Integrand size = 37, antiderivative size = 256 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=\frac {4 a (c+d) (B c-9 A d-8 B d) \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{315 d f \sqrt {a+a \sin (e+f x)}}+\frac {8 (5 c-d) (c+d) (B c-9 A d-8 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}+\frac {4 d (c+d) (B c-9 A d-8 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 a f}+\frac {2 a (B c-9 A d-8 B d) \cos (e+f x) (c+d \sin (e+f x))^3}{63 d f \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a+a \sin (e+f x)}} \]
4/105*d*(c+d)*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/a/f+4/3 15*a*(c+d)*(-9*A*d+B*c-8*B*d)*(15*c^2+10*c*d+7*d^2)*cos(f*x+e)/d/f/(a+a*si n(f*x+e))^(1/2)+2/63*a*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(c+d*sin(f*x+e))^3/d/ f/(a+a*sin(f*x+e))^(1/2)-2/9*a*B*cos(f*x+e)*(c+d*sin(f*x+e))^4/d/f/(a+a*si n(f*x+e))^(1/2)+8/315*(5*c-d)*(c+d)*(-9*A*d+B*c-8*B*d)*cos(f*x+e)*(a+a*sin (f*x+e))^(1/2)/f
Time = 1.30 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.19 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (2520 A c^3+1680 B c^3+5040 A c^2 d+4788 B c^2 d+4788 A c d^2+4104 B c d^2+1368 A d^3+1321 B d^3-4 d \left (27 A d (7 c+2 d)+B \left (189 c^2+162 c d+83 d^2\right )\right ) \cos (2 (e+f x))+35 B d^3 \cos (4 (e+f x))+840 B c^3 \sin (e+f x)+2520 A c^2 d \sin (e+f x)+2016 B c^2 d \sin (e+f x)+2016 A c d^2 \sin (e+f x)+2538 B c d^2 \sin (e+f x)+846 A d^3 \sin (e+f x)+752 B d^3 \sin (e+f x)-270 B c d^2 \sin (3 (e+f x))-90 A d^3 \sin (3 (e+f x))-80 B d^3 \sin (3 (e+f x))\right )}{1260 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
-1/1260*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]* (2520*A*c^3 + 1680*B*c^3 + 5040*A*c^2*d + 4788*B*c^2*d + 4788*A*c*d^2 + 41 04*B*c*d^2 + 1368*A*d^3 + 1321*B*d^3 - 4*d*(27*A*d*(7*c + 2*d) + B*(189*c^ 2 + 162*c*d + 83*d^2))*Cos[2*(e + f*x)] + 35*B*d^3*Cos[4*(e + f*x)] + 840* B*c^3*Sin[e + f*x] + 2520*A*c^2*d*Sin[e + f*x] + 2016*B*c^2*d*Sin[e + f*x] + 2016*A*c*d^2*Sin[e + f*x] + 2538*B*c*d^2*Sin[e + f*x] + 846*A*d^3*Sin[e + f*x] + 752*B*d^3*Sin[e + f*x] - 270*B*c*d^2*Sin[3*(e + f*x)] - 90*A*d^3 *Sin[3*(e + f*x)] - 80*B*d^3*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin [(e + f*x)/2]))
Time = 1.04 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3460, 3042, 3249, 3042, 3240, 27, 3042, 3230, 3042, 3125}
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.
