\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\) [374]

Optimal result

Integrand size = 30, antiderivative size = 88 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}} \] Output:

1/8*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(9/2)+1/48*cos(f* 
x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c*sin(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 5.92 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (5-3 \cos (2 (e+f x))+4 \sin (e+f x))}{12 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(9/2),x]
 

Output:

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(5 - 
 3*Cos[2*(e + f*x)] + 4*Sin[e + f*x]))/(12*c^4*f*(Cos[(e + f*x)/2] + Sin[( 
e + f*x)/2])*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3222, 3042, 3221}

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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3221

\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

Input:

Int[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(9/2),x]
 

Output:

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^(9/2)) 
 + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(48*c*f*(c - c*Sin[e + f*x])^ 
(7/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3221

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] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne 
Q[m, -2^(-1)]
 

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])
 

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2} \left (\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5}+3 \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {4}}{96 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4}}\)	\(96\)

Input:

int((a+sin(f*x+e)*a)^(5/2)/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

1/96/f*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*a^2/(c*cos(1/4*Pi+1/2*f*x+1/2 
*e)^2)^(1/2)/c^4*(tan(1/4*Pi+1/2*f*x+1/2*e)^5+3*tan(1/4*Pi+1/2*f*x+1/2*e)^ 
5*sec(1/4*Pi+1/2*f*x+1/2*e)^2)*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.51 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 4 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxi 
ma")
 

Output:

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(9/2), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (6 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 8 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{48 \, c^{\frac {9}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac 
")
 

Output:

-1/48*(6*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 - 8*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 + 3*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^(9/2)*f 
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)
 

Mupad [B] (verification not implemented)

Time = 23.79 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.75 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {40\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}-\frac {8\,a^2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^5\,f}\right )}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \] Input:

int((a + a*sin(e + f*x))^(5/2)/(c - c*sin(e + f*x))^(9/2),x)
 

Output:

((c - c*sin(e + f*x))^(1/2)*((40*a^2*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x 
))^(1/2))/(3*c^5*f) + (32*a^2*exp(e*5i + f*x*5i)*sin(e + f*x)*(a + a*sin(e 
 + f*x))^(1/2))/(3*c^5*f) - (8*a^2*exp(e*5i + f*x*5i)*cos(2*e + 2*f*x)*(a 
+ a*sin(e + f*x))^(1/2))/(c^5*f)))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 5 
4*exp(e*5i + f*x*5i)*cos(3*e + 3*f*x) + 2*exp(e*5i + f*x*5i)*cos(5*e + 5*f 
*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x) + 16*exp(e*5i + f*x*5i)*sin(4 
*e + 4*f*x))
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )-2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{5}} \] Input:

int((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(9/2),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)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f 
*x)**3 - 10*sin(e + f*x)**2 + 5*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)**5 - 5*si 
n(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 
 1),x) - int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f 
*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*s 
in(e + f*x) - 1),x)))/c**5