Integrand size = 35, antiderivative size = 212 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {64 a^3 (21 A c+15 B c+15 A d+13 B d) \cos (e+f x)}{315 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{315 f}-\frac {2 a (21 A c+15 B c+15 A d+13 B d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{105 f}-\frac {2 (9 B c+9 A d-2 B d) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{63 f}-\frac {2 B d \cos (e+f x) (a+a \sin (e+f x))^{7/2}}{9 a f} \] Output:
-64/315*a^3*(21*A*c+15*A*d+15*B*c+13*B*d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1 /2)-16/315*a^2*(21*A*c+15*A*d+15*B*c+13*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^( 1/2)/f-2/105*a*(21*A*c+15*A*d+15*B*c+13*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^( 3/2)/f-2/63*(9*A*d+9*B*c-2*B*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f-2/9*B* d*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/f
Time = 8.01 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.95 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (7476 A c+6240 B c+6240 A d+5653 B d-4 (63 A c+180 B c+180 A d+254 B d) \cos (2 (e+f x))+35 B d \cos (4 (e+f x))+2352 A c \sin (e+f x)+3030 B c \sin (e+f x)+3030 A d \sin (e+f x)+3116 B d \sin (e+f x)-90 B c \sin (3 (e+f x))-90 A d \sin (3 (e+f x))-260 B d \sin (3 (e+f x)))}{1260 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x])*(c + d*Sin[e + f *x]),x]
Output:
-1/1260*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x ])]*(7476*A*c + 6240*B*c + 6240*A*d + 5653*B*d - 4*(63*A*c + 180*B*c + 180 *A*d + 254*B*d)*Cos[2*(e + f*x)] + 35*B*d*Cos[4*(e + f*x)] + 2352*A*c*Sin[ e + f*x] + 3030*B*c*Sin[e + f*x] + 3030*A*d*Sin[e + f*x] + 3116*B*d*Sin[e + f*x] - 90*B*c*Sin[3*(e + f*x)] - 90*A*d*Sin[3*(e + f*x)] - 260*B*d*Sin[3 *(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))
Time = 0.94 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3126, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \left ((A d+B c) \sin (e+f x)+A c+B d \sin ^2(e+f x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} \left ((A d+B c) \sin (e+f x)+A c+B d \sin (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2 \int \frac {1}{2} (\sin (e+f x) a+a)^{5/2} (a (9 A c+7 B d)+a (9 B c+9 A d-2 B d) \sin (e+f x))dx}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (a (9 A c+7 B d)+a (9 B c+9 A d-2 B d) \sin (e+f x))dx}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (a (9 A c+7 B d)+a (9 B c+9 A d-2 B d) \sin (e+f x))dx}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \int (\sin (e+f x) a+a)^{5/2}dx-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \int (\sin (e+f x) a+a)^{5/2}dx-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \left (\frac {8}{5} a \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \left (\frac {8}{5} a \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A c+15 A d+15 B c+13 B d) \left (\frac {8}{5} a \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 a (9 A d+9 B c-2 B d) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}}{9 a}-\frac {2 B d \cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{9 a f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(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])^(7/2))/(9*a*f) + ((-2*a*(9*B*c + 9*A*d - 2*B*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*f) + (3*a*(21* A*c + 15*B*c + 15*A*d + 13*B*d)*((-2*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^( 3/2))/(5*f) + (8*a*((-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)/(9*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 = 7.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (35 B d \cos \left (f x +e \right )^{4}+\left (-45 A d -45 B c -130 B d \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-63 A c -180 A d -180 B c -289 B d \right ) \cos \left (f x +e \right )^{2}+\left (294 A c +390 A d +390 B c +422 B d \right ) \sin \left (f x +e \right )+966 A c +870 A d +870 B c +838 B d \right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(152\) |
| parts | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (A d +B c \right ) \left (3 \sin \left (f x +e \right )^{3}+12 \sin \left (f x +e \right )^{2}+23 \sin \left (f x +e \right )+46\right )}{21 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 A c \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (3 \sin \left (f x +e \right )^{2}+14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B d \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (35 \sin \left (f x +e \right )^{4}+130 \sin \left (f x +e \right )^{3}+219 \sin \left (f x +e \right )^{2}+292 \sin \left (f x +e \right )+584\right )}{315 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(235\) |
Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x,method=_RET URNVERBOSE)
Output:
