\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [137]

Optimal result

Integrand size = 40, antiderivative size = 92 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a (A+B) \cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}-\frac {a B \cos (e+f x)}{c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 4.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {a (1+\sin (e+f x))} (A-B+2 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{2 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f 
*x])^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3450, 3042, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

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

Input:

Int[(Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^( 
5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3217

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

rule 3450

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[B/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] - 
Simp[(B*c - A*d)/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(82)=164\).

Time = 8.05 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.91

method	result	size
default	\(\frac {\left (\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) A \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{2}\right ) \left (1+\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-4 B \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{2 \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, f \,c^{2} \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(176\)
parts	\(\frac {A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}{16 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2}}-\frac {2 B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}}{c^{2} f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}}\)	\(215\)

Input:

int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

1/2/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*((cos(1/2*f*x+1/2*e)^2-1/2)*A*(s 
in(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1/2)*(1+sec(1/4*Pi+1/2*f*x+1/2*e)^2)* 
tan(1/4*Pi+1/2*f*x+1/2*e)-4*B*sin(1/2*f*x+1/2*e)^2*cos(1/2*f*x+1/2*e)^2)*( 
a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/f/c^2/(-1+2*cos(1/2*f*x+1/2*e)^2)/(2* 
sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {{\left (2 \, B \sin \left (f x + e\right ) + A - B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 2 \, c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, c^{3} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x 
, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+a*sin(f*x+e))**(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2) 
,x)
 

Output:

Integral(sqrt(a*(sin(e + f*x) + 1))*(A + B*sin(e + f*x))/(-c*(sin(e + f*x) 
 - 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x 
, algorithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)/(-c*sin(f*x + e) + 
 c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x 
, algorithm="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. \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2))/(c - c*sin(e + f*x)) 
^(5/2),x)
 

Output:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2))/(c - c*sin(e + f*x)) 
^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a \right )}{c^{3}} \] Input:

int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

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