\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx\) [120]

Optimal result

Integrand size = 38, antiderivative size = 135 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} a^2 \sqrt {c} f}-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{2 a^2 c f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c^2 f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.66 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (-A+B)-3 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(3+3 i) \sqrt [4]{-1} (A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{6 a^2 f (1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*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 
])*(2*(-A + B) - 3*(A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (3 + 
3*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])^3))/(6*a^2*f*(1 + Sin[e + f*x])^ 
2*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3042, 3446, 3042, 3334, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^2 \sqrt {c-c \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3154

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]]) 
,x]
 

Output:

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

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 = 0.36 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )-1\right ) \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c A -6 A \,c^{\frac {5}{2}} \sin \left (f x +e \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c B -6 B \,c^{\frac {5}{2}} \sin \left (f x +e \right )-10 A \,c^{\frac {5}{2}}-2 B \,c^{\frac {5}{2}}\right )}{12 a^{2} c^{\frac {5}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(168\)

Input:

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

Output:

-1/12*(sin(f*x+e)-1)*(3*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)))^(3/2)*c*A-6*A*c^(5/2)*sin(f*x+e)+3*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)))^( 
3/2)*c*B-6*B*c^(5/2)*sin(f*x+e)-10*A*c^(5/2)-2*B*c^(5/2))/a^2/c^(5/2)/(1+s 
in(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.61 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (3 \, {\left (A + B\right )} \sin \left (f x + e\right ) + 5 \, A + B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{24 \, {\left (a^{2} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} c f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

1/24*(3*sqrt(2)*((A + B)*cos(f*x + e)*sin(f*x + e) + (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*(3*(A + B)*sin(f*x + e) + 5*A + B)*sqrt(-c*si 
n(f*x + e) + c))/(a^2*c*f*cos(f*x + e)*sin(f*x + e) + a^2*c*f*cos(f*x + e) 
)
 

Sympy [F]

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

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

Output:

(Integral(A/(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 + 2*sqrt(-c*sin(e + 
 f*x) + c)*sin(e + f*x) + sqrt(-c*sin(e + f*x) + c)), x) + Integral(B*sin( 
e + f*x)/(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 + 2*sqrt(-c*sin(e + f* 
x) + c)*sin(e + f*x) + sqrt(-c*sin(e + f*x) + c)), x))/a**2
 

Maxima [F]

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

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(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. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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