\(\int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\) [360]

Optimal result

Integrand size = 30, antiderivative size = 96 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.18 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2} \left (4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 3219, 3042, 3216, 3042, 3146, 16}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

\(\displaystyle -\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 3042

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

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3216

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x 
]]*Sqrt[c + d*Sin[e + f*x]]))   Int[Cos[e + f*x]/(c + d*Sin[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 3219

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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m]) &&  !( 
ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
 

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.62

method	result	size
default	\(-\frac {\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (\sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )-2 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a \sqrt {4}}{f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}\)	\(156\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, a^{\frac {3}{2}} \sqrt {c} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

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

Output:

2*a^(3/2)*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e)^2/(c*sgn(sin(-1/4*pi + 1 
/2*f*x + 1/2*e))) + log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c*sgn(sin( 
-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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