\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx\) [128]

Optimal result

Integrand size = 38, antiderivative size = 174 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f} \] Output:

1/8*(A+B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2 
^(1/2)/a^3/c^(1/2)/f-1/4*(A+B)*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^3/c/f-1 
/6*(A+B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/c^2/f-1/5*(A-B)*sec(f*x+e 
)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c^3/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.30 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.17 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (-A+B)-10 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-15 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-(15+15 i) \sqrt [4]{-1} (A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{60 a^3 f (1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*Sqrt[c - c*Sin[e + 
f*x]]),x]
 

Output:

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 
])*(12*(-A + B) - 10*(A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 15* 
(A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (15 + 15*I)*(-1)^(1/4)*( 
A + B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x 
)/2] + Sin[(e + f*x)/2])^5))/(60*a^3*f*(1 + Sin[e + f*x])^3*Sqrt[c - c*Sin 
[e + f*x]])
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3446, 3042, 3334, 3042, 3154, 3042, 3154, 3042, 3128, 219}

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.

\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 \sqrt {c-c \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 \sqrt {c-c \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {1}{2} c (A+B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c (A+B) \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3154

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \int \frac {\sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^2}dx-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3154

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \left (-\frac {c \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} c (A+B) \left (\frac {1}{2} c \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 f}}{a^3 c^3}\)

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]]) 
,x]
 

Output:

(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/f + ((A + B)*c*( 
-1/3*(Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/f + (c*((Sqrt[c]*ArcTanh[ 
(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[2]*f) - 
(Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/f))/2))/2)/(a^3*c^3)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

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

rule 3154

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

rule 3334

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

rule 3446

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )-1\right ) \left (-30 A \,c^{\frac {9}{2}} \sin \left (f x +e \right )^{2}-30 B \,c^{\frac {9}{2}} \sin \left (f x +e \right )^{2}-80 A \,c^{\frac {9}{2}} \sin \left (f x +e \right )-80 B \,c^{\frac {9}{2}} \sin \left (f x +e \right )+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{2} A -74 c^{\frac {9}{2}} A +15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{2} B -26 c^{\frac {9}{2}} B \right )}{120 a^{3} c^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(200\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-1/120*(sin(f*x+e)-1)*(-30*A*c^(9/2)*sin(f*x+e)^2-30*B*c^(9/2)*sin(f*x+e)^ 
2-80*A*c^(9/2)*sin(f*x+e)-80*B*c^(9/2)*sin(f*x+e)+15*2^(1/2)*arctanh(1/2*( 
c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*c^2*A-74 
*c^(9/2)*A+15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2) 
)*(c*(1+sin(f*x+e)))^(5/2)*c^2*B-26*c^(9/2)*B)/a^3/c^(9/2)/(1+sin(f*x+e))^ 
2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.51 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {15 \, \sqrt {2} {\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, {\left (A + B\right )} \cos \left (f x + e\right )^{2} - 40 \, {\left (A + B\right )} \sin \left (f x + e\right ) - 52 \, A - 28 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{240 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/240*(15*sqrt(2)*((A + B)*cos(f*x + e)^3 - 2*(A + B)*cos(f*x + e)*sin(f*x 
 + e) - 2*(A + B)*cos(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2) 
*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c 
*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 
 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(15*(A + B)*co 
s(f*x + e)^2 - 40*(A + B)*sin(f*x + e) - 52*A - 28*B)*sqrt(-c*sin(f*x + e) 
 + c))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a 
^3*c*f*cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^3*sqrt(-c*sin(f*x + e 
) + c)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(1/2 
)),x)
 

Output:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(1/2 
)), x)
 

Reduce [F]

\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \left (\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4}+2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )-1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4}+2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )-1}d x \right ) b \right )}{a^{3} c} \] Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x)
 

Output:

( - sqrt(c)*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**4 + 2*sin(e + f* 
x)**3 - 2*sin(e + f*x) - 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e + 
f*x))/(sin(e + f*x)**4 + 2*sin(e + f*x)**3 - 2*sin(e + f*x) - 1),x)*b))/(a 
**3*c)