\(\int \frac {1}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx\) [521]

Optimal result

Integrand size = 27, antiderivative size = 257 \[ \int \frac {1}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {(c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d)^2 f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 (c-d) f (a+a \sin (e+f x))^2}-\frac {(c-3 d) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c-2 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 a^2 (c-d) f \sqrt {c+d \sin (e+f x)}} \] Output:

-1/3*(c-3*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a^2/(c-d)^2/f/(1+sin(f*x+e) 
)-1/3*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)/f/(a+a*sin(f*x+e))^2+1/3*(c- 
3*d)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin 
(f*x+e))^(1/2)/a^2/(c-d)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)+1/3*(c-2*d)*In 
verseJacobiAM(1/2*e-1/4*Pi+1/2*f*x,2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+ 
e))/(c+d))^(1/2)/a^2/(c-d)/f/(c+d*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 5.72 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-((c-3 d) (c+d \sin (e+f x)))-\frac {\left (2 d \cos \left (\frac {1}{2} (e+f x)\right )+(c-3 d) \cos \left (\frac {3}{2} (e+f x)\right )+(-3 c+7 d) \sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-2 d^2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+(c-3 d) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{3 a^2 (c-d)^2 f (1+\sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-((c - 3*d)*(c + d*Sin[e + f*x]) 
) - ((2*d*Cos[(e + f*x)/2] + (c - 3*d)*Cos[(3*(e + f*x))/2] + (-3*c + 7*d) 
*Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x) 
/2])^3 - 2*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d 
*Sin[e + f*x])/(c + d)] + (c - 3*d)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x) 
/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sq 
rt[(c + d*Sin[e + f*x])/(c + d)]))/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])^2 
*Sqrt[c + d*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3245, 27, 3042, 3457, 25, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3245

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (2 c-5 d)+a d \sin (e+f x)}{(\sin (e+f x) a+a) \sqrt {c+d \sin (e+f x)}}dx}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (2 c-5 d)+a d \sin (e+f x)}{(\sin (e+f x) a+a) \sqrt {c+d \sin (e+f x)}}dx}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {-\frac {\int -\frac {2 a^2 d^2-a^2 (c-3 d) d \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 a^2 d^2-a^2 (c-3 d) d \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2 d^2-a^2 (c-3 d) d \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {a^2 (c-2 d) (c-d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-a^2 (c-3 d) \int \sqrt {c+d \sin (e+f x)}dx}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (c-2 d) (c-d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-a^2 (c-3 d) \int \sqrt {c+d \sin (e+f x)}dx}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {a^2 (c-2 d) (c-d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (c-2 d) (c-d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {a^2 (c-2 d) (c-d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {2 a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {\frac {a^2 (c-2 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^2 (c-2 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {\frac {2 a^2 (c-2 d) (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{f \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 (c-3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2 (c-d)}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (c-d) (\sin (e+f x)+1)}}{6 a^2 (c-d)}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

Input:

Int[1/((a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]),x]
 

Output:

-1/3*(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((c - d)*f*(a + a*Sin[e + f*x 
])^2) + ((-2*(c - 3*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((c - d)*f*( 
1 + Sin[e + f*x])) + ((-2*a^2*(c - 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d 
)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[(c + d*Sin[e + f*x])/(c + d)] 
) + (2*a^2*(c - 2*d)*(c - d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]* 
Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(f*Sqrt[c + d*Sin[e + f*x]]))/(a^2*(c 
- d)))/(6*a^2*(c - d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3245

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(504\) vs. \(2(242)=484\).

