Integrand size = 29, antiderivative size = 242 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {(5 c-3 d) d^{3/2} \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{3 \sqrt {3} f}-\frac {(c-d)^{3/2} (c+9 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{6 \sqrt {6} f}+\frac {(c-3 d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{6 f \sqrt {3+3 \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (3+3 \sin (e+f x))^{3/2}} \]
-(5*c-3*d)*d^(3/2)*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2 )/(c+d*sin(f*x+e))^(1/2))/a^(3/2)/f-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^ (3/2)/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)^(3/2)*(c+9*d)*arctanh(1/2*cos(f*x +e)*a^(1/2)*(c-d)^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1 /2))/a^(3/2)/f*2^(1/2)+1/2*(c-3*d)*d*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/f /(a+a*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 15.35 (sec) , antiderivative size = 1844, normalized size of antiderivative = 7.62 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx =\text {Too large to display} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-(d^2*Cos[(e + f*x)/2]) + d^2*Si n[(e + f*x)/2] - (c - d)^2/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (c^ 2*Sin[(e + f*x)/2] - 2*c*d*Sin[(e + f*x)/2] + d^2*Sin[(e + f*x)/2])/(Cos[( e + f*x)/2] + Sin[(e + f*x)/2])^2)*Sqrt[c + d*Sin[e + f*x]])/(f*(3 + 3*Sin [e + f*x])^(3/2)) + ((((c - d)^(3/2)*(c + 9*d)*Log[1 + Tan[(e + f*x)/2]])/ Sqrt[2] + I*(5*c - 3*d)*d^(3/2)*Log[((-I)*((-I)*c + d + (1 - I)*Sqrt[2]*Sq rt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c - I*d)*T an[(e + f*x)/2]))/(d^(5/2)*(-5*c + 3*d)*(-I + Tan[(e + f*x)/2]))] + I*d^(3 /2)*(-5*c + 3*d)*Log[(I*(I*c + d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (c + I*d)*Tan[(e + f*x)/2]))/(d^ (5/2)*(-5*c + 3*d)*(I + Tan[(e + f*x)/2]))] - ((c - d)^(3/2)*(c + 9*d)*Log [c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f* x]] + (-c + d)*Tan[(e + f*x)/2]])/Sqrt[2])*(Cos[(e + f*x)/2] + Sin[(e + f* x)/2])^3*(c^3/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (7*c^2*d)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin [e + f*x]]) - (7*c*d^2)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (3*d^3)/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (5*c*d^2*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) - (3*d^3*Sin[e + f*x])/(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(3 + 3*Sin[e...
Time = 1.43 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3244, 27, 3042, 3462, 3042, 3461, 3042, 3254, 218, 3261, 221}
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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int -\frac {\sqrt {c+d \sin (e+f x)} \left (a \left (c^2+6 d c-3 d^2\right )-2 a (c-3 d) d \sin (e+f x)\right )}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (c^2+6 d c-3 d^2\right )-2 a (c-3 d) d \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (c^2+6 d c-3 d^2\right )-2 a (c-3 d) d \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \frac {\frac {\int \frac {\left (c^3+7 d c^2-7 d^2 c+3 d^3\right ) a^2+2 (5 c-3 d) d^2 \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (c^3+7 d c^2-7 d^2 c+3 d^3\right ) a^2+2 (5 c-3 d) d^2 \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3461 |
| \(\displaystyle \frac {\frac {a^2 (c+9 d) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+2 a d^2 (5 c-3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (c+9 d) (c-d)^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+2 a d^2 (5 c-3 d) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {\frac {a^2 (c-d)^2 (c+9 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {4 a^2 d^2 (5 c-3 d) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {a^2 (c-d)^2 (c+9 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {-\frac {2 a^3 (c-d)^2 (c+9 d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {-\frac {4 a^{3/2} d^{3/2} (5 c-3 d) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {\sqrt {2} a^{3/2} (c+9 d) (c-d)^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{a}+\frac {2 a d (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
-1/2*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f *x])^(3/2)) + (((-4*a^(3/2)*(5*c - 3*d)*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Co s[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (Sqr t[2]*a^(3/2)*(c - d)^(3/2)*(c + 9*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f)/a + (2*a*(c - 3*d)*d*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin [e + f*x]]))/(4*a^2)
3.6.94.3.1 Defintions of rubi rules used
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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*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]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[B/b Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] , x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 - b^2, 0] && NeQ[c^2 - d^2, 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[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
\[\int \frac {\left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (210) = 420\).
Time = 0.82 (sec) , antiderivative size = 3420, normalized size of antiderivative = 14.13 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
[1/8*(2*sqrt(1/2)*(2*a*c^2 + 16*a*c*d - 18*a*d^2 - (a*c^2 + 8*a*c*d - 9*a* d^2)*cos(f*x + e)^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e) + (2*a*c^2 + 16*a*c*d - 18*a*d^2 + (a*c^2 + 8*a*c*d - 9*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt((c - d)/a)*log(-(4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*si n(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) + (c - 3 *d)*cos(f*x + e)^2 + (3*c - d)*cos(f*x + e) - ((c - 3*d)*cos(f*x + e) - 2* c - 2*d)*sin(f*x + e) + 2*c + 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*si n(f*x + e) - cos(f*x + e) - 2)) + (10*a*c*d - 6*a*d^2 - (5*a*c*d - 3*a*d^2 )*cos(f*x + e)^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e) + (10*a*c*d - 6*a*d^2 + (5*a*c*d - 3*a*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/a)*log((128*d^4* cos(f*x + e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^ 2*d^2 + 4*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 32*(c^3*d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 + 8*(16*d^3*cos( f*x + e)^4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 31*c*d^2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2* d + 59*c*d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 14*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sq rt(d*sin(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c* d^3 + 289*d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d +...
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]