Integrand size = 40, antiderivative size = 134 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a^2 A \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 f \sqrt {a+a \sin (e+f x)}}-\frac {a A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f} \]
-1/4*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2)/f-1/3*a^2* A*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(1/2)-1/3*a*A*cos(f *x+e)*(c-c*sin(f*x+e))^(3/2)*(a+a*sin(f*x+e))^(1/2)/f
Time = 1.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.74 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {c \sec ^3(e+f x) (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (9 B+12 B \cos (2 (e+f x))+3 B \cos (4 (e+f x))-72 A \sin (e+f x)-8 A \sin (3 (e+f x)))}{96 f} \]
(c*Sec[e + f*x]^3*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(9*B + 12*B*Cos[2*(e + f*x)] + 3*B*Cos[4*(e + f*x)] - 72 *A*Sin[e + f*x] - 8*A*Sin[3*(e + f*x)]))/(96*f)
Time = 0.71 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {2}{3} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {2}{3} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle A \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}{3 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}{4 f}\) |
-1/4*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2) )/f + A*(-1/3*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(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])^(3/2))/(3*f))
3.2.41.3.1 Defintions of rubi rules used
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 = 2.84 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.60
| method | result | size |
| default | \(\frac {a c \tan \left (f x +e \right ) \left (3 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 A \left (\cos ^{2}\left (f x +e \right )\right )+3 B \sin \left (f x +e \right )+8 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{12 f}\) | \(80\) |
| parts | \(\frac {A \tan \left (f x +e \right ) c a \left (\cos ^{2}\left (f x +e \right )+2\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{3 f}+\frac {B \sin \left (f x +e \right ) \tan \left (f x +e \right ) c a \left (\cos ^{2}\left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{4 f}\) | \(106\) |
1/12*a*c/f*tan(f*x+e)*(3*B*cos(f*x+e)^2*sin(f*x+e)+4*A*cos(f*x+e)^2+3*B*si n(f*x+e)+8*A)*(-c*(sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.62 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (3 \, B a c \cos \left (f x + e\right )^{4} - 3 \, B a c - 4 \, {\left (A a c \cos \left (f x + e\right )^{2} + 2 \, A a c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="fricas")
-1/12*(3*B*a*c*cos(f*x + e)^4 - 3*B*a*c - 4*(A*a*c*cos(f*x + e)^2 + 2*A*a* 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. \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (116) = 232\).
Time = 0.50 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.77 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {4 \, {\left (3 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 2 \, A a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 6 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3 \, A a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 3 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="giac")
4/3*(3*B*a*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 - 2*A*a*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 - 6*B*a*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 + 3*A*a*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 + 3*B*a*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 = 2.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,B\,\cos \left (e+f\,x\right )+15\,B\,\cos \left (3\,e+3\,f\,x\right )+3\,B\,\cos \left (5\,e+5\,f\,x\right )-80\,A\,\sin \left (2\,e+2\,f\,x\right )-8\,A\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]