Integrand size = 40, antiderivative size = 94 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f \sqrt {a+a \sin (e+f x)}}+\frac {a B \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{5 c f \sqrt {a+a \sin (e+f x)}} \] Output:
-1/4*a*(A+B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)+1/ 5*a*B*cos(f*x+e)*(c-c*sin(f*x+e))^(9/2)/c/f/(a+a*sin(f*x+e))^(1/2)
Time = 3.73 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.26 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (4 (-60 A+23 B) \sin (e+f x)+4 \cos (2 (e+f x)) (-35 A+25 B+4 (5 A-6 B) \sin (e+f x))+\cos (4 (e+f x)) (5 A-15 B+4 B \sin (e+f x)))}{160 f} \] Input:
Integrate[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x ])^(7/2),x]
Output:
-1/160*(c^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] + 4*Cos[2*(e + f*x)]*(-35*A + 25*B + 4*( 5*A - 6*B)*Sin[e + f*x]) + Cos[4*(e + f*x)]*(5*A - 15*B + 4*B*Sin[e + f*x] )))/f
Time = 0.61 (sec) , antiderivative size = 94, 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.
| \(\displaystyle \int \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3450 |
| \(\displaystyle (A+B) \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {B \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{9/2}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A+B) \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {B \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{9/2}dx}{c}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {a B \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{5 c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f \sqrt {a \sin (e+f x)+a}}\) |
Input:
Int[Sqrt[a + a*Sin[e + f*x]]*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/ 2),x]
Output:
-1/4*(a*(A + B)*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(f*Sqrt[a + a*Sin [e + f*x]]) + (a*B*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(5*c*f*Sqrt[a + a*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]
Leaf count of result is larger than twice the leaf count of optimal. \(237\) vs. \(2(82)=164\).
Time = 8.32 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.53
| method | result | size |
| default | \(\frac {40 c^{3} \left (A \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+1\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {8 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8}-\frac {5 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{4}+\frac {15 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8}+\frac {5}{16}\right )}{5}\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{10 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-5 f}\) | \(238\) |
| parts | \(\frac {2 A \sqrt {4}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+1\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{3} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 B \,c^{3} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \left (\left (16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-20 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(271\) |
Input:
int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x,metho d=_RETURNVERBOSE)
Output:
40*c^3*(A*(cos(1/4*Pi+1/2*f*x+1/2*e)^2+1)*(cos(1/2*f*x+1/2*e)^2-1/2)*(cos( 1/4*Pi+1/2*f*x+1/2*e)^4+1)*tan(1/4*Pi+1/2*f*x+1/2*e)+8/5*sin(1/2*f*x+1/2*e )^2*B*cos(1/2*f*x+1/2*e)^2*(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))*(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)/(10*f*cos(1 /2*f*x+1/2*e)^2-5*f)
Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.49 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {{\left (5 \, {\left (A - 3 \, B\right )} c^{3} \cos \left (f x + e\right )^{4} - 40 \, {\left (A - B\right )} c^{3} \cos \left (f x + e\right )^{2} + 5 \, {\left (7 \, A - 5 \, B\right )} c^{3} + 4 \, {\left (B c^{3} \cos \left (f x + e\right )^{4} + {\left (5 \, A - 7 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A - 3 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{20 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="fricas")
Output:
-1/20*(5*(A - 3*B)*c^3*cos(f*x + e)^4 - 40*(A - B)*c^3*cos(f*x + e)^2 + 5* (7*A - 5*B)*c^3 + 4*(B*c^3*cos(f*x + e)^4 + (5*A - 7*B)*c^3*cos(f*x + e)^2 - 2*(5*A - 3*B)*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f *x + e) + c)/(f*cos(f*x + e))
Timed out. \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2) ,x)
Output:
Timed out
\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(7/2), x)
Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.60 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {4 \, {\left (8 \, B c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 5 \, A c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 5 \, B c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{5 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="giac")
Output:
-4/5*(8*B*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f* x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 - 5*A*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f* x + 1/2*e)^8 - 5*B*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)*sqrt(a)*sqrt(c)/f
Time = 40.07 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.84 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (100\,B\,\cos \left (e+f\,x\right )-140\,A\,\cos \left (e+f\,x\right )-135\,A\,\cos \left (3\,e+3\,f\,x\right )+5\,A\,\cos \left (5\,e+5\,f\,x\right )+85\,B\,\cos \left (3\,e+3\,f\,x\right )-15\,B\,\cos \left (5\,e+5\,f\,x\right )-240\,A\,\sin \left (2\,e+2\,f\,x\right )+40\,A\,\sin \left (4\,e+4\,f\,x\right )+90\,B\,\sin \left (2\,e+2\,f\,x\right )-48\,B\,\sin \left (4\,e+4\,f\,x\right )+2\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{160\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^( 7/2),x)
Output:
-(c^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(100*B*co s(e + f*x) - 140*A*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))
\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\sqrt {c}\, \sqrt {a}\, c^{3} \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b +3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a -3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b -3 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x)
Output:
sqrt(c)*sqrt(a)*c**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( - sin(e + f*x) + 1),x)*a)