\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) [372]

Optimal result

Integrand size = 30, antiderivative size = 147 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

1/2*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(5/2)-a^2*cos(f 
*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(3/2)-a^3*cos(f*x+e)*ln( 
1-sin(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 = 8.64 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.29 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (-2-3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 \left (1+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

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 = 2.00 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (4 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) a^{2} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {4}}{4 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2}}\)	\(236\)

Input:

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

Output:

-1/4/f*sec(1/4*Pi+1/2*f*x+1/2*e)^3*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(2/(cos(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+4*cos(1/4*Pi+1/2*f*x+1/2*e)^2-1)*a 
^2*(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 [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {8 \, a^{\frac {5}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (c^{3} - \frac {4 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} + \frac {a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}}}{f} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.85 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {{\left (4 \, a^{2} \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )} \sqrt {a}}{2 \, c^{\frac {5}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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