\(\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) [375]

Optimal result

Integrand size = 30, antiderivative size = 133 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}} \] Output:

1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(11/2)+1/40*cos( 
f*x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c*sin(f*x+e))^(9/2)+1/240*cos(f*x+e)* 
(a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(7/2)
 

Mathematica [A] (verified)

Time = 7.83 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (9-5 \cos (2 (e+f x))+10 \sin (e+f x))}{30 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(11/2),x]
 

Output:

-1/30*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x]) 
]*(9 - 5*Cos[2*(e + f*x)] + 10*Sin[e + f*x]))/(c^5*f*(Cos[(e + f*x)/2] + S 
in[(e + f*x)/2])*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3222, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{11/2}}dx\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\)

\(\Big \downarrow \) 3221

\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}\)

Input:

Int[(a + a*Sin[e + f*x])^(5/2)/(c - c*Sin[e + f*x])^(11/2),x]
 

Output:

(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2 
)) + ((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)))/(5*c)
 

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 = 1.85 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+6\right ) a^{2} \sqrt {4}}{480 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{5}}\)	\(111\)

Input:

int((a+sin(f*x+e)*a)^(5/2)/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE 
)
 

Output:

1/480/f*tan(1/4*Pi+1/2*f*x+1/2*e)^5*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+3*cos(1/4*Pi+1/2* 
f*x+1/2*e)^2+6)*a^2/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c^5*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} \sin \left (f x + e\right ) - 7 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="fri 
cas")
 

Output:

-1/15*(5*a^2*cos(f*x + e)^2 - 5*a^2*sin(f*x + e) - 7*a^2)*sqrt(a*sin(f*x + 
 e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos( 
f*x + e)^3 + 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*cos( 
f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="max 
ima")
 

Output:

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(11/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),x, algorithm="gia 
c")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 23.79 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.05 \[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,96{}\mathrm {i}}{5\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,64{}\mathrm {i}}{3\,c^6\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,32{}\mathrm {i}}{3\,c^6\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \] Input:

int((a + a*sin(e + f*x))^(5/2)/(c - c*sin(e + f*x))^(11/2),x)
 

Output:

((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^ 
(1/2)*96i)/(5*c^6*f) + (a^2*exp(e*6i + f*x*6i)*sin(e + f*x)*(a + a*sin(e + 
 f*x))^(1/2)*64i)/(3*c^6*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a 
+ a*sin(e + f*x))^(1/2)*32i)/(3*c^6*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)* 
264i - exp(e*6i + f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5 
*e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i + f* 
x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x \right )}{c^{6}} \] Input:

int((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(11/2),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)**6 - 6*sin(e + f*x)**5 + 15*sin(e + f*x) 
**4 - 20*sin(e + f*x)**3 + 15*sin(e + f*x)**2 - 6*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)**6 - 6*sin(e + f*x)**5 + 15*sin(e + f*x)**4 - 20*sin(e + f*x)**3 
+ 15*sin(e + f*x)**2 - 6*sin(e + f*x) + 1),x) + int((sqrt(sin(e + f*x) + 1 
)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**6 - 6*sin(e + f*x)**5 + 15*sin 
(e + f*x)**4 - 20*sin(e + f*x)**3 + 15*sin(e + f*x)**2 - 6*sin(e + f*x) + 
1),x)))/c**6