Time = 0.67 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {\sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}\, \left (-\frac {\sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}{3 \left (c -d \right ) \left (1+\sin \left (f x +e \right )\right )^{2}}-\frac {\left (-\sin \left (f x +e \right )^{2} d -c \sin \left (f x +e \right )+d \sin \left (f x +e \right )+c \right ) \left (c -3 d \right )}{3 \left (c -d \right )^{2} \sqrt {\left (1+\sin \left (f x +e \right )\right ) \left (-1+\sin \left (f x +e \right )\right ) \left (-c -d \sin \left (f x +e \right )\right )}}+\frac {2 d^{2} \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (3 c^{2}-6 c d +3 d^{2}\right ) \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}-\frac {d \left (c -3 d \right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) \operatorname {EllipticE}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+\operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{3 \left (c -d \right )^{2} \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}\right )}{a^{2} \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(505\)

Input:

int(1/(a+sin(f*x+e)*a)^2/(c+d*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)/a^2*(-1/3/(c-d)*(-(-c-d*sin(f*x+e) 
)*cos(f*x+e)^2)^(1/2)/(1+sin(f*x+e))^2-1/3*(-sin(f*x+e)^2*d-c*sin(f*x+e)+d 
*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((1+sin(f*x+e))*(-1+sin(f*x+e))*(-c-d*sin(f 
*x+e)))^(1/2)+2*d^2/(3*c^2-6*c*d+3*d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^( 
1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)/(-(-c- 
d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2) 
,((c-d)/(c+d))^(1/2))-1/3*d*(c-3*d)/(c-d)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d 
))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)/(- 
(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e) 
)/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/ 
2),((c-d)/(c+d))^(1/2))))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 967, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/9*(2*((c^2 - 3*c*d + 3*d^2)*cos(f*x + e)^2 - 2*c^2 + 6*c*d - 6*d^2 - (c^ 
2 - 3*c*d + 3*d^2)*cos(f*x + e) - (2*c^2 - 6*c*d + 6*d^2 + (c^2 - 3*c*d + 
3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(1/2*I*d)*weierstrassPInverse(-4/3* 
(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e 
) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + 2*((c^2 - 3*c*d + 3*d^2)*cos(f*x + e) 
^2 - 2*c^2 + 6*c*d - 6*d^2 - (c^2 - 3*c*d + 3*d^2)*cos(f*x + e) - (2*c^2 - 
 6*c*d + 6*d^2 + (c^2 - 3*c*d + 3*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-1 
/2*I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9* 
I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) + 3*( 
(I*c*d - 3*I*d^2)*cos(f*x + e)^2 - 2*I*c*d + 6*I*d^2 + (-I*c*d + 3*I*d^2)* 
cos(f*x + e) + (-2*I*c*d + 6*I*d^2 + (-I*c*d + 3*I*d^2)*cos(f*x + e))*sin( 
f*x + e))*sqrt(1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8 
*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/ 
27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 
 2*I*c)/d)) + 3*((-I*c*d + 3*I*d^2)*cos(f*x + e)^2 + 2*I*c*d - 6*I*d^2 + ( 
I*c*d - 3*I*d^2)*cos(f*x + e) + (2*I*c*d - 6*I*d^2 + (I*c*d - 3*I*d^2)*cos 
(f*x + e))*sin(f*x + e))*sqrt(-1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^ 
2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 
- 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3* 
I*d*sin(f*x + e) + 2*I*c)/d)) + 3*((c*d - 3*d^2)*cos(f*x + e)^2 + c*d -...
 

Sympy [F]

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

integrate(1/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**(1/2),x)
 

Output:

Integral(1/(sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2 + 2*sqrt(c + d*sin(e 
+ f*x))*sin(e + f*x) + sqrt(c + d*sin(e + f*x))), x)/a**2
 

Maxima [F]

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

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

Output:

integrate(1/((a*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e) + c)), x)
 

Giac [F]

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

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

Output:

integrate(1/((a*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e) + c)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sin(e + f*x)*d + c)/(sin(e + f*x)**3*d + sin(e + f*x)**2*c + 2*si 
n(e + f*x)**2*d + 2*sin(e + f*x)*c + sin(e + f*x)*d + c),x)/a**2