\(\int \frac {1}{(g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\) [1435]

Optimal result

Integrand size = 37, antiderivative size = 380 \[ \int \frac {1}{(g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {2} b^2 \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} b^2 \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}+\frac {2 a \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) d f g \sqrt {g \cos (e+f x)}}-\frac {2 b (d \sin (e+f x))^{3/2}}{\left (a^2-b^2\right ) d^2 f g \sqrt {g \cos (e+f x)}}+\frac {2 b \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) d f g^2 \sqrt {\sin (2 e+2 f x)}} \] Output:

2*2^(1/2)*b^2*EllipticPi((g*cos(f*x+e))^(1/2)/g^(1/2)/(1+sin(f*x+e))^(1/2) 
,-(-a+b)^(1/2)/(a+b)^(1/2),I)*sin(f*x+e)^(1/2)/(-a+b)^(3/2)/(a+b)^(3/2)/f/ 
g^(3/2)/(d*sin(f*x+e))^(1/2)-2*2^(1/2)*b^2*EllipticPi((g*cos(f*x+e))^(1/2) 
/g^(1/2)/(1+sin(f*x+e))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*sin(f*x+e)^(1/2) 
/(-a+b)^(3/2)/(a+b)^(3/2)/f/g^(3/2)/(d*sin(f*x+e))^(1/2)+2*a*(d*sin(f*x+e) 
)^(1/2)/(a^2-b^2)/d/f/g/(g*cos(f*x+e))^(1/2)-2*b*(d*sin(f*x+e))^(3/2)/(a^2 
-b^2)/d^2/f/g/(g*cos(f*x+e))^(1/2)-2*b*(g*cos(f*x+e))^(1/2)*EllipticE(cos( 
e+1/4*Pi+f*x),2^(1/2))*(d*sin(f*x+e))^(1/2)/(a^2-b^2)/d/f/g^2/sin(2*f*x+2* 
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 = 26.17 (sec) , antiderivative size = 1279, normalized size of antiderivative = 3.37 \[ \int \frac {1}{(g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*Cos[e + f*x]*Sin[e + f*x]*(a - b*Sin[e + f*x]))/((a^2 - b^2)*f*(g*Cos[e 
 + f*x])^(3/2)*Sqrt[d*Sin[e + f*x]]) + (b*Cos[e + f*x]^(3/2)*Sqrt[Sin[e + 
f*x]]*((4*a*(-(b*AppellF1[3/4, -1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + 
f*x]^2)/(-a^2 + b^2)]) + a*AppellF1[3/4, 1/4, 1, 7/4, Cos[e + f*x]^2, (b^2 
*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^(3/2)*(a + b*Sqrt[1 - Cos[e + 
 f*x]^2])*Sin[e + f*x]^(3/2))/(3*(a^2 - b^2)*(1 - Cos[e + f*x]^2)^(3/4)*(a 
 + b*Sin[e + f*x])) + (Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]]*(b*Tan[e + f*x] 
 + a*Sqrt[1 + Tan[e + f*x]^2])*(56*b*(-3*a^2 + b^2)*AppellF1[3/4, 1/2, 1, 
7/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(3/2) + 
24*b*(-a^2 + b^2)*AppellF1[7/4, 1/2, 1, 11/4, -Tan[e + f*x]^2, (-1 + b^2/a 
^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(7/2) + 21*a^(3/2)*(4*Sqrt[2]*a^(3/2)*Arc 
Tan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 4*Sqrt[2]*a^(3/2)*ArcTan[1 + Sqrt[2] 
*Sqrt[Tan[e + f*x]]] - (4*Sqrt[2]*a^2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4 
)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) + (2*Sqrt[2]*b^2*ArcTan[ 
1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^( 
1/4) + (4*Sqrt[2]*a^2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f 
*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) - (2*Sqrt[2]*b^2*ArcTan[1 + (Sqrt[2]*(a^ 
2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]])/(a^2 - b^2)^(1/4) + 2*Sqrt[2] 
*a^(3/2)*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 2*Sqrt[2]*a^ 
(3/2)*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - (2*Sqrt[2]*a...
 

Rubi [A] (verified)

Time = 2.20 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.92, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {3042, 3383, 3042, 3317, 3042, 3043, 3051, 3042, 3052, 3042, 3119, 3385, 3042, 3384, 993, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3383

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3317

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3043

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {2 a \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {b \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}}{a^2-b^2}-\frac {b^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3385

\(\displaystyle \frac {\frac {2 a \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {b \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}}{a^2-b^2}-\frac {b^2 \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {b \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}}{a^2-b^2}-\frac {b^2 \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}\)

\(\Big \downarrow \) 3384

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

\(\Big \downarrow \) 993

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

\(\Big \downarrow \) 1542

\(\displaystyle \frac {4 \sqrt {2} b^2 \sqrt {\sin (e+f x)} \left (\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}-\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}\right )}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}+\frac {\frac {2 a \sqrt {d \sin (e+f x)}}{d f g \sqrt {g \cos (e+f x)}}-\frac {b \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{d}}{a^2-b^2}\)

Input:

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

Output:

(4*Sqrt[2]*b^2*(-1/2*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqrt[g 
*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(Sqrt[-a + b]*Sqrt[a 
 + b]*Sqrt[g]) + EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Cos[e 
+ f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(2*Sqrt[-a + b]*Sqrt[a + b] 
*Sqrt[g]))*Sqrt[Sin[e + f*x]])/((a^2 - b^2)*f*g*Sqrt[d*Sin[e + f*x]]) + (( 
2*a*Sqrt[d*Sin[e + f*x]])/(d*f*g*Sqrt[g*Cos[e + f*x]]) - (b*((2*(d*Sin[e + 
 f*x])^(3/2))/(d*f*g*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Cos[e + f*x]]*Ellip 
ticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(f*g^2*Sqrt[Sin[2*e + 2*f*x] 
])))/d)/(a^2 - b^2)
 

Defintions of rubi rules used

rule 993

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> 
With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* 
b)   Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b)   Int[1/((r 
 - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
a*d, 0]
 

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 3051

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

rule 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, 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 3317

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

rule 3383

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

rule 3384

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_ 
) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[-4*Sqrt[2]*(g/f)   S 
ubst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sqrt[g 
*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 3385

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]] 
*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Sqrt[Sin[e + f* 
x]]/Sqrt[d*Sin[e + f*x]]   Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[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. \(1147\) vs. \(2(323)=646\).

Time = 3.81 (sec) , antiderivative size = 1148, normalized size of antiderivative = 3.02

method	result	size
default	\(\text {Expression too large to display}\)	\(1148\)

Input:

int(1/(g*cos(f*x+e))^(3/2)/(d*sin(f*x+e))^(1/2)/(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*co 
t(f*x+e)+2)^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(-a^2+b^2)^(1/2)*a*b*Elli 
pticF((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1/2))-4*(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*b*EllipticE((csc(f*x+e)-cot(f*x+e)+1)^(1/2),1/2*2^(1 
/2))*(-a^2+b^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((c 
sc(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((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...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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