\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \sin (e+f x)+a} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3dx\) |
\(\Big \downarrow \) 3460 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^3dx}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^3dx}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2dx-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \int \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2dx-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3240 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {2 \int \frac {1}{2} \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {\int \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {\int \sqrt {\sin (e+f x) a+a} \left (a \left (5 c^2+3 d^2\right )+2 a (5 c-d) d \sin (e+f x)\right )dx}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {\frac {1}{3} a \left (15 c^2+10 c d+7 d^2\right ) \int \sqrt {\sin (e+f x) a+a}dx-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {\frac {1}{3} a \left (15 c^2+10 c d+7 d^2\right ) \int \sqrt {\sin (e+f x) a+a}dx-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle -\frac {(-9 A d+B c-8 B d) \left (\frac {6}{7} (c+d) \left (\frac {-\frac {2 a^2 \left (15 c^2+10 c d+7 d^2\right ) \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {4 a d (5 c-d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}}{5 a}-\frac {2 d^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 a f}\right )-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^3}{7 f \sqrt {a \sin (e+f x)+a}}\right )}{9 d}-\frac {2 a B \cos (e+f x) (c+d \sin (e+f x))^4}{9 d f \sqrt {a \sin (e+f x)+a}}\) |
(-2*a*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(9*d*f*Sqrt[a + a*Sin[e + f*x ]]) - ((B*c - 9*A*d - 8*B*d)*((-2*a*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/( 7*f*Sqrt[a + a*Sin[e + f*x]]) + (6*(c + d)*((-2*d^2*Cos[e + f*x]*(a + a*Si n[e + f*x])^(3/2))/(5*a*f) + ((-2*a^2*(15*c^2 + 10*c*d + 7*d^2)*Cos[e + f* x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (4*a*(5*c - d)*d*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f))/(5*a)))/7))/(9*d)
3.3.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^2, x_Symbol] :> Simp[(-d^2)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ (m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^ m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && !LtQ[m, -1]
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] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) 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] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Time = 2.19 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (35 B \left (\cos ^{4}\left (f x +e \right )\right ) d^{3}+\left (-45 A \,d^{3}-135 d^{2} c B -40 d^{3} B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-189 d^{2} c A -54 A \,d^{3}-189 c^{2} d B -162 d^{2} c B -118 d^{3} B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (315 c^{2} d A +252 d^{2} c A +117 A \,d^{3}+105 B \,c^{3}+252 c^{2} d B +351 d^{2} c B +104 d^{3} B \right ) \sin \left (f x +e \right )+315 A \,c^{3}+630 c^{2} d A +693 d^{2} c A +198 A \,d^{3}+210 B \,c^{3}+693 c^{2} d B +594 d^{2} c B +211 d^{3} B \right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(242\) |
parts | \(\frac {2 A \,c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) a}{\cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 c^{2} \left (3 d A +B c \right ) \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+2\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 d^{2} \left (d A +3 B c \right ) \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (5 \left (\sin ^{3}\left (f x +e \right )\right )+6 \left (\sin ^{2}\left (f x +e \right )\right )+8 \sin \left (f x +e \right )+16\right )}{35 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 d^{3} B \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (35 \left (\sin ^{4}\left (f x +e \right )\right )+40 \left (\sin ^{3}\left (f x +e \right )\right )+48 \left (\sin ^{2}\left (f x +e \right )\right )+64 \sin \left (f x +e \right )+128\right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 c d \left (d A +B c \right ) \left (1+\sin \left (f x +e \right )\right ) a \left (\sin \left (f x +e \right )-1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )+4 \sin \left (f x +e \right )+8\right )}{5 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(349\) |
