Integrand size = 36, antiderivative size = 157 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a (3 A-B) c^4 \cos ^3(e+f x)}{315 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a (3 A-B) c^3 \cos ^3(e+f x)}{105 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a (3 A-B) c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{21 f}-\frac {2 a B c \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f} \] Output:
64/315*a*(3*A-B)*c^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+16/105*a*(3*A-B )*c^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(1/2)+2/21*a*(3*A-B)*c^2*cos(f*x+e)^ 3*(c-c*sin(f*x+e))^(1/2)/f-2/9*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/f
Time = 4.71 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)} (-942 A+664 B+30 (3 A-8 B) \cos (2 (e+f x))+(648 A-741 B) \sin (e+f x)+35 B \sin (3 (e+f x)))}{630 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 5/2),x]
Output:
-1/630*(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*Sqrt[c - c*Sin[e + f *x]]*(-942*A + 664*B + 30*(3*A - 8*B)*Cos[2*(e + f*x)] + (648*A - 741*B)*S in[e + f*x] + 35*B*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x) /2]))
Time = 0.88 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3446, 3042, 3335, 3042, 3153, 3042, 3153, 3042, 3152}
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.
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a c \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \cos (e+f x)^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3335 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \int \cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}dx-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \left (\frac {8}{7} c \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \left (\frac {8}{7} c \int \cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\cos (e+f x)^2}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle a c \left (\frac {1}{3} (3 A-B) \left (\frac {8}{7} c \left (\frac {8 c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )-\frac {2 B \cos ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{9 f}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2),x ]
Output:
a*c*((-2*B*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(9*f) + ((3*A - B)*( (2*c*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(7*f) + (8*c*((8*c^2*Cos[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^(3/2)) + (2*c*Cos[e + f*x]^3)/(5*f*Sqr t[c - c*Sin[e + f*x]])))/7))/3)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 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] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 6.96 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66
| method | result | size |
| default | \(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{2} a \left (-35 B \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-45 A +120 B \right ) \cos \left (f x +e \right )^{2}+\left (-162 A +194 B \right ) \sin \left (f x +e \right )+258 A -226 B \right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(103\) |
| parts | \(-\frac {2 A a \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{2}-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (35 \sin \left (f x +e \right )^{4}-130 \sin \left (f x +e \right )^{3}+219 \sin \left (f x +e \right )^{2}-292 \sin \left (f x +e \right )+584\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{3}-12 \sin \left (f x +e \right )^{2}+23 \sin \left (f x +e \right )-46\right )}{21 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(235\) |
Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x,method=_RET URNVERBOSE)
Output:
-2/315*(sin(f*x+e)-1)*c^3*(1+sin(f*x+e))^2*a*(-35*B*cos(f*x+e)^2*sin(f*x+e )+(-45*A+120*B)*cos(f*x+e)^2+(-162*A+194*B)*sin(f*x+e)+258*A-226*B)/cos(f* x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.55 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {2 \, {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{5} + 5 \, {\left (9 \, A - 10 \, B\right )} a c^{2} \cos \left (f x + e\right )^{4} + {\left (117 \, A - 109 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} - 8 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2} + {\left (35 \, B a c^{2} \cos \left (f x + e\right )^{4} - 5 \, {\left (9 \, A - 17 \, B\right )} a c^{2} \cos \left (f x + e\right )^{3} + 24 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right )^{2} + 32 \, {\left (3 \, A - B\right )} a c^{2} \cos \left (f x + e\right ) + 64 \, {\left (3 \, A - B\right )} a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algo rithm="fricas")
Output:
2/315*(35*B*a*c^2*cos(f*x + e)^5 + 5*(9*A - 10*B)*a*c^2*cos(f*x + e)^4 + ( 117*A - 109*B)*a*c^2*cos(f*x + e)^3 - 8*(3*A - B)*a*c^2*cos(f*x + e)^2 + 3 2*(3*A - B)*a*c^2*cos(f*x + e) + 64*(3*A - B)*a*c^2 + (35*B*a*c^2*cos(f*x + e)^4 - 5*(9*A - 17*B)*a*c^2*cos(f*x + e)^3 + 24*(3*A - B)*a*c^2*cos(f*x + e)^2 + 32*(3*A - B)*a*c^2*cos(f*x + e) + 64*(3*A - B)*a*c^2)*sin(f*x + e ))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=a \left (\int A c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int \left (- A c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\right )\, dx + \int \left (- A c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int A c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\, dx + \int B c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- B c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- B c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int B c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2),x)
Output:
a*(Integral(A*c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(-A*c**2*sqrt(- c*sin(e + f*x) + c)*sin(e + f*x), x) + Integral(-A*c**2*sqrt(-c*sin(e + f* x) + c)*sin(e + f*x)**2, x) + Integral(A*c**2*sqrt(-c*sin(e + f*x) + c)*si n(e + f*x)**3, x) + Integral(B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) , x) + Integral(-B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + In tegral(-B*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3, x) + Integral(B* c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4, x))
\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algo rithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^ (5/2), x)
Time = 0.35 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.67 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (35 \, B a c^{2} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 630 \, {\left (5 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 210 \, {\left (A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) - 126 \, {\left (3 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 45 \, {\left (2 \, A a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B a c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right )\right )} \sqrt {c}}{2520 \, f} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x, algo rithm="giac")
Output:
-1/2520*sqrt(2)*(35*B*a*c^2*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 630*(5*A*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*B*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x + 1 /2*e) + 210*(A*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a*c^2*sgn(sin (-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-3/4*pi + 3/2*f*x + 3/2*e) - 126*(3*A*a* c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*a*c^2*sgn(sin(-1/4*pi + 1/2*f* x + 1/2*e)))*cos(-5/4*pi + 5/2*f*x + 5/2*e) + 45*(2*A*a*c^2*sgn(sin(-1/4*p i + 1/2*f*x + 1/2*e)) - 3*B*a*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos (-7/4*pi + 7/2*f*x + 7/2*e))*sqrt(c)/f
Timed out. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2),x )
Output:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2), x)
\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, a \,c^{2} \left (\left (\int \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a +\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b +\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b -\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b \right ) \] Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2),x)
Output:
sqrt(c)*a*c**2*(int(sqrt( - sin(e + f*x) + 1),x)*a + int(sqrt( - sin(e + f *x) + 1)*sin(e + f*x)**4,x)*b + int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x) **3,x)*a - int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3,x)*b - int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2,x)*a - int(sqrt( - sin(e + f*x) + 1)*s in(e + f*x)**2,x)*b - int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x)*a + in t(sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x)*b)