Integrand size = 27, antiderivative size = 247 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=-\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{27 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (3+3 \sin (e+f x))^2}-\frac {\left (c^2+5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{27 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+5 d) \left (c^2-d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{27 f \sqrt {c+d \sin (e+f x)}} \]
-1/3*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^2-1/3*(c-d )*(c+5*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a^2/f/(1+sin(f*x+e))+1/3*(c^2+ 5*c*d-12*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x )*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f* x+e))^(1/2)/a^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-1/3*(c+5*d)*(c^2-d^2)*(si n(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1 /2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/ 2)/a^2/f/(c+d*sin(f*x+e))^(1/2)
Time = 1.89 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\left (\left (c^2+5 c d-6 d^2\right ) (c+d \sin (e+f x))\right )+\frac {(c-d) \left (7 d \cos \left (\frac {1}{2} (e+f x)\right )-(c+6 d) \cos \left (\frac {3}{2} (e+f x)\right )+(3 c+11 d) \sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+d^2 (-11 c+5 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (c^2+5 c d-12 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{27 f (1+\sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-((c^2 + 5*c*d - 6*d^2)*(c + d*S in[e + f*x])) + ((c - d)*(7*d*Cos[(e + f*x)/2] - (c + 6*d)*Cos[(3*(e + f*x ))/2] + (3*c + 11*d)*Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x )/2] + Sin[(e + f*x)/2])^3 + d^2*(-11*c + 5*d)*EllipticF[(-2*e + Pi - 2*f* x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + (c^2 + 5*c*d - 1 2*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*Ellipt icF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(27*f*(1 + Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]])
Time = 1.33 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 3244, 27, 3042, 3456, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^2}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle -\frac {\int -\frac {\sqrt {c+d \sin (e+f x)} \left (a \left (2 c^2+7 d c-3 d^2\right )-a (c-7 d) d \sin (e+f x)\right )}{2 (\sin (e+f x) a+a)}dx}{3 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (2 c^2+7 d c-3 d^2\right )-a (c-7 d) d \sin (e+f x)\right )}{\sin (e+f x) a+a}dx}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (a \left (2 c^2+7 d c-3 d^2\right )-a (c-7 d) d \sin (e+f x)\right )}{\sin (e+f x) a+a}dx}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (11 c-5 d) d^2-a^2 d \left (c^2+5 d c-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (11 c-5 d) d^2-a^2 d \left (c^2+5 d c-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-a^2 \left (c^2+5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)}dx}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-a^2 \left (c^2+5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)}dx}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx-\frac {2 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {\frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^2 (c+5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (c+5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{f \sqrt {c+d \sin (e+f x)}}-\frac {2 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{a^2}-\frac {2 (c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{f (\sin (e+f x)+1)}}{6 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2}\) |
-1/3*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f *x])^2) + ((-2*(c - d)*(c + 5*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f *(1 + Sin[e + f*x])) + ((-2*a^2*(c^2 + 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a^2*(c + 5*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x) /2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(f*Sqrt[c + d*Sin[e + f*x]]))/a^2)/(6*a^2)
3.6.10.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b , c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
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_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 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 - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1371\) vs. \(2(302)=604\).
Time = 21.83 (sec) , antiderivative size = 1372, normalized size of antiderivative = 5.55
(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^2*(-4*d^3*(c/d-1)*((c+d*sin(f*x+ e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d )^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e) )/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2*d^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d)) ^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(- (-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e) )/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/ 2),((c-d)/(c+d))^(1/2)))+6*c*d^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d *(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f *x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c- d)/(c+d))^(1/2))+3*d*(c^2-2*c*d+d^2)*(-(-d*sin(f*x+e)^2-c*sin(f*x+e)+d*sin (f*x+e)+c)/(c-d)/((sin(f*x+e)+1)*(sin(f*x+e)-1)*(-d*sin(f*x+e)-c))^(1/2)-2 *d/(2*c-2*d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d ))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^ 2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-d/( c-d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2) *(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2) *((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+E llipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+(c^3-3*c^2* d+3*c*d^2-d^3)*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 1152, normalized size of antiderivative = 4.66 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\text {Too large to display} \]
1/18*((sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e)^2 - sqrt (2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) - (sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) + 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*sin(f*x + e) - 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*sqrt(I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8 *I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I* c)/d) + (sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e)^2 - sq rt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) - (sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) + 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*sin(f*x + e) - 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27 *(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) + 3*(sqrt(2)*(I*c^2*d + 5*I*c*d^2 - 12*I*d^3)*cos(f*x + e)^2 + s qrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3)*cos(f*x + e) + (sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3)*cos(f*x + e) + 2*sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3))*sin(f*x + e) + 2*sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3))*sqr t(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^ 2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I *c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)) + 3*( sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3)*cos(f*x + e)^2 + sqrt(2)*(I*c...
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]