\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx\) [396]

Optimal result

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

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

Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \left (10-3 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \left (-2+\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+3 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (-5+2 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-2 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{8 c^2 f (-1+\sin (e+f x))^2 \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.72 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3220

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

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	\(\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right )-4 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sqrt {4}}{32 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2}}\)	\(242\)

Input:

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

Output:

1/32/f*tan(1/4*Pi+1/2*f*x+1/2*e)*sec(1/4*Pi+1/2*f*x+1/2*e)^2*(4*cos(1/4*Pi 
+1/2*f*x+1/2*e)^4*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)) 
-4*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)-1)*cos(1/4*Pi+1 
/2*f*x+1/2*e)^4-4*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+ 
1)*cos(1/4*Pi+1/2*f*x+1/2*e)^4-3*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+1)/(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)/c^2*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 374, normalized size of antiderivative = 2.67 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {{\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) - 2\right )}}{8 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}, -\frac {{\left (\cos \left (f x + e\right )^{3} + 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )}\right ) - \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\sin \left (f x + e\right ) - 2\right )}}{4 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}}\right ] \] Input:

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2),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*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*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*c^3*f*cos(f*x + e)^3 + 
 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

int(1/(a+a*sin(f*x+e))^(1/2)/(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)**4 - 2*sin(e + f*x)**3 + 2*sin(e + f*x) - 1),x))/(a*c**3)