\(\int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) [1430]

Optimal result

Integrand size = 37, antiderivative size = 320 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} b^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 \sqrt {-a^2+b^2} d^{3/2} f \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {g \cos (e+f x)}}{a d f g \sqrt {d \sin (e+f x)}}-\frac {b \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \] Output:

-2*2^(1/2)*b^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^(3/2) 
/f/(g*cos(f*x+e))^(1/2)+2*2^(1/2)*b^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^(3/2)/f/(g*cos(f*x+e))^(1/2)-2*(g*cos(f*x+e))^(1/2)/a/d/ 
f/g/(d*sin(f*x+e))^(1/2)-b*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 = 18.27 (sec) , antiderivative size = 715, normalized size of antiderivative = 2.23 \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \cos (e+f x) \sin (e+f x)}{a f \sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}+\frac {2 b \sqrt {\cos (e+f x)} \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sin ^2(e+f x)}{a f \sqrt {g \cos (e+f x)} \sqrt [4]{1-\cos ^2(e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \] Input:

Integrate[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f* 
x])),x]
 

Output:

(-2*Cos[e + f*x]*Sin[e + f*x])/(a*f*Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^ 
(3/2)) + (2*b*Sqrt[Cos[e + f*x]]*(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)*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[Co 
s[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)))*Sin[e + 
 f*x]^2)/(a*f*Sqrt[g*Cos[e + f*x]]*(1 - Cos[e + f*x]^2)^(1/4)*(d*Sin[e + f 
*x])^(3/2)*(a + b*Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 3389, 3042, 3043, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3389

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3043

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

\(\Big \downarrow \) 3389

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3387

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3386

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

\(\Big \downarrow \) 1542

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

Input:

Int[1/(Sqrt[g*Cos[e + f*x]]*(d*Sin[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x 
]
 

Output:

(-2*Sqrt[g*Cos[e + f*x]])/(a*d*f*g*Sqrt[d*Sin[e + f*x]]) - (b*(-((b*Sqrt[C 
os[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*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]])))/(a*d)
 

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 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. \(1023\) vs. \(2(274)=548\).

Time = 3.38 (sec) , antiderivative size = 1024, normalized size of antiderivative = 3.20

method	result	size
default	\(\text {Expression too large to display}\)	\(1024\)

Input:

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

Output:

1/f*(2*(1+cos(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)*(-a^2+b^2)^(1/2)*a*b*Ellip 
ticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))-4*cos(f*x+e)*(-a^2+b^2)^ 
(1/2)*a*b+(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)*(-1-cos(f*x+e))*a*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^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)*(1+cos(f*x+e))*a*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^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)*( 
1+cos(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))*(-a^2+b^2)^(1/2)*b^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)*(1+co 
s(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))*b^3+(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)*(1+cos(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))*(-a 
^2+b^2)^(1/2)*b^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)*(-1-cos(f*x+e))*EllipticPi((cs 
c(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...
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\left (d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}} \sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {1}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {1}{\sqrt {g\,\cos \left (e+f\,x\right )}\,{\left (d\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g}\, \sqrt {d}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\cos \left (f x +e \right )}}{\cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a}d x \right )}{d^{2} g} \] Input:

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