Integrand size = 23, antiderivative size = 203 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {284 a^3 \cos (c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}-\frac {710 a^3 \cos (c+d x) \sin ^3(c+d x)}{693 d \sqrt {a+a \sin (c+d x)}}-\frac {46 a^3 \cos (c+d x) \sin ^4(c+d x)}{99 d \sqrt {a+a \sin (c+d x)}}+\frac {568 a^2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{693 d}-\frac {2 a^2 \cos (c+d x) \sin ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{11 d}-\frac {284 a \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{231 d} \] Output:
-284/99*a^3*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-710/693*a^3*cos(d*x+c)*sin (d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)-46/99*a^3*cos(d*x+c)*sin(d*x+c)^4/d/(a+ a*sin(d*x+c))^(1/2)+568/693*a^2*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-2/11*a ^2*cos(d*x+c)*sin(d*x+c)^4*(a+a*sin(d*x+c))^(1/2)/d-284/231*a*cos(d*x+c)*( a+a*sin(d*x+c))^(3/2)/d
Time = 1.54 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.92 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-31878-40656 \cos (c+d x)+36352 \sqrt {2} \sqrt {1+\cos (c+d x)}-5313 \cos (2 (c+d x))+4752 \cos (3 (c+d x))+902 \cos (4 (c+d x))-448 \cos (5 (c+d x))-63 \cos (6 (c+d x))+23100 \sin (c+d x)-12243 \sin (2 (c+d x))-2178 \sin (3 (c+d x))+1672 \sin (4 (c+d x))+322 \sin (5 (c+d x))-63 \sin (6 (c+d x))\right )}{22176 d \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:
Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(a^2*Sec[(c + d*x)/2]^2*Sqrt[a*(1 + Sin[c + d*x])]*(-31878 - 40656*Cos[c + d*x] + 36352*Sqrt[2]*Sqrt[1 + Cos[c + d*x]] - 5313*Cos[2*(c + d*x)] + 475 2*Cos[3*(c + d*x)] + 902*Cos[4*(c + d*x)] - 448*Cos[5*(c + d*x)] - 63*Cos[ 6*(c + d*x)] + 23100*Sin[c + d*x] - 12243*Sin[2*(c + d*x)] - 2178*Sin[3*(c + d*x)] + 1672*Sin[4*(c + d*x)] + 322*Sin[5*(c + d*x)] - 63*Sin[6*(c + d* x)]))/(22176*d*(1 + Tan[(c + d*x)/2]))
Time = 1.06 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 3242, 27, 3042, 3460, 3042, 3249, 3042, 3238, 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.
| \(\displaystyle \int \sin ^3(c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^3 (a \sin (c+d x)+a)^{5/2}dx\) |
| \(\Big \downarrow \) 3242 |
| \(\displaystyle \frac {2}{11} \int \frac {1}{2} \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a} \left (23 \sin (c+d x) a^2+19 a^2\right )dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a} \left (23 \sin (c+d x) a^2+19 a^2\right )dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a} \left (23 \sin (c+d x) a^2+19 a^2\right )dx-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \int \sin ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \int \sin (c+d x)^3 \sqrt {\sin (c+d x) a+a}dx-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3249 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3238 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {1}{11} \left (\frac {355}{9} a^2 \left (\frac {6}{7} \left (\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\right )-\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {46 a^3 \sin ^4(c+d x) \cos (c+d x)}{9 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a^2 \sin ^4(c+d x) \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{11 d}\) |
Input:
Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2),x]
Output:
(-2*a^2*Cos[c + d*x]*Sin[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(11*d) + ((- 46*a^3*Cos[c + d*x]*Sin[c + d*x]^4)/(9*d*Sqrt[a + a*Sin[c + d*x]]) + (355* a^2*((-2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) + ( 6*((-2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*d) + ((-14*a^2*Cos[c + d*x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4*a*Cos[c + d*x]*Sqrt[a + a*Sin[ c + d*x]])/(3*d))/(5*a)))/7))/9)/11
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[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-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*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
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 = 3.74 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.47
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{3} \left (\sin \left (d x +c \right )-1\right ) \left (63 \sin \left (d x +c \right )^{5}+224 \sin \left (d x +c \right )^{4}+355 \sin \left (d x +c \right )^{3}+426 \sin \left (d x +c \right )^{2}+568 \sin \left (d x +c \right )+1136\right )}{693 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(95\) |
Input:
int(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/693*(1+sin(d*x+c))*a^3*(sin(d*x+c)-1)*(63*sin(d*x+c)^5+224*sin(d*x+c)^4+ 355*sin(d*x+c)^3+426*sin(d*x+c)^2+568*sin(d*x+c)+1136)/cos(d*x+c)/(a+a*sin (d*x+c))^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.95 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (63 \, a^{2} \cos \left (d x + c\right )^{6} + 224 \, a^{2} \cos \left (d x + c\right )^{5} - 320 \, a^{2} \cos \left (d x + c\right )^{4} - 874 \, a^{2} \cos \left (d x + c\right )^{3} + 593 \, a^{2} \cos \left (d x + c\right )^{2} + 1786 \, a^{2} \cos \left (d x + c\right ) + 800 \, a^{2} + {\left (63 \, a^{2} \cos \left (d x + c\right )^{5} - 161 \, a^{2} \cos \left (d x + c\right )^{4} - 481 \, a^{2} \cos \left (d x + c\right )^{3} + 393 \, a^{2} \cos \left (d x + c\right )^{2} + 986 \, a^{2} \cos \left (d x + c\right ) - 800 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{693 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:
integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/693*(63*a^2*cos(d*x + c)^6 + 224*a^2*cos(d*x + c)^5 - 320*a^2*cos(d*x + c)^4 - 874*a^2*cos(d*x + c)^3 + 593*a^2*cos(d*x + c)^2 + 1786*a^2*cos(d*x + c) + 800*a^2 + (63*a^2*cos(d*x + c)^5 - 161*a^2*cos(d*x + c)^4 - 481*a^ 2*cos(d*x + c)^3 + 393*a^2*cos(d*x + c)^2 + 986*a^2*cos(d*x + c) - 800*a^2 )*sin(d*x + c))*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3*(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sin \left (d x + c\right )^{3} \,d x } \] Input:
integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(d*x + c) + a)^(5/2)*sin(d*x + c)^3, x)
Time = 0.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.95 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {2} {\left (31878 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8778 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3465 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1287 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 385 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 63 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, d x + \frac {11}{2} \, c\right )\right )} \sqrt {a}}{11088 \, d} \] Input:
integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
1/11088*sqrt(2)*(31878*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) + 8778*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-3/ 4*pi + 3/2*d*x + 3/2*c) + 3465*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin (-5/4*pi + 5/2*d*x + 5/2*c) + 1287*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) *sin(-7/4*pi + 7/2*d*x + 7/2*c) + 385*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2* c))*sin(-9/4*pi + 9/2*d*x + 9/2*c) + 63*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/ 2*c))*sin(-11/4*pi + 11/2*d*x + 11/2*c))*sqrt(a)/d
Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:
int(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2),x)
Output:
int(sin(c + d*x)^3*(a + a*sin(c + d*x))^(5/2), x)
\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{5}d x +2 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{4}d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{3}d x \right ) \] Input:
int(sin(d*x+c)^3*(a+a*sin(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(c + d*x) + 1)*sin(c + d*x)**5,x) + 2*int(sqrt(s in(c + d*x) + 1)*sin(c + d*x)**4,x) + int(sqrt(sin(c + d*x) + 1)*sin(c + d *x)**3,x))