+e))^(1/2)*(d*si 
n(f*x+e))^(1/2)/a/d/f/(a+b*sin(f*x+e))+1/2*g^2*InverseJacobiAM(e-1/4*Pi+f* 
x,2^(1/2))*sin(2*f*x+2*e)^(1/2)/a^2/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 = 18.27 (sec) , antiderivative size = 717, normalized size of antiderivative = 2.17 \[ \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^2} \, dx=-\frac {(g \cos (e+f x))^{3/2} \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sin (e+f x)}{a f \cos ^{\frac {3}{2}}(e+f x) \sqrt [4]{1-\cos ^2(e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}+\frac {(g \cos (e+f x))^{3/2} \tan (e+f x)}{a f \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \] Input:

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

Output:

-(((g*Cos[e + f*x])^(3/2)*(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*Cos[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)*Appe 
llF1[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 + C 
os[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)))*Sin[e + f*x])/ 
(a*f*Cos[e + f*x]^(3/2)*(1 - Cos[e + f*x]^2)^(1/4)*Sqrt[d*Sin[e + f*x]]*(a 
 + b*Sin[e + f*x]))) + ((g*Cos[e + f*x])^(3/2)*Tan[e + f*x])/(a*f*Sqrt[d*S 
in[e + f*x]]*(a + b*Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 3366, 3042, 3389, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3366

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3389

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3387

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3386

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

\(\Big \downarrow \) 1542

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

Input:

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

Output:

(g*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]])/(a*d*f*(a + b*Sin[e + f*x])) 
 + (g^2*(-((b*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]]/(S 
qrt[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*d*Sqrt[g*Cos[e + f*x]])) + (EllipticF[e - Pi/ 
4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e 
 + f*x]])))/(2*a)
 

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 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 3366

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-g)*(g* 
Cos[e + f*x])^(p - 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^(m + 1)/(a 
*d*f*(m + 1))), x] + Simp[g^2*((2*m + 3)/(2*a*(m + 1)))   Int[(g*Cos[e + f* 
x])^(p - 2)*((a + b*Sin[e + f*x])^(m + 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; 
FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && EqQ[m + 
p + 1/2, 0]
 

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]
 

rule 3389

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1683\) vs. \(2(283)=566\).

Time = 3.48 (sec) , antiderivative size = 1684, normalized size of antiderivative = 5.09

method	result	size
default	\(\text {Expression too large to display}\)	\(1684\)

Input:

int((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^2,x,method= 
_RETURNVERBOSE)
 

Output:

-1/2/f*(g*cos(f*x+e))^(1/2)*g/(a+b*sin(f*x+e))/(d*sin(f*x+e))^(1/2)*(b*(-a 
^2+b^2)^(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))*(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)*a*(sec(f*x+e)+1)+(-a^2+b^2 
)^(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))*(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)*b^2*(sin(f*x+e)+tan(f*x+e))+(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^2*b*(-sec(f*x+e)-1)+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))*(csc(f*x+e)-co 
t(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)*a*b^2*(-sin(f*x+e)-tan(f*x+e)+1+sec(f*x+e))+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))*(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)+co 
t(f*x+e))^(1/2)*b^3*(sin(f*x+e)+tan(f*x+e))+b*(-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)*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*(sec(f*x+e)+1)+(-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*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

integrate((g*cos(f*x+e))**(3/2)/(d*sin(f*x+e))**(1/2)/(a+b*sin(f*x+e))**2, 
x)
 

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^2) 
,x)
                                                                                    
                                                                                    
 

Output:

int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^2) 
, x)
 

Reduce [F]

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

int((g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^2,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**2 + 2*sin(e + f*x)**2*a*b + sin(e + f*x)*a**2),x)*g)/d