\(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [406]

Optimal result

Integrand size = 30, antiderivative size = 143 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (a+a \sin (e+f x))^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 3.86 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.20 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (2+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-\cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 \left (1+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3218, 3042, 3218, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

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 3218

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

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (32 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-32 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-13 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+11\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2} \sqrt {4}}{32 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2}}\)	\(190\)

Input:

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

Output:

1/32/f*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)^3*(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(2/(cos(1/4*Pi+1/2*f*x+1/2*e)+1))*sin(1/4*Pi+1/2*f*x+1/2*e)^4-13*co 
s(1/4*Pi+1/2*f*x+1/2*e)^4-6*cos(1/4*Pi+1/2*f*x+1/2*e)^2+11)*(c*cos(1/4*Pi+ 
1/2*f*x+1/2*e)^2)^(1/2)*c^2/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a^2*4^(1 
/2)
 

Fricas [F]

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

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

Output:

integral((c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2)*sqrt(a*sin(f*x 
+ e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*c 
os(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.28 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\frac {8 \, \sqrt {a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (a^{3} + \frac {4 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {4 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{\frac {5}{2}}} + \frac {c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{a^{\frac {5}{2}}}}{f} \] Input:

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

Output:

(8*sqrt(a)*c^(5/2)*sin(f*x + e)^2/((a^3 + 4*a^3*sin(f*x + e)/(cos(f*x + e) 
 + 1) + 6*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^3*sin(f*x + e)^3/( 
cos(f*x + e) + 1)^3 + a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*x + 
e) + 1)^2) - 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(5/2) + 
c^(5/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/a^(5/2))/f
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} c^{\frac {5}{2}} {\left (\frac {2 \, \sqrt {2} \log \left (-2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{4 \, f} \] Input:

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

Output:

-1/4*sqrt(2)*sqrt(a)*c^(5/2)*(2*sqrt(2)*log(-2*sin(-1/4*pi + 1/2*f*x + 1/2 
*e)^2 + 2)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(4*sin(-1/4 
*pi + 1/2*f*x + 1/2*e)^2 - 3)/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^ 
3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e) 
)/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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