Integrand size = 40, antiderivative size = 142 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {(5 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{30 f \sqrt {c-c \sin (e+f x)}}+\frac {(5 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{20 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f} \] Output:
1/30*(5*A+B)*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(1/2 )+1/20*(5*A+B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2)/ f-1/5*B*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(3/2)/f
Time = 9.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.16 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {c (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)} (4 (100 A+11 B) \sin (e+f x)+4 \cos (2 (e+f x)) (-15 (A+B)+4 (5 A-2 B) \sin (e+f x))-3 \cos (4 (e+f x)) (5 (A+B)+4 B \sin (e+f x)))}{480 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f *x])^(3/2),x]
Output:
-1/480*(c*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[ e + f*x]]*(4*(100*A + 11*B)*Sin[e + f*x] + 4*Cos[2*(e + f*x)]*(-15*(A + B) + 4*(5*A - 2*B)*Sin[e + f*x]) - 3*Cos[4*(e + f*x)]*(5*(A + B) + 4*B*Sin[e + f*x])))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
Time = 0.74 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3452, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle \frac {1}{5} (5 A+B) \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+B) \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {1}{5} (5 A+B) \left (\frac {1}{2} c \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+B) \left (\frac {1}{2} c \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {1}{5} (5 A+B) \left (\frac {c^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}{5 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 3/2),x]
Output:
-1/5*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2) )/f + ((5*A + B)*((c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*f*Sqrt[ c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]])/(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 = 7.54 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.67
| method | result | size |
| default | \(\frac {120 \left (\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) A \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {1}{3}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {8 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{8}+\frac {5 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{12}-\frac {5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{8}+\frac {5}{16}\right )}{5}\right ) c \,a^{2} \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{30 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-15 f}\) | \(237\) |
| parts | \(\frac {2 A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{2} c \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{3 f}+\frac {2 B c \,a^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \left (\left (48 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-48 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+20 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-30 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+15\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(273\) |
Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x,metho d=_RETURNVERBOSE)
Output:
120*((cos(1/2*f*x+1/2*e)^2-1/2)*A*(cos(1/4*Pi+1/2*f*x+1/2*e)^2+1/3)*tan(1/ 4*Pi+1/2*f*x+1/2*e)*sin(1/4*Pi+1/2*f*x+1/2*e)^4+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)+5/8*cos(1/2*f*x+1/2*e)^4+5/12*sin(1/2*f*x+1/2 *e)*cos(1/2*f*x+1/2*e)-5/8*cos(1/2*f*x+1/2*e)^2+5/16))*c*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 = 119, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (15 \, {\left (A + B\right )} a^{2} c \cos \left (f x + e\right )^{4} - 15 \, {\left (A + B\right )} a^{2} c + 4 \, {\left (3 \, B a^{2} c \cos \left (f x + e\right )^{4} - {\left (5 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A + B\right )} a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{60 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="fricas")
Output:
-1/60*(15*(A + B)*a^2*c*cos(f*x + e)^4 - 15*(A + B)*a^2*c + 4*(3*B*a^2*c*c os(f*x + e)^4 - (5*A + B)*a^2*c*cos(f*x + e)^2 - 2*(5*A + B)*a^2*c)*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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2) ,x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),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)^(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (124) = 248\).
Time = 0.32 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.40 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="giac")
Output:
-4/15*(24*B*a^2*c*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*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 - 75*B*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(s in(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8 + 40*A*a^2 *c*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 + 80*B*a^2*c*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 - 30*A*a^2*c*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)^4 - 30*B*a^2*c*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)^4)*sqrt(a)*sqrt(c)/f
Time = 38.58 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.23 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a^2\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (60\,A\,\cos \left (e+f\,x\right )+60\,B\,\cos \left (e+f\,x\right )+75\,A\,\cos \left (3\,e+3\,f\,x\right )+15\,A\,\cos \left (5\,e+5\,f\,x\right )+75\,B\,\cos \left (3\,e+3\,f\,x\right )+15\,B\,\cos \left (5\,e+5\,f\,x\right )-400\,A\,\sin \left (2\,e+2\,f\,x\right )-40\,A\,\sin \left (4\,e+4\,f\,x\right )-50\,B\,\sin \left (2\,e+2\,f\,x\right )+16\,B\,\sin \left (4\,e+4\,f\,x\right )+6\,B\,\sin \left (6\,e+6\,f\,x\right )\right )}{480\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^( 3/2),x)
Output:
-(a^2*c*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(60*A*c os(e + f*x) + 60*B*cos(e + f*x) + 75*A*cos(3*e + 3*f*x) + 15*A*cos(5*e + 5 *f*x) + 75*B*cos(3*e + 3*f*x) + 15*B*cos(5*e + 5*f*x) - 400*A*sin(2*e + 2* f*x) - 40*A*sin(4*e + 4*f*x) - 50*B*sin(2*e + 2*f*x) + 16*B*sin(4*e + 4*f* x) + 6*B*sin(6*e + 6*f*x)))/(480*f*(cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} c \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x)
Output:
sqrt(c)*sqrt(a)*a**2*c*( - 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 - 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 + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin( e + f*x) + 1)*sin(e + f*x)**2,x)*b + int(sqrt(sin(e + f*x) + 1)*sqrt( - si n(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)