\(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\) [364]

Optimal result

Integrand size = 30, antiderivative size = 92 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{12 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}} \] Output:

1/4*a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f/(c-c*sin(f*x+e))^(9/2)-1/12*a^2* 
cos(f*x+e)/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 4.47 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (1+2 \sin (e+f x))}{6 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(9/2),x]
 

Output:

(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(1 + 2 
*Sin[e + f*x]))/(6*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 = 0.45 (sec) , antiderivative size = 92, 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, 3218, 3042, 3217}

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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2}}{(c-c \sin (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 3218

\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 \cos (e+f x)}{12 c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}\)

Input:

Int[(a + a*Sin[e + f*x])^(3/2)/(c - c*Sin[e + f*x])^(9/2),x]
 

Output:

(a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(4*f*(c - c*Sin[e + f*x])^(9/2)) 
 - (a^2*Cos[e + f*x])/(12*c*f*Sqrt[a + a*Sin[e + f*x]]*(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 3217

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

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 = 1.86 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) a \sqrt {4}}{192 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4}}\)	\(109\)

Input:

int((a+sin(f*x+e)*a)^(3/2)/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

1/192/f*tan(1/4*Pi+1/2*f*x+1/2*e)^3*sec(1/4*Pi+1/2*f*x+1/2*e)^4*(a*sin(1/4 
*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(cos(1/4*Pi+1/2*f*x+1/2*e)^4+2*cos(1/4*Pi+1/2* 
f*x+1/2*e)^2+3)*a/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c^4*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {{\left (2 \, a \sin \left (f x + e\right ) + a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="fric 
as")
 

Output:

1/6*(2*a*sin(f*x + e) + a)*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. \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxi 
ma")
 

Output:

integrate((a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(9/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {{\left (4 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{96 \, c^{\frac {9}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac 
")
 

Output:

1/96*(4*a*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*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 = 21.75 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.12 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\left (\frac {16\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \] Input:

int((a + a*sin(e + f*x))^(3/2)/(c - c*sin(e + f*x))^(9/2),x)
 

Output:

(((16*a*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f) + (32*a*e 
xp(e*5i + f*x*5i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(3*c^5*f))*(c - 
 c*sin(e + f*x))^(1/2))/(84*cos(e + f*x)*exp(e*5i + f*x*5i) - 54*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*ex 
p(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]

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

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