Integrand size = 23, antiderivative size = 145 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {75 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac {13 \cos (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {9 \cos (c+d x)}{4 a^2 d \sqrt {a+a \sin (c+d x)}} \] Output:
75/32*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/ 2)/a^(5/2)/d+1/4*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(5/2)-13/16*co s(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)-9/4*cos(d*x+c)/a^2/d/(a+a*sin(d*x+c))^ (1/2)
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (-45 \cos \left (\frac {1}{2} (c+d x)\right )-69 \cos \left (\frac {3}{2} (c+d x)\right )+16 \cos \left (\frac {5}{2} (c+d x)\right )+45 \sin \left (\frac {1}{2} (c+d x)\right )-(150+150 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4-69 \sin \left (\frac {3}{2} (c+d x)\right )-16 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{32 d (a (1+\sin (c+d x)))^{5/2}} \] Input:
Integrate[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]
Output:
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(-45*Cos[(c + d*x)/2] - 69*Cos[(3*( c + d*x))/2] + 16*Cos[(5*(c + d*x))/2] + 45*Sin[(c + d*x)/2] - (150 + 150* I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos [(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 69*Sin[(3*(c + d*x))/2] - 16*Sin[(5* (c + d*x))/2]))/(32*d*(a*(1 + Sin[c + d*x]))^(5/2))
Time = 0.77 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3498, 27, 3042, 3230, 3042, 3128, 219}
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.
| \(\displaystyle \int \frac {\sin ^3(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{(a \sin (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\int \frac {\sin (c+d x) (4 a-9 a \sin (c+d x))}{2 (\sin (c+d x) a+a)^{3/2}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\int \frac {\sin (c+d x) (4 a-9 a \sin (c+d x))}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\int \frac {\sin (c+d x) (4 a-9 a \sin (c+d x))}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\int \frac {4 a \sin (c+d x)-9 a \sin ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\int \frac {4 a \sin (c+d x)-9 a \sin (c+d x)^2}{(\sin (c+d x) a+a)^{3/2}}dx}{8 a^2}\) |
| \(\Big \downarrow \) 3498 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int -\frac {3 \left (13 a^2-12 a^2 \sin (c+d x)\right )}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}}{8 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \int \frac {13 a^2-12 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \int \frac {13 a^2-12 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \left (25 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {24 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \left (25 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {24 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \left (\frac {24 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {50 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}\right )}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}-\frac {\frac {3 \left (\frac {24 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {25 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}\right )}{4 a^2}+\frac {13 a \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}}{8 a^2}\) |
Input:
Int[Sin[c + d*x]^3/(a + a*Sin[c + d*x])^(5/2),x]
Output:
(Cos[c + d*x]*Sin[c + d*x]^2)/(4*d*(a + a*Sin[c + d*x])^(5/2)) - ((13*a*Co s[c + d*x])/(2*d*(a + a*Sin[c + d*x])^(3/2)) + (3*((-25*Sqrt[2]*a^(3/2)*Ar cTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (24* a^2*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]])))/(4*a^2))/(8*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
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[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.37 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.68
| method | result | size |
| default | \(\frac {\left (-75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \cos \left (d x +c \right )^{2}+64 a^{\frac {3}{2}} \sqrt {a -a \sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}+150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )-128 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )+150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+42 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-204 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(243\) |
Input:
int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/32/a^(9/2)*(-75*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/ 2))*a^2*cos(d*x+c)^2+64*a^(3/2)*(a-a*sin(d*x+c))^(1/2)*cos(d*x+c)^2+150*2^ (1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(d*x+c)-1 28*(a-a*sin(d*x+c))^(1/2)*a^(3/2)*sin(d*x+c)+150*2^(1/2)*arctanh(1/2*(a-a* sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+42*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-2 04*(a-a*sin(d*x+c))^(1/2)*a^(3/2))*(-a*(sin(d*x+c)-1))^(1/2)/(1+sin(d*x+c) )/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (122) = 244\).
Time = 0.10 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.35 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (32 \, \cos \left (d x + c\right )^{3} - 53 \, \cos \left (d x + c\right )^{2} - {\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 4\right )} \sin \left (d x + c\right ) - 81 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/64*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (cos(d*x + c)^2 - 2* cos(d*x + c) - 4)*sin(d*x + c) - 2*cos(d*x + c) - 4)*sqrt(a)*log(-(a*cos(d *x + c)^2 + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin (d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2) ) - 4*(32*cos(d*x + c)^3 - 53*cos(d*x + c)^2 - (32*cos(d*x + c)^2 + 85*cos (d*x + c) + 4)*sin(d*x + c) - 81*cos(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a ))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3/(a+a*sin(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate(sin(d*x + c)^3/(a*sin(d*x + c) + a)^(5/2), x)
Time = 0.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {128 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (21 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 19 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \] Input:
integrate(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/64*(75*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos (-1/4*pi + 1/2*d*x + 1/2*c))) - 75*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/ 2*c) + 1)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 128*sqrt(2)*sin( -1/4*pi + 1/2*d*x + 1/2*c)/(a^(5/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 2*sqrt(2)*(21*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 19*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^(5/2)*sgn(cos(-1/4* pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] 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)
\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sin \left (d x +c \right )^{3}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(sin(d*x+c)^3/(a+a*sin(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sin(c + d*x)**3)/(sin(c + d*x)**3 + 3 *sin(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3