\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\) [157]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-7 B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}} \] Output:

1/8*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(9/2)+1/48* 
(A-7*B)*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 = 13.78 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (5 A-5 B-3 (A-B) \cos (2 (e+f x))+(4 A+17 B) \sin (e+f x)-3 B \sin (3 (e+f x)))}{12 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(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*A 
 - 5*B - 3*(A - B)*Cos[2*(e + f*x)] + (4*A + 17*B)*Sin[e + f*x] - 3*B*Sin[ 
3*(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 = 0.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 3451, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3451

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3221

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

Input:

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

Output:

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x] 
)^(9/2)) + ((A - 7*B)*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 3451

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(A*b - a*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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[ 
{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2* 
m + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(343\) vs. \(2(84)=168\).

Time = 8.06 (sec) , antiderivative size = 344, normalized size of antiderivative = 3.58

method	result	size
parts	\(\frac {A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2} \left (\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5}+3 \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}{96 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4}}+\frac {2 B \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a^{2}}{3 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c^{4}}\)	\(344\)
default	\(\frac {a^{2} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2}-\frac {3 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{4}+\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {1}{8}\right ) A \left (\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {1}{3}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{5}+\frac {8 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {3}{4}\right )}{3}\right )}{\sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4} f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right )}\)	\(365\)

Input:

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

Output:

1/96*A/f*4^(1/2)*(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)+2/3*B/f*((2*sin(1/2*f*x+1/2*e)*co 
s(1/2*f*x+1/2*e)+1)*a)^(1/2)*cos(1/2*f*x+1/2*e)^2*(-4*cos(1/2*f*x+1/2*e)^4 
+4*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+4*cos(1/2*f*x+1/2*e)^2+3)*sin(1/2 
*f*x+1/2*e)^2*a^2/(-1+2*cos(1/2*f*x+1/2*e)^2)/(8*cos(1/2*f*x+1/2*e)^5*sin( 
1/2*f*x+1/2*e)-8*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)-12*cos(1/2*f*x+1/ 
2*e)^4-6*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+12*cos(1/2*f*x+1/2*e)^2+1)/ 
(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)*c)^(1/2)/c^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.72 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (3 \, {\left (A - B\right )} a^{2} \cos \left (f x + e\right )^{2} - 4 \, {\left (A - B\right )} a^{2} + 2 \, {\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - {\left (A + 5 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

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

Output:

-1/6*(3*(A - B)*a^2*cos(f*x + e)^2 - 4*(A - B)*a^2 + 2*(3*B*a^2*cos(f*x + 
e)^2 - (A + 5*B)*a^2)*sin(f*x + e))*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. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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 [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(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 )^{3}}{\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 ) b -\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 ) 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 )^{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 ) b -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 ) 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 \right ) b -\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 ) a \right )}{c^{5}} \] Input:

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(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)**3)/(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)*b - 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)*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 - 1 
0*sin(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*b - 2*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)*a 
 - int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(si 
n(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)*b - 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*sin(e + f*x) - 1),x)*a))/c**5