Integrand size = 29, antiderivative size = 243 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {\sqrt {d} \left (15 c^2-10 c d+7 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt {d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{4 \sqrt {3} f}-\frac {\sqrt {\frac {2}{3}} (c-d)^{5/2} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \sqrt {c-d} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {(7 c-d) d \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {3+3 \sin (e+f x)}}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {3+3 \sin (e+f x)}} \]
-(c-d)^(5/2)*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))*2^(1/2)/f/a^(1/2)-1/4*(15*c^2-10*c*d+ 7*d^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))*d^(1/2)/f/a^(1/2)-1/2*d*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f/ (a+a*sin(f*x+e))^(1/2)-1/4*(7*c-d)*d*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/( a+a*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 15.47 (sec) , antiderivative size = 1893, normalized size of antiderivative = 7.79 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx =\text {Too large to display} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*((d*(-9*c + 2*d)*Cos[(e + f*x)/2])/4 - (d^2*Cos[(3*(e + f*x))/2])/4 - (d*(-9*c + 2*d )*Sin[(e + f*x)/2])/4 - (d^2*Sin[(3*(e + f*x))/2])/4))/(f*Sqrt[3 + 3*Sin[e + f*x]]) + ((Sqrt[2]*(c - d)^(5/2)*Log[1 + Tan[(e + f*x)/2]] - Sqrt[2]*(c - d)^(5/2)*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]] - (I/8)*Sqrt[d]*(15*c^2 - 10*c*d + 7*d^2)*Log[(2*(c - I*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x)/2])) /(d^(3/2)*(15*c^2 - 10*c*d + 7*d^2)*(I + Tan[(e + f*x)/2]))] + (I/8)*Sqrt[ d]*(15*c^2 - 10*c*d + 7*d^2)*Log[(2*(c + I*d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqr t[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d)*Tan[(e + f *x)/2]))/(d^(3/2)*(15*c^2 - 10*c*d + 7*d^2)*(-I + Tan[(e + f*x)/2]))])*(Co s[(e + f*x)/2] + Sin[(e + f*x)/2])*(c^3/((Cos[(e + f*x)/2] + Sin[(e + f*x) /2])*Sqrt[c + d*Sin[e + f*x]]) - (9*c^2*d)/(8*(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]]) - d^3/(8*(Cos[(e + f*x)/2] + Sin[ (e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]) + (15*c^2*d*Sin[e + f*x])/(8*(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])/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x] ]) + (7*d^3*Sin[e + f*x])/(8*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt...
Time = 1.42 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 3257, 25, 3042, 3462, 27, 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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {a \sin (e+f x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {a \sin (e+f x)+a}}dx\) |
| \(\Big \downarrow \) 3257 |
| \(\displaystyle -\frac {\int -\frac {\sqrt {c+d \sin (e+f x)} \left (a \left (4 c^2-d c+3 d^2\right )+a (7 c-d) d \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (4 c^2-d c+3 d^2\right )+a (7 c-d) d \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (4 c^2-d c+3 d^2\right )+a (7 c-d) d \sin (e+f x)\right )}{\sqrt {\sin (e+f x) a+a}}dx}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \frac {\frac {\int \frac {\left (8 c^3-9 d c^2+14 d^2 c-d^3\right ) a^2+d \left (15 c^2-10 d c+7 d^2\right ) \sin (e+f x) a^2}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (8 c^3-9 d c^2+14 d^2 c-d^3\right ) a^2+d \left (15 c^2-10 d c+7 d^2\right ) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (8 c^3-9 d c^2+14 d^2 c-d^3\right ) a^2+d \left (15 c^2-10 d c+7 d^2\right ) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3461 |
| \(\displaystyle \frac {\frac {8 a^2 (c-d)^3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+a d \left (15 c^2-10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {8 a^2 (c-d)^3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx+a d \left (15 c^2-10 c d+7 d^2\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {\frac {8 a^2 (c-d)^3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 a^2 d \left (15 c^2-10 c d+7 d^2\right ) \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {8 a^2 (c-d)^3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {2 a^{3/2} \sqrt {d} \left (15 c^2-10 c d+7 d^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {-\frac {16 a^3 (c-d)^3 \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {2 a^{3/2} \sqrt {d} \left (15 c^2-10 c d+7 d^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {-\frac {2 a^{3/2} \sqrt {d} \left (15 c^2-10 c d+7 d^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {8 \sqrt {2} a^{3/2} (c-d)^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{2 a}-\frac {a d (7 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}}{4 a}-\frac {d \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\) |
-1/2*(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*Sqrt[a + a*Sin[e + f*x ]]) + (((-2*a^(3/2)*Sqrt[d]*(15*c^2 - 10*c*d + 7*d^2)*ArcTan[(Sqrt[a]*Sqrt [d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (8*Sqrt[2]*a^(3/2)*(c - d)^(5/2)*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)/(2*a) - (a*(7*c - d)*d*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[a + a*Sin [e + f*x]]))/(4*a)
3.6.88.3.1 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[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[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*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] && GtQ[n, 1] && IntegerQ[2*n]
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])
\[\int \frac {\left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {a +a \sin \left (f x +e \right )}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (208) = 416\).
Time = 0.79 (sec) , antiderivative size = 2925, normalized size of antiderivative = 12.04 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\text {Too large to display} \]
[1/32*(16*sqrt(2)*(a*c^2 - 2*a*c*d + a*d^2 + (a*c^2 - 2*a*c*d + a*d^2)*cos (f*x + e) + (a*c^2 - 2*a*c*d + a*d^2)*sin(f*x + e))*sqrt((c - d)/a)*log((2 *sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(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)*sin(f*x + e) - cos(f*x + e) - 2)) + (15*a*c^2 - 10*a*c*d + 7*a*d^2 + (15*a*c^2 - 10*a*c*d + 7*a*d^2)*cos(f* x + e) + (15*a*c^2 - 10*a*c*d + 7*a*d^2)*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)*sqrt(d*sin(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 4 76*c*d^3 + 289*d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^ 2*d^2 - 6*c*d^3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*...
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]