\(\int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\) [382]

Optimal result

Integrand size = 30, antiderivative size = 184 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {8 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {4 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{f \sqrt {c-c \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{f \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 8.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \sqrt {a (1+\sin (e+f x))} \left (-12 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+87 \sin (e+f x)-\sin (3 (e+f x))\right )}{12 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(7/2)/Sqrt[c - c*Sin[e + f*x]],x]
 

Output:

-1/12*(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*Sqrt 
[a*(1 + Sin[e + f*x])]*(-12*Cos[2*(e + f*x)] + 192*Log[Cos[(e + f*x)/2] - 
Sin[(e + f*x)/2]] + 87*Sin[e + f*x] - Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x) 
/2] + Sin[(e + f*x)/2])^7*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 3219, 3042, 3219, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{\sqrt {c-c \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

Int[(a + a*Sin[e + f*x])^(7/2)/Sqrt[c - c*Sin[e + f*x]],x]
 

Output:

-1/3*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(f*Sqrt[c - c*Sin[e + f*x 
]]) + 2*a*(-1/2*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f*Sqrt[c - c* 
Sin[e + f*x]]) + 2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(f*Sqr 
t[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*Sqrt[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 = 1.86 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {2 \cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}-9 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+18 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-12 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )-12 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )+12 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )-11\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{3} \sqrt {4}}{3 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}\)	\(193\)

Input:

int((a+sin(f*x+e)*a)^(7/2)/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2/3/f*cot(1/4*Pi+1/2*f*x+1/2*e)*(2*cos(1/4*Pi+1/2*f*x+1/2*e)^6-9*cos(1/4*P 
i+1/2*f*x+1/2*e)^4+18*cos(1/4*Pi+1/2*f*x+1/2*e)^2-12*ln(-cot(1/4*Pi+1/2*f* 
x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)-12*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc 
(1/4*Pi+1/2*f*x+1/2*e)-1)+12*ln(2/(cos(1/4*Pi+1/2*f*x+1/2*e)+1))-11)*(a*si 
n(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*a^3/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2) 
*4^(1/2)
 

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxi 
ma")
 

Output:

integrate((a*sin(f*x + e) + a)^(7/2)/sqrt(-c*sin(f*x + e) + c), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.77 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \, a^{\frac {7}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 6 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{3 \, f} \] Input:

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac 
")
 

Output:

4/3*a^(7/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c*sgn(s 
in(-1/4*pi + 1/2*f*x + 1/2*e))) + (2*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6 
+ 3*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 6*c^2*cos(-1/4*pi + 1/2*f*x + 1 
/2*e)^2)/(c^3*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. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(1/2),x)
 

Output:

int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{3} \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )-1}d x \right )\right )}{c} \] Input:

int((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(1/2),x)
 

Output:

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