\(\int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx\) [409]

Optimal result

Integrand size = 30, antiderivative size = 140 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \left (10+3 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-3 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (-2-\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+2 \left (5+2 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{8 f (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x]
 

Output:

(Cos[(e + f*x)/2]^2*(10 + 3*Log[1 - Tan[(e + f*x)/2]] - 3*Log[1 + Tan[(e + 
 f*x)/2]] + Cos[2*(e + f*x)]*(-2 - Log[1 - Tan[(e + f*x)/2]] + Log[1 + Tan 
[(e + f*x)/2]]) + 2*(5 + 2*Log[1 - Tan[(e + f*x)/2]] - 2*Log[1 + Tan[(e + 
f*x)/2]])*Sin[e + f*x])*(-1 + Tan[(e + f*x)/2])*(1 + Tan[(e + f*x)/2]))/(8 
*f*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3222, 3042, 3222, 3042, 3220, 3042, 4257}

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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3220

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\cos (e+f x) \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{2 a \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}}{2 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{2 a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}}{2 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}\)

Input:

Int[1/((a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]),x]
 

Output:

-1/4*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]) 
+ (-1/2*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x] 
]) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*a*f*Sqrt[a + a*Sin[e + f*x]]* 
Sqrt[c - c*Sin[e + f*x]]))/(2*a)
 

Defintions of rubi rules used

rule 3042

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

rule 3220

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

rule 3222

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( 
(c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) 
)   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free 
Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && 
 ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || 
!SumSimplerQ[n, 1])
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (32 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-32 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-32 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+11 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-13\right ) \sqrt {4}}{256 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}\)	\(242\)

Input:

int(1/(a+sin(f*x+e)*a)^(5/2)/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/256/f*cot(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)^2*(32*ln(-cot( 
1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e))*sin(1/4*Pi+1/2*f*x+1/2*e) 
^4-32*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)*sin(1/4*P 
i+1/2*f*x+1/2*e)^4-32*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2 
*e)-1)*sin(1/4*Pi+1/2*f*x+1/2*e)^4+11*cos(1/4*Pi+1/2*f*x+1/2*e)^4-6*cos(1/ 
4*Pi+1/2*f*x+1/2*e)^2-13)/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*cos(1/4 
*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a^2*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.67 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\left [\frac {{\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) + 2\right )}}{8 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}, -\frac {{\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )}\right ) - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) + 2\right )}}{4 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}}\right ] \] Input:

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

Output:

[1/8*((cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f*x + e))*sqrt 
(a*c)*log(-(a*c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*s 
in(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/cos(f*x + e)^3) + 
 2*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*(sin(f*x + e) + 2))/ 
(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)), -1/4*((cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2*cos 
(f*x + e))*sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*s 
in(f*x + e) + c)*sin(f*x + e)/(a*c*cos(f*x + e))) - sqrt(a*sin(f*x + e) + 
a)*sqrt(-c*sin(f*x + e) + c)*(sin(f*x + e) + 2))/(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4}+2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )-1}d x \right )}{a^{3} c} \] Input:

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

Output:

( - sqrt(c)*sqrt(a)*int((sqrt(sin(e + f*x) + 1)*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**3*c)