2/315*(1+sin(f*x+e))*a*(sin(f*x+e)-1)*(35*B*cos(f*x+e)^4*d^3+(-45*A*d^3-13 5*B*c*d^2-40*B*d^3)*cos(f*x+e)^2*sin(f*x+e)+(-189*A*c*d^2-54*A*d^3-189*B*c ^2*d-162*B*c*d^2-118*B*d^3)*cos(f*x+e)^2+(315*A*c^2*d+252*A*c*d^2+117*A*d^ 3+105*B*c^3+252*B*c^2*d+351*B*c*d^2+104*B*d^3)*sin(f*x+e)+315*A*c^3+630*c^ 2*d*A+693*d^2*c*A+198*A*d^3+210*B*c^3+693*c^2*d*B+594*d^2*c*B+211*d^3*B)/c os(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Time = 0.27 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.82 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=-\frac {2 \, {\left (35 \, B d^{3} \cos \left (f x + e\right )^{5} - 5 \, {\left (27 \, B c d^{2} + {\left (9 \, A + B\right )} d^{3}\right )} \cos \left (f x + e\right )^{4} + 105 \, {\left (3 \, A + B\right )} c^{3} + 63 \, {\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \, {\left (49 \, A + 27 \, B\right )} c d^{2} + {\left (81 \, A + 107 \, B\right )} d^{3} - {\left (189 \, B c^{2} d + 27 \, {\left (7 \, A + 6 \, B\right )} c d^{2} + 2 \, {\left (27 \, A + 59 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (105 \, B c^{3} + 63 \, {\left (5 \, A + B\right )} c^{2} d + 9 \, {\left (7 \, A + 36 \, B\right )} c d^{2} + 2 \, {\left (54 \, A + 13 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (105 \, {\left (3 \, A + 2 \, B\right )} c^{3} + 63 \, {\left (10 \, A + 11 \, B\right )} c^{2} d + 99 \, {\left (7 \, A + 6 \, B\right )} c d^{2} + {\left (198 \, A + 211 \, B\right )} d^{3}\right )} \cos \left (f x + e\right ) - {\left (35 \, B d^{3} \cos \left (f x + e\right )^{4} + 105 \, {\left (3 \, A + B\right )} c^{3} + 63 \, {\left (5 \, A + 7 \, B\right )} c^{2} d + 9 \, {\left (49 \, A + 27 \, B\right )} c d^{2} + {\left (81 \, A + 107 \, B\right )} d^{3} + 5 \, {\left (27 \, B c d^{2} + {\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (63 \, B c^{2} d + 9 \, {\left (7 \, A + B\right )} c d^{2} + {\left (3 \, A + 26 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (105 \, B c^{3} + 63 \, {\left (5 \, A + 4 \, B\right )} c^{2} d + 9 \, {\left (28 \, A + 39 \, B\right )} c d^{2} + 13 \, {\left (9 \, A + 8 \, B\right )} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{315 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]
-2/315*(35*B*d^3*cos(f*x + e)^5 - 5*(27*B*c*d^2 + (9*A + B)*d^3)*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A + 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A + 107*B)*d^3 - (189*B*c^2*d + 27*(7*A + 6*B)*c*d^2 + 2*(27*A + 59*B )*d^3)*cos(f*x + e)^3 + (105*B*c^3 + 63*(5*A + B)*c^2*d + 9*(7*A + 36*B)*c *d^2 + 2*(54*A + 13*B)*d^3)*cos(f*x + e)^2 + (105*(3*A + 2*B)*c^3 + 63*(10 *A + 11*B)*c^2*d + 99*(7*A + 6*B)*c*d^2 + (198*A + 211*B)*d^3)*cos(f*x + e ) - (35*B*d^3*cos(f*x + e)^4 + 105*(3*A + B)*c^3 + 63*(5*A + 7*B)*c^2*d + 9*(49*A + 27*B)*c*d^2 + (81*A + 107*B)*d^3 + 5*(27*B*c*d^2 + (9*A + 8*B)*d ^3)*cos(f*x + e)^3 - 3*(63*B*c^2*d + 9*(7*A + B)*c*d^2 + (3*A + 26*B)*d^3) *cos(f*x + e)^2 - (105*B*c^3 + 63*(5*A + 4*B)*c^2*d + 9*(28*A + 39*B)*c*d^ 2 + 13*(9*A + 8*B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)
\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{3}\, dx \]
\[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}^{3} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (236) = 472\).
Time = 0.40 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.15 \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=\text {Too large to display} \]
1/2520*sqrt(2)*(35*B*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-9/4*pi + 9/2*f*x + 9/2*e) + 630*(8*A*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B *c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*A*c^2*d*sgn(cos(-1/4*pi + 1/ 2*f*x + 1/2*e)) + 12*B*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*A*c* d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*B*c*d^2*sgn(cos(-1/4*pi + 1/2* f*x + 1/2*e)) + 3*A*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*d^3*sgn( cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 210*(4*B *c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*A*c^2*d*sgn(cos(-1/4*pi + 1/ 2*f*x + 1/2*e)) + 6*B*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 6*A*c*d^ 2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 9*B*c*d^2*sgn(cos(-1/4*pi + 1/2*f* x + 1/2*e)) + 3*A*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*d^3*sgn(co s(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e) + 126*(6*B*c ^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 6*A*c*d^2*sgn(cos(-1/4*pi + 1/2 *f*x + 1/2*e)) + 3*B*c*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + A*d^3*sgn (cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 *e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e) + 45*(6*B*c*d^2*sgn(cos(-1/4*pi + 1/2 *f*x + 1/2*e)) + 2*A*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + B*d^3*sgn(c os(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-7/4*pi + 7/2*f*x + 7/2*e))*sqrt(a)/f
Timed out. \[ \int \sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3 \,d x \]