\(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [395]

Optimal result

Integrand size = 30, antiderivative size = 95 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x)}{2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{2 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \left (-\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+\left (1+\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{2 c f (-1+\sin (e+f x)) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3220

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

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]])
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(210\) vs. \(2(83)=166\).

Time = 1.76 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.22

method	result	size
default	\(-\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )-2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right )+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )-\sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {4}}{8 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c}\)	\(211\)

Input:

int(1/(a+sin(f*x+e)*a)^(1/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

-1/8/f*tan(1/4*Pi+1/2*f*x+1/2*e)*(2*cos(1/4*Pi+1/2*f*x+1/2*e)^2*ln(-cot(1/ 
4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)-2*cos(1/4*Pi+1/2*f*x+1/2* 
e)^2*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e))+2*cos(1/4*Pi 
+1/2*f*x+1/2*e)^2*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)^2)/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*c 
os(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.25 \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\left [\frac {\sqrt {a c} {\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) - \cos \left (f x + e\right )\right )} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}}, -\frac {\sqrt {-a c} {\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) - \cos \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )}\right ) + \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{2 \, {\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \left (f x + e\right )\right )}}\right ] \] Input:

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

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

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

Output:

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