Integrand size = 40, antiderivative size = 180 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 a^3 A \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f} \] Output:
-2/15*a^3*A*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(1/2)-1/5 *a^2*A*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(5/2)/f-1/5*a*A* cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/f-1/6*B*cos(f*x+e )*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(5/2)/f
Time = 5.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.64 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 c^2 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (50 B+75 B \cos (2 (e+f x))+30 B \cos (4 (e+f x))+5 B \cos (6 (e+f x))-600 A \sin (e+f x)-100 A \sin (3 (e+f x))-12 A \sin (5 (e+f x)))}{960 f} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f *x])^(5/2),x]
Output:
-1/960*(a^2*c^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(50*B + 75*B*Cos[2*(e + f*x)] + 30*B*Cos[4*(e + f*x)] + 5*B*Cos[6*( e + f*x)] - 600*A*Sin[e + f*x] - 100*A*Sin[3*(e + f*x)] - 12*A*Sin[5*(e + f*x)]))/f
Time = 0.94 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3452, 3042, 3219, 3042, 3219, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{5/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{5/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle A \left (\frac {4}{5} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}{6 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 5/2),x]
Output:
-1/6*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2) )/f + A*(-1/5*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f* x])^(5/2))/f + (4*a*(-1/6*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f *Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2))/(4*f)))/5)
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[((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 - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
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[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)) Int[( a + b*Sin[e + f*x])^m*(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)] && NeQ[m + n + 1, 0]
Time = 79.08 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.20
| method | result | size |
| default | \(\frac {192 c^{2} \left (A \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {1}{6}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {5 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {1}{4}\right ) B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {3}{4}\right )}{3}\right ) a^{2} \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{30 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-15 f}\) | \(216\) |
| parts | \(\frac {8 A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{2} c^{2} \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{15 f}+\frac {2 B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \left (16 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}-32 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+28 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a^{2} c^{2}}{3 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(271\) |
Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x,metho d=_RETURNVERBOSE)
Output:
192*c^2*(A*(cos(1/2*f*x+1/2*e)^2-1/2)*(cos(1/4*Pi+1/2*f*x+1/2*e)^4+1/2*cos (1/4*Pi+1/2*f*x+1/2*e)^2+1/6)*tan(1/4*Pi+1/2*f*x+1/2*e)*sin(1/4*Pi+1/2*f*x +1/2*e)^4+5/3*sin(1/2*f*x+1/2*e)^2*(cos(1/2*f*x+1/2*e)^4-1/2*cos(1/2*f*x+1 /2*e)^2+1/4)*B*cos(1/2*f*x+1/2*e)^2*(cos(1/2*f*x+1/2*e)^4-3/2*cos(1/2*f*x+ 1/2*e)^2+3/4))*a^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)/(30*f*cos(1/2*f*x+1/2*e)^2-15*f)
Time = 0.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.65 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {{\left (5 \, B a^{2} c^{2} \cos \left (f x + e\right )^{6} - 5 \, B a^{2} c^{2} - 2 \, {\left (3 \, A a^{2} c^{2} \cos \left (f x + e\right )^{4} + 4 \, A a^{2} c^{2} \cos \left (f x + e\right )^{2} + 8 \, A a^{2} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x , algorithm="fricas")
Output:
-1/30*(5*B*a^2*c^2*cos(f*x + e)^6 - 5*B*a^2*c^2 - 2*(3*A*a^2*c^2*cos(f*x + e)^4 + 4*A*a^2*c^2*cos(f*x + e)^2 + 8*A*a^2*c^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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2) ,x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),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)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (156) = 312\).
Time = 0.25 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.97 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x , algorithm="giac")
Output:
-16/15*(10*B*a^2*c^2*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)^12 - 6*A*a^2*c^2*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 - 30*B*a^2*c^2*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 + 15*A*a^2*c^2*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 + 30*B*a^2*c^2*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 - 10*A*a^2*c^2*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)^6 - 10*B*a^2*c ^2*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)^6)*sqrt(a)*sqrt(c)/f
Time = 38.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.73 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,B\,\cos \left (e+f\,x\right )+105\,B\,\cos \left (3\,e+3\,f\,x\right )+35\,B\,\cos \left (5\,e+5\,f\,x\right )+5\,B\,\cos \left (7\,e+7\,f\,x\right )-700\,A\,\sin \left (2\,e+2\,f\,x\right )-112\,A\,\sin \left (4\,e+4\,f\,x\right )-12\,A\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^( 5/2),x)
Output:
-(a^2*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(75*B *cos(e + f*x) + 105*B*cos(3*e + 3*f*x) + 35*B*cos(5*e + 5*f*x) + 5*B*cos(7 *e + 7*f*x) - 700*A*sin(2*e + 2*f*x) - 112*A*sin(4*e + 4*f*x) - 12*A*sin(6 *e + 6*f*x)))/(960*f*(cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} c^{2} \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) a -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x)
Output:
sqrt(c)*sqrt(a)*a**2*c**2*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**5,x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* x) + 1)*sin(e + f*x)**4,x)*a - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3,x)*b - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - si n(e + f*x) + 1)*sin(e + f*x)**2,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)