Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}} \] Output:
1/2*(A-B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(1/2)+1/3 *B*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f/(c-c*sin(f*x+e))^(1/2)
Time = 2.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-2 (6 A+B) \sin (e+f x)+\cos (2 (e+f x)) (3 (A+B)+2 B \sin (e+f x)))}{12 f} \] Input:
Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]
Output:
-1/12*(a*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]* (-2*(6*A + B)*Sin[e + f*x] + Cos[2*(e + f*x)]*(3*(A + B) + 2*B*Sin[e + f*x ])))/f
Time = 0.64 (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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3450 |
| \(\displaystyle (A-B) \int (\sin (e+f x) a+a)^{3/2} \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-B) \int (\sin (e+f x) a+a)^{3/2} \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx}{a}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 a f \sqrt {c-c \sin (e+f x)}}\) |
Input:
Int[(a + a*Sin[e + f*x])^(3/2)*(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])^(3/2))/(2*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*a*f*Sqrt[c - c *Sin[e + f*x]])
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)]
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]
Time = 7.50 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {12 \left (A \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {4 B \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+B \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) a \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{6 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-3 f}\) | \(151\) |
| parts | \(\frac {A \sqrt {4}\, a \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 B a \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \left (4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(194\) |
Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x,metho d=_RETURNVERBOSE)
Output:
12*(A*tan(1/4*Pi+1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2-1/2)*sin(1/4*Pi+1/2* f*x+1/2*e)^2+4/3*B*sin(1/2*f*x+1/2*e)^3*cos(1/2*f*x+1/2*e)^3+B*sin(1/2*f*x +1/2*e)^2*cos(1/2*f*x+1/2*e)^2)*a*(c*cos(1/4*Pi+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)/(6*f*cos(1/2*f*x+1/2*e)^2-3*f)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (3 \, {\left (A + B\right )} a \cos \left (f x + e\right )^{2} - 3 \, {\left (A + B\right )} a + 2 \, {\left (B a \cos \left (f x + e\right )^{2} - {\left (3 \, A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x , algorithm="fricas")
Output:
-1/6*(3*(A + B)*a*cos(f*x + e)^2 - 3*(A + B)*a + 2*(B*a*cos(f*x + e)^2 - ( 3*A + B)*a)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \] Input:
integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2) ,x)
Output:
Integral((a*(sin(e + f*x) + 1))**(3/2)*sqrt(-c*(sin(e + f*x) - 1))*(A + B* sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(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)^(3/2)*sqrt(-c*sin(f*x + e) + c), x)
Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.50 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (4 \, B a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, A a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{3 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x , algorithm="giac")
Output:
-2/3*(4*B*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^6*sgn(cos(-1/4*pi + 1/2*f*x + 1 /2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*A*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*a*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*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
Time = 37.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (3\,A\,\cos \left (e+f\,x\right )+3\,B\,\cos \left (e+f\,x\right )+3\,A\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (3\,e+3\,f\,x\right )-12\,A\,\sin \left (2\,e+2\,f\,x\right )-2\,B\,\sin \left (2\,e+2\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\right )}{12\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^( 1/2),x)
Output:
-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(3*A*cos(e + f*x) + 3*B*cos(e + f*x) + 3*A*cos(3*e + 3*f*x) + 3*B*cos(3*e + 3*f*x) - 12*A*sin(2*e + 2*f*x) - 2*B*sin(2*e + 2*f*x) + B*sin(4*e + 4*f*x)))/(12*f* (cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\sqrt {c}\, \sqrt {a}\, a \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)
Output:
sqrt(c)*sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*si n(e + f*x)**2,x)*b + 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)*s in(e + f*x),x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1),x) *a)