2/315*(1+sin(f*x+e))*a^3*(sin(f*x+e)-1)*(35*B*d*cos(f*x+e)^4+(-45*A*d-45*B *c-130*B*d)*cos(f*x+e)^2*sin(f*x+e)+(-63*A*c-180*A*d-180*B*c-289*B*d)*cos( f*x+e)^2+(294*A*c+390*A*d+390*B*c+422*B*d)*sin(f*x+e)+966*A*c+870*A*d+870* B*c+838*B*d)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Time = 0.11 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.70 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=-\frac {2 \, {\left (35 \, B a^{2} d \cos \left (f x + e\right )^{5} - 5 \, {\left (9 \, B a^{2} c + {\left (9 \, A + 19 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{4} + 96 \, {\left (7 \, A + 5 \, B\right )} a^{2} c + 32 \, {\left (15 \, A + 13 \, B\right )} a^{2} d - {\left (9 \, {\left (7 \, A + 20 \, B\right )} a^{2} c + {\left (180 \, A + 289 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, {\left (77 \, A + 85 \, B\right )} a^{2} c + {\left (255 \, A + 263 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (3 \, {\left (161 \, A + 145 \, B\right )} a^{2} c + {\left (435 \, A + 419 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right ) - {\left (35 \, B a^{2} d \cos \left (f x + e\right )^{4} + 96 \, {\left (7 \, A + 5 \, B\right )} a^{2} c + 32 \, {\left (15 \, A + 13 \, B\right )} a^{2} d + 5 \, {\left (9 \, B a^{2} c + {\left (9 \, A + 26 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, {\left (7 \, A + 15 \, B\right )} a^{2} c + {\left (45 \, A + 53 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, {\left (49 \, A + 65 \, B\right )} a^{2} c + {\left (195 \, A + 211 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{315 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algo rithm="fricas")
Output:
-2/315*(35*B*a^2*d*cos(f*x + e)^5 - 5*(9*B*a^2*c + (9*A + 19*B)*a^2*d)*cos (f*x + e)^4 + 96*(7*A + 5*B)*a^2*c + 32*(15*A + 13*B)*a^2*d - (9*(7*A + 20 *B)*a^2*c + (180*A + 289*B)*a^2*d)*cos(f*x + e)^3 + (3*(77*A + 85*B)*a^2*c + (255*A + 263*B)*a^2*d)*cos(f*x + e)^2 + 2*(3*(161*A + 145*B)*a^2*c + (4 35*A + 419*B)*a^2*d)*cos(f*x + e) - (35*B*a^2*d*cos(f*x + e)^4 + 96*(7*A + 5*B)*a^2*c + 32*(15*A + 13*B)*a^2*d + 5*(9*B*a^2*c + (9*A + 26*B)*a^2*d)* cos(f*x + e)^3 - 3*(3*(7*A + 15*B)*a^2*c + (45*A + 53*B)*a^2*d)*cos(f*x + e)^2 - 2*(3*(49*A + 65*B)*a^2*c + (195*A + 211*B)*a^2*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)
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (A + B \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )\, dx \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)
Output:
Integral((a*(sin(e + f*x) + 1))**(5/2)*(A + B*sin(e + f*x))*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(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)^(5/2)*(d*sin(f*x + e) + c), x)
Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (192) = 384\).
Time = 0.30 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.91 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algo rithm="giac")
Output:
1/2520*sqrt(2)*(35*B*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-9/4*pi + 9/2*f*x + 9/2*e) + 630*(20*A*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 15*B*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 15*A*a^2*d*sgn(cos(-1/4 *pi + 1/2*f*x + 1/2*e)) + 13*B*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))* sin(-1/4*pi + 1/2*f*x + 1/2*e) + 210*(10*A*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 11*B*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 11*A*a^2*d*s gn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 10*B*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e) + 126*(2*A*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*B*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*A *a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 6*B*a^2*d*sgn(cos(-1/4*pi + 1 /2*f*x + 1/2*e)))*sin(-5/4*pi + 5/2*f*x + 5/2*e) + 45*(2*B*a^2*c*sgn(cos(- 1/4*pi + 1/2*f*x + 1/2*e)) + 2*A*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*B*a^2*d*sgn(cos(-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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)),x )
Output:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}d x \right ) a c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right ) b d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) a d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b c +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) b d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a c +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) a d +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) b d +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) a d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) b c \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1),x)*a*c + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**4,x)*b*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**3,x)* a*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**3,x)*b*c + 2*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*c + 2*int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*a*d + 2*in t(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 + 2*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)