\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx\) [113]

Optimal result

Integrand size = 38, antiderivative size = 136 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(3 A-B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a c^{3/2} f}+\frac {(3 A-B) \cos (e+f x)}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {(A-B) \sec (e+f x)}{a c f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.56 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-(1+i) \sqrt [4]{-1} (3 A-B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{4 a f (1+\sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^ 
(3/2)),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)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (A + B)*(Cos[(e + 
 f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (1 + 
I)*(-1)^(1/4)*(3*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])^2*(Cos[(e + f*x)/2] + Sin[(e + f 
*x)/2]) + 2*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) 
))/(4*a*f*(1 + Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96, 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, 3129, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3129

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 3129

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

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 = 0.32 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.65

method	result	size
default	\(-\frac {\sin \left (f x +e \right ) \left (3 A \sqrt {c +c \sin \left (f x +e \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c -B \sqrt {c +c \sin \left (f x +e \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c -6 A \,c^{\frac {3}{2}}+2 B \,c^{\frac {3}{2}}\right )-3 A \sqrt {c +c \sin \left (f x +e \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c +B \sqrt {c +c \sin \left (f x +e \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c +2 A \,c^{\frac {3}{2}}-6 B \,c^{\frac {3}{2}}}{8 c^{\frac {5}{2}} a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(225\)

Input:

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

Output:

-1/8/c^(5/2)/a*(sin(f*x+e)*(3*A*(c+c*sin(f*x+e))^(1/2)*arctanh(1/2*(c+c*si 
n(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c-B*(c+c*sin(f*x+e))^(1/2)*arctan 
h(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c-6*A*c^(3/2)+2*B*c^ 
(3/2))-3*A*(c+c*sin(f*x+e))^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/ 
2)/c^(1/2))*2^(1/2)*c+B*(c+c*sin(f*x+e))^(1/2)*arctanh(1/2*(c+c*sin(f*x+e) 
)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c+2*A*c^(3/2)-6*B*c^(3/2))/cos(f*x+e)/(c- 
c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.70 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A - B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (3 \, A - B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (3 \, A - B\right )} \sin \left (f x + e\right ) - A + 3 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

-1/16*(sqrt(2)*((3*A - B)*cos(f*x + e)*sin(f*x + e) - (3*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)*s 
qrt(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) - A + 3*B)*sqrt(-c 
*sin(f*x + e) + c))/(a*c^2*f*cos(f*x + e)*sin(f*x + e) - a*c^2*f*cos(f*x + 
 e))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="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. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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