Integrand size = 35, antiderivative size = 165 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {8 a^2 (35 A c+21 B c+21 A d+19 B d) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a (35 A c+21 B c+21 A d+19 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 (7 B c+7 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 B d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 a f} \] Output:
-8/105*a^2*(35*A*c+21*A*d+21*B*c+19*B*d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/ 2)-2/105*a*(35*A*c+21*A*d+21*B*c+19*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2) /f-2/35*(7*A*d+7*B*c-2*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/7*B*d*co s(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f
Time = 3.65 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.87 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (700 A c+546 B c+546 A d+494 B d-6 (7 B c+7 A d+13 B d) \cos (2 (e+f x))+(140 A c+252 B c+252 A d+253 B d) \sin (e+f x)-15 B d \sin (3 (e+f x)))}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c + d*Sin[e + f *x]),x]
Output:
-1/210*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])] *(700*A*c + 546*B*c + 546*A*d + 494*B*d - 6*(7*B*c + 7*A*d + 13*B*d)*Cos[2 *(e + f*x)] + (140*A*c + 252*B*c + 252*A*d + 253*B*d)*Sin[e + f*x] - 15*B* d*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))
Time = 0.83 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3126, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))dx\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} \left ((A d+B c) \sin (e+f x)+A c+B d \sin ^2(e+f x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} \left ((A d+B c) \sin (e+f x)+A c+B d \sin (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {2 \int \frac {1}{2} (\sin (e+f x) a+a)^{3/2} (a (7 A c+5 B d)+a (7 B c+7 A d-2 B d) \sin (e+f x))dx}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (a (7 A c+5 B d)+a (7 B c+7 A d-2 B d) \sin (e+f x))dx}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (a (7 A c+5 B d)+a (7 B c+7 A d-2 B d) \sin (e+f x))dx}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {1}{5} a (35 A c+21 A d+21 B c+19 B d) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} a (35 A c+21 A d+21 B c+19 B d) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {\frac {1}{5} a (35 A c+21 A d+21 B c+19 B d) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{5} a (35 A c+21 A d+21 B c+19 B d) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {\frac {1}{5} a (35 A c+21 A d+21 B c+19 B d) \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a (7 A d+7 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}}{7 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 a f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]),x ]
Output:
(-2*B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*a*f) + ((-2*a*(7*B*c + 7*A*d - 2*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f) + (a*(35*A* c + 21*B*c + 21*A*d + 19*B*d)*((-8*a^2*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f)))/5)/(7*a)
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 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_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.97 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.91
method | result | size |
default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (15 B \sin \left (f x +e \right )^{3} d +21 A \sin \left (f x +e \right )^{2} d +21 B \sin \left (f x +e \right )^{2} c +39 B \sin \left (f x +e \right )^{2} d +35 A c \sin \left (f x +e \right )+63 A \sin \left (f x +e \right ) d +63 B \sin \left (f x +e \right ) c +52 B \sin \left (f x +e \right ) d +175 A c +126 A d +126 B c +104 B d \right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(150\) |
parts | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (A d +B c \right ) \left (\sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 A c \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+5\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B d \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (15 \sin \left (f x +e \right )^{3}+39 \sin \left (f x +e \right )^{2}+52 \sin \left (f x +e \right )+104\right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(201\) |
Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x,method=_RET URNVERBOSE)
Output:
2/105*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(15*B*sin(f*x+e)^3*d+21*A*sin(f*x+ e)^2*d+21*B*sin(f*x+e)^2*c+39*B*sin(f*x+e)^2*d+35*A*c*sin(f*x+e)+63*A*sin( f*x+e)*d+63*B*sin(f*x+e)*c+52*B*sin(f*x+e)*d+175*A*c+126*A*d+126*B*c+104*B *d)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.56 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {2 \, {\left (15 \, B a d \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, B a c + {\left (7 \, A + 13 \, B\right )} a d\right )} \cos \left (f x + e\right )^{3} - 28 \, {\left (5 \, A + 3 \, B\right )} a c - 4 \, {\left (21 \, A + 19 \, B\right )} a d - {\left (7 \, {\left (5 \, A + 6 \, B\right )} a c + {\left (42 \, A + 43 \, B\right )} a d\right )} \cos \left (f x + e\right )^{2} - {\left (7 \, {\left (25 \, A + 21 \, B\right )} a c + {\left (147 \, A + 143 \, B\right )} a d\right )} \cos \left (f x + e\right ) + {\left (15 \, B a d \cos \left (f x + e\right )^{3} + 28 \, {\left (5 \, A + 3 \, B\right )} a c + 4 \, {\left (21 \, A + 19 \, B\right )} a d - 3 \, {\left (7 \, B a c + {\left (7 \, A + 8 \, B\right )} a d\right )} \cos \left (f x + e\right )^{2} - {\left (7 \, {\left (5 \, A + 9 \, B\right )} a c + {\left (63 \, A + 67 \, B\right )} a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algo rithm="fricas")
Output:
2/105*(15*B*a*d*cos(f*x + e)^4 + 3*(7*B*a*c + (7*A + 13*B)*a*d)*cos(f*x + e)^3 - 28*(5*A + 3*B)*a*c - 4*(21*A + 19*B)*a*d - (7*(5*A + 6*B)*a*c + (42 *A + 43*B)*a*d)*cos(f*x + e)^2 - (7*(25*A + 21*B)*a*c + (147*A + 143*B)*a* d)*cos(f*x + e) + (15*B*a*d*cos(f*x + e)^3 + 28*(5*A + 3*B)*a*c + 4*(21*A + 19*B)*a*d - 3*(7*B*a*c + (7*A + 8*B)*a*d)*cos(f*x + e)^2 - (7*(5*A + 9*B )*a*c + (63*A + 67*B)*a*d)*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)
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )\, dx \] Input:
integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)
Output:
Integral((a*(sin(e + f*x) + 1))**(3/2)*(A + B*sin(e + f*x))*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algo rithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c), x)
Time = 0.21 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.73 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (15 \, B a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 105 \, {\left (12 \, A a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, A a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, B a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (4 \, A a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 6 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 6 \, A a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, B a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (2 \, B a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, A a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algo rithm="giac")
Output:
1/420*sqrt(2)*(15*B*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 7/2*f*x + 7/2*e) + 105*(12*A*a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 8*B *a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 8*A*a*d*sgn(cos(-1/4*pi + 1/2*f *x + 1/2*e)) + 7*B*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 35*(4*A*a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 6*B*a *c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 6*A*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*B*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/ 2*f*x + 3/2*e) + 21*(2*B*a*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*a*d *sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e))*sqrt(a)/f
Timed out. \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x)),x )
Output:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}d x \right ) a c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b c \right ) \] Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)
Output:
sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1),x)*a*c + int(sqrt(sin(e + f*x) + 1)* sin(e + f*x)**3,x)*b*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*a*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*b*c + int(sqrt(sin(e + f* x) + 1)*sin(e + f*x)**2,x)*b*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x),x )*a*c + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*a*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*b*c)