\(\int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx\) [164]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}} \] Output:

1/4*(A-B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(1/2)+1/5 
*B*c*cos(f*x+e)*(a+a*sin(f*x+e))^(9/2)/a/f/(c-c*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 3.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.26 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {a^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (4 (60 A+23 B) \sin (e+f x)+\cos (4 (e+f x)) (5 A+15 B+4 B \sin (e+f x))-4 \cos (2 (e+f x)) (5 (7 A+5 B)+4 (5 A+6 B) \sin (e+f x)))}{160 f} \] Input:

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

Output:

(a^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(4*( 
60*A + 23*B)*Sin[e + f*x] + Cos[4*(e + f*x)]*(5*A + 15*B + 4*B*Sin[e + f*x 
]) - 4*Cos[2*(e + f*x)]*(5*(7*A + 5*B) + 4*(5*A + 6*B)*Sin[e + f*x])))/(16 
0*f)
 

Rubi [A] (verified)

Time = 0.67 (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, 3450, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3450

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

\(\displaystyle \frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{9/2}}{5 a f \sqrt {c-c \sin (e+f x)}}\)

Input:

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

Output:

((A - B)*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*f*Sqrt[c - c*Sin[e 
+ f*x]]) + (B*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(9/2))/(5*a*f*Sqrt[c - c 
*Sin[e + f*x]])
 

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 3450
Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[B/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] - 
Simp[(B*c - A*d)/d   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x 
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[ 
a^2 - b^2, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(237\) vs. \(2(84)=168\).

Time = 9.57 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.48

method result size
default \(-\frac {40 \left (\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) A \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {8 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8}-\frac {5 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{4}-\frac {15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8}-\frac {5}{16}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5}\right ) a^{3} \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{10 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5 f}\) \(238\)
parts \(-\frac {2 A \sqrt {4}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3} a^{3} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 B \,a^{3} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \left (\left (16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-20 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) \(271\)

Input:

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

Output:

-40*((cos(1/2*f*x+1/2*e)^2-1/2)*A*(cos(1/4*Pi+1/2*f*x+1/2*e)+1)^3*(cos(1/4 
*Pi+1/2*f*x+1/2*e)-1)^3*tan(1/4*Pi+1/2*f*x+1/2*e)+8/5*(cos(1/2*f*x+1/2*e)^ 
5*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)+15/8*cos(1/2* 
f*x+1/2*e)^4-5/4*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-15/8*cos(1/2*f*x+1/ 
2*e)^2-5/16)*sin(1/2*f*x+1/2*e)^2*B*cos(1/2*f*x+1/2*e)^2)*a^3*(c*cos(1/4*P 
i+1/2*f*x+1/2*e)^2)^(1/2)*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(10*f*cos( 
1/2*f*x+1/2*e)^2-5*f)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.45 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {{\left (5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{4} - 40 \, {\left (A + B\right )} a^{3} \cos \left (f x + e\right )^{2} + 5 \, {\left (7 \, A + 5 \, B\right )} a^{3} + 4 \, {\left (B a^{3} \cos \left (f x + e\right )^{4} - {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, f \cos \left (f x + e\right )} \] Input:

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

Output:

1/20*(5*(A + 3*B)*a^3*cos(f*x + e)^4 - 40*(A + B)*a^3*cos(f*x + e)^2 + 5*( 
7*A + 5*B)*a^3 + 4*(B*a^3*cos(f*x + e)^4 - (5*A + 7*B)*a^3*cos(f*x + e)^2 
+ 2*(5*A + 3*B)*a^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f* 
x + e) + c)/(f*cos(f*x + e))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.56 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {4 \, {\left (8 \, B a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, A a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, B a^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{5 \, f} \] Input:

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

Output:

-4/5*(8*B*a^3*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x 
+ 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*A*a^3*cos(-1/4*pi + 1/2* 
f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f 
*x + 1/2*e)) - 5*B*a^3*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 
1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)*sqrt(c)/f
 

Mupad [B] (verification not implemented)

Time = 38.81 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.80 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (140\,A\,\cos \left (e+f\,x\right )+100\,B\,\cos \left (e+f\,x\right )+135\,A\,\cos \left (3\,e+3\,f\,x\right )-5\,A\,\cos \left (5\,e+5\,f\,x\right )+85\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )-240\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )-90\,B\,\sin \left (2\,e+2\,f\,x\right )+48\,B\,\sin \left (4\,e+4\,f\,x\right )-2\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{160\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

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

Output:

-(a^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(140*A*co 
s(e + f*x) + 100*B*cos(e + f*x) + 135*A*cos(3*e + 3*f*x) - 5*A*cos(5*e + 5 
*f*x) + 85*B*cos(3*e + 3*f*x) - 15*B*cos(5*e + 5*f*x) - 240*A*sin(2*e + 2* 
f*x) + 40*A*sin(4*e + 4*f*x) - 90*B*sin(2*e + 2*f*x) + 48*B*sin(4*e + 4*f* 
x) - 2*B*sin(6*e + 6*f*x)))/(160*f*(cos(2*e + 2*f*x) + 1))
 

Reduce [F]

\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\sqrt {c}\, \sqrt {a}\, a^{3} \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:

int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(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)**4,x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 
1)*sin(e + f*x)**3,x)*a + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x 
) + 1)*sin(e + f*x)**3,x)*b + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + 
 f*x) + 1)*sin(e + f*x)**2,x)*a + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin 
(e + f*x) + 1)*sin(e + f*x)**2,x)*b + 3*int(sqrt(sin(e + f*x) + 1)*sqrt( - 
 sin(e + f*x) + 1)*sin(e + f*x),x)*a + int(sqrt(sin(e + f*x) + 1)*sqrt( - 
sin(e + f*x) + 1)*sin(e + f*x),x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - s 
in(e + f*x) + 1),x)*a)