Integrand size = 40, antiderivative size = 142 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {(3 A+B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{30 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 A+B) c \cos (e+f x) (a+a \sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)}}{15 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f} \] Output:
1/30*(3*A+B)*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/f/(c-c*sin(f*x+e))^(1/2 )+1/15*(3*A+B)*c*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(1/2)/ f-1/6*B*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(3/2)/f
Time = 10.34 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.49 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a^3 c (-1+\sin (e+f x)) (1+\sin (e+f x))^3 \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-15 (16 A+11 B) \cos (2 (e+f x))-30 (2 A+B) \cos (4 (e+f x))+5 B \cos (6 (e+f x))+840 A \sin (e+f x)+240 B \sin (e+f x)+60 A \sin (3 (e+f x))-40 B \sin (3 (e+f x))-12 A \sin (5 (e+f x))-24 B \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \] Input:
Integrate[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f *x])^(3/2),x]
Output:
-1/960*(a^3*c*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-15*(16*A + 11*B)*Cos[2*(e + f*x)] - 30* (2*A + B)*Cos[4*(e + f*x)] + 5*B*Cos[6*(e + f*x)] + 840*A*Sin[e + f*x] + 2 40*B*Sin[e + f*x] + 60*A*Sin[3*(e + f*x)] - 40*B*Sin[3*(e + f*x)] - 12*A*S in[5*(e + f*x)] - 24*B*Sin[5*(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])^7)
Time = 0.75 (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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle \frac {1}{3} (3 A+B) \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} (3 A+B) \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{3/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {1}{3} (3 A+B) \left (\frac {2}{5} c \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} (3 A+B) \left (\frac {2}{5} c \int (\sin (e+f x) a+a)^{7/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {1}{3} (3 A+B) \left (\frac {c^2 \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{7/2} \sqrt {c-c \sin (e+f x)}}{5 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{3/2}}{6 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(7/2)*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^( 3/2),x]
Output:
-1/6*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(3/2) )/f + ((3*A + B)*((c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*Sqrt [c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*Sqrt[c - c*Sin[e + f*x]])/(5*f)))/3
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]
Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(124)=248\).
Time = 8.22 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.76
| method | result | size |
| default | \(\frac {192 c \,a^{3} \left (A \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {1}{4}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} B \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (-\frac {5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{6}+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3}-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {5 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{6}+\frac {5 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{12}+\frac {5}{32}\right )\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{30 f \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-15 f}\) | \(250\) |
| parts | \(\frac {4 A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{3} c \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )}{5 f}+\frac {2 B \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\left (96 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-96 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+40 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-80 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+160 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-80 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+15\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a^{3} c}{15 f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(286\) |
Input:
int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x,metho d=_RETURNVERBOSE)
Output:
192*c*a^3*(A*(cos(1/4*Pi+1/2*f*x+1/2*e)^2+1/4)*(cos(1/2*f*x+1/2*e)^2-1/2)* tan(1/4*Pi+1/2*f*x+1/2*e)*sin(1/4*Pi+1/2*f*x+1/2*e)^6+2*sin(1/2*f*x+1/2*e) ^2*B*cos(1/2*f*x+1/2*e)^2*(-5/6*cos(1/2*f*x+1/2*e)^8+cos(1/2*f*x+1/2*e)^5* sin(1/2*f*x+1/2*e)+5/3*cos(1/2*f*x+1/2*e)^6-cos(1/2*f*x+1/2*e)^3*sin(1/2*f *x+1/2*e)-5/6*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)+5/32))*(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 = 142, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {{\left (5 \, B a^{3} c \cos \left (f x + e\right )^{6} - 15 \, {\left (A + B\right )} a^{3} c \cos \left (f x + e\right )^{4} + 5 \, {\left (3 \, A + 2 \, B\right )} a^{3} c - 2 \, {\left (3 \, {\left (A + 2 \, B\right )} a^{3} c \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, A + B\right )} a^{3} c \cos \left (f x + e\right )^{2} - 4 \, {\left (3 \, A + B\right )} a^{3} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="fricas")
Output:
1/30*(5*B*a^3*c*cos(f*x + e)^6 - 15*(A + B)*a^3*c*cos(f*x + e)^4 + 5*(3*A + 2*B)*a^3*c - 2*(3*(A + 2*B)*a^3*c*cos(f*x + e)^4 - 2*(3*A + B)*a^3*c*cos (f*x + e)^2 - 4*(3*A + B)*a^3*c)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq rt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))
Timed out. \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(3/2) ,x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(7/2)*(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)^(7/2)*(-c*sin(f*x + e) + c)^(3/2), x)
Time = 0.27 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.74 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 \, {\left (20 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 12 \, A a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 36 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 15 \, A a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 15 \, B a^{3} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{15 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x , algorithm="giac")
Output:
8/15*(20*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^12*sgn(cos(-1/4*pi + 1/2*f *x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 12*A*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 36*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^10*sgn(cos (-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 15*A*a^ 3*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 15*B*a^3*c*cos(-1/4*pi + 1/2*f*x + 1/ 2*e)^8*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2 *e)))*sqrt(a)*sqrt(c)/f
Time = 41.28 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.26 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\left (2\,A+B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{16\,f}-\frac {B\,a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (A\,7{}\mathrm {i}+B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (16\,A+11\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\left (A\,3{}\mathrm {i}-B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{24\,f}-\frac {a^3\,c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\left (A\,1{}\mathrm {i}+B\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{40\,f}\right )}{2\,\cos \left (e+f\,x\right )} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(7/2)*(c - c*sin(e + f*x))^( 3/2),x)
Output:
-(exp(- e*6i - f*x*6i)*(c - c*sin(e + f*x))^(1/2)*((a^3*c*exp(e*6i + f*x*6 i)*cos(4*e + 4*f*x)*(2*A + B)*(a + a*sin(e + f*x))^(1/2))/(16*f) - (B*a^3* c*exp(e*6i + f*x*6i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2))/(96*f) + (a^3*c*exp(e*6i + f*x*6i)*sin(e + f*x)*(A*7i + B*2i)*(a + a*sin(e + f*x)) ^(1/2)*1i)/(4*f) + (a^3*c*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(16*A + 11*B )*(a + a*sin(e + f*x))^(1/2))/(32*f) + (a^3*c*exp(e*6i + f*x*6i)*sin(3*e + 3*f*x)*(A*3i - B*2i)*(a + a*sin(e + f*x))^(1/2)*1i)/(24*f) - (a^3*c*exp(e *6i + f*x*6i)*sin(5*e + 5*f*x)*(A*1i + B*2i)*(a + a*sin(e + f*x))^(1/2)*1i )/(40*f)))/(2*cos(e + f*x))
\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{3} c \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}d x \right ) b -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) a -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b -2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) a \right ) \] Input:
int((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(3/2),x)
Output:
sqrt(c)*sqrt(a)*a**3*c*( - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**5,x)*b - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* x) + 1)*sin(e + f*x)**4,x)*a - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**4,x)*b - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - si n(e + f*x) + 1)*sin(e + f*x)**3,x)*a + 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2,x)*b + 2*int(sqrt(sin(e + f*x) + 1)*sq rt( - sin(e + f*x) + 1)*sin(e + f*x),x)*a + int(sqrt(sin(e + f*x) + 1)*sqr t( - 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)