Integrand size = 40, antiderivative size = 96 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {(A-B) c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}} \] Output:
1/3*(A-B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(1/2)+1/4 *B*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f/(c-c*sin(f*x+e))^(1/2)
Time = 2.84 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {a^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (3 B \cos (4 (e+f x))+16 (7 A+2 B) \sin (e+f x)-4 \cos (2 (e+f x)) (12 A+9 B+4 (A+2 B) \sin (e+f x)))}{96 f} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]],x]
Output:
(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(3*B *Cos[4*(e + f*x)] + 16*(7*A + 2*B)*Sin[e + f*x] - 4*Cos[2*(e + f*x)]*(12*A + 9*B + 4*(A + 2*B)*Sin[e + f*x])))/(96*f)
Time = 0.66 (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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3450 |
| \(\displaystyle (A-B) \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (A-B) \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {B \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx}{a}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {c (A-B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}+\frac {B c \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{4 a f \sqrt {c-c \sin (e+f x)}}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(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])^(5/2))/(3*f*Sqrt[c - c*Sin[e + f*x]]) + (B*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(4*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]
Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(84)=168\).
Time = 8.51 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.83
| method | result | size |
| default | \(\frac {16 a^{2} \left (A \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (-\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{4}+\sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{4}+\frac {3}{8}\right ) B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{6 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-3 f}\) | \(176\) |
| parts | \(\frac {4 A \sqrt {4}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2} a^{2} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{3 f}+\frac {2 B \,a^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (-6 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+6 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+8 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(243\) |
Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x,metho d=_RETURNVERBOSE)
Output:
16*a^2*(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)^4+2*sin(1/2*f*x+1/2*e)^2*(-3/4*cos(1/2*f*x+1/2*e)^4+sin(1/2 *f*x+1/2*e)*cos(1/2*f*x+1/2*e)+3/4*cos(1/2*f*x+1/2*e)^2+3/8)*B*cos(1/2*f*x +1/2*e)^2)*(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 = 0.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.22 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {{\left (3 \, B a^{2} \cos \left (f x + e\right )^{4} - 12 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right )^{2} + 3 \, {\left (4 \, A + 3 \, B\right )} a^{2} - 4 \, {\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x , algorithm="fricas")
Output:
1/12*(3*B*a^2*cos(f*x + e)^4 - 12*(A + B)*a^2*cos(f*x + e)^2 + 3*(4*A + 3* B)*a^2 - 4*((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(2*A + B)*a^2)*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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2) ,x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(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)^(5/2)*sqrt(-c*sin(f*x + e) + c), x)
Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.56 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {4 \, {\left (3 \, B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{3 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x , algorithm="giac")
Output:
-4/3*(3*B*a^2*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)) + 2*A*a^2*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)) - 2*B*a^2*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)))*sqrt(a)*sqrt(c)/f
Time = 3.15 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.55 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (48\,A\,\cos \left (e+f\,x\right )+36\,B\,\cos \left (e+f\,x\right )+48\,A\,\cos \left (3\,e+3\,f\,x\right )+33\,B\,\cos \left (3\,e+3\,f\,x\right )-3\,B\,\cos \left (5\,e+5\,f\,x\right )-112\,A\,\sin \left (2\,e+2\,f\,x\right )+8\,A\,\sin \left (4\,e+4\,f\,x\right )-32\,B\,\sin \left (2\,e+2\,f\,x\right )+16\,B\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^( 1/2),x)
Output:
-(a^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(48*A*cos (e + f*x) + 36*B*cos(e + f*x) + 48*A*cos(3*e + 3*f*x) + 33*B*cos(3*e + 3*f *x) - 3*B*cos(5*e + 5*f*x) - 112*A*sin(2*e + 2*f*x) + 8*A*sin(4*e + 4*f*x) - 32*B*sin(2*e + 2*f*x) + 16*B*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),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,x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2,x)*a + 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x ) + 1)*sin(e + f*x)**2,x)*b + 2*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)