\(\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [191]

Optimal result

Integrand size = 40, antiderivative size = 94 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {(A-B) c \cos (e+f x)}{2 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {B c \cos (e+f x)}{a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a (1+\sin (e+f x))} (A+B+2 B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{2 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 94, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

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

Input:

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

Output:

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

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. \(205\) vs. \(2(84)=168\).

Time = 9.23 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.19

method	result	size
default	\(\frac {3 \left (A \left (\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{2}\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {5}{3}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}+\frac {32 B \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{16 \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, f \,a^{2} \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\)	\(206\)
parts	\(\frac {A \sqrt {4}\, \left (3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}}{128 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}+\frac {2 B \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a^{2} f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}}\)	\(235\)

Input:

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

Output:

3/16/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(A*(sin(1/2*f*x+1/2*e)*cos(1/2* 
f*x+1/2*e)+1/2)*(cos(1/4*Pi+1/2*f*x+1/2*e)^4-2*cos(1/4*Pi+1/2*f*x+1/2*e)^2 
-5/3)*(cos(1/2*f*x+1/2*e)^2-1/2)*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2* 
f*x+1/2*e)^3+32/3*B*sin(1/2*f*x+1/2*e)^2*cos(1/2*f*x+1/2*e)^2)*(c*cos(1/4* 
Pi+1/2*f*x+1/2*e)^2)^(1/2)/f/a^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 = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (2 \, B \sin \left (f x + e\right ) + A + B\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2)/(a+a*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 [B] (verification not implemented)

Time = 37.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {2\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (A\,\sin \left (2\,e+2\,f\,x\right )+3\,B\,\sin \left (2\,e+2\,f\,x\right )-2\,A\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )-3\,B\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+B\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )\right )}{a^2\,f\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-8\,{\sin \left (e+f\,x\right )}^2+4\,\sin \left (e+f\,x\right )+2\,{\sin \left (2\,e+2\,f\,x\right )}^2+4\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \] Input:

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

Output:

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

Reduce [F]

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

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