Integrand size = 25, antiderivative size = 178 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=-\frac {2 d^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{27 (c-d)^3 \sqrt {c^2-d^2} f}-\frac {\cos (e+f x)}{5 (c-d) f (3+3 \sin (e+f x))^3}-\frac {(2 c-7 d) \cos (e+f x)}{45 (c-d)^2 f (3+3 \sin (e+f x))^2}-\frac {\left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f (27+27 \sin (e+f x))} \]
-1/5*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^3-1/15*(2*c-7*d)*cos(f*x+e)/a/(c- d)^2/f/(a+a*sin(f*x+e))^2-1/15*(2*c^2-9*c*d+22*d^2)*cos(f*x+e)/(c-d)^3/f/( a^3+a^3*sin(f*x+e))-2*d^3*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2)) /a^3/(c-d)^3/f/(c^2-d^2)^(1/2)
Time = 2.53 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (2 c-7 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+(c-d) (-2 c+7 d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 \left (2 c^2-9 c d+22 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-\frac {30 d^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\sqrt {c^2-d^2}}\right )}{405 (c-d)^3 f (1+\sin (e+f x))^3} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(c - d)^2*Sin[(e + f*x)/2] - 3*( c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(2*c - 7*d)*(c - d)*Sin [(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (c - d)*(-2*c + 7* d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 2*(2*c^2 - 9*c*d + 22*d^2)*Si n[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (30*d^3*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/ 2])^5)/Sqrt[c^2 - d^2]))/(405*(c - d)^3*f*(1 + Sin[e + f*x])^3)
Time = 0.95 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.21, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3245, 25, 3042, 3457, 25, 3042, 3457, 27, 3042, 3139, 1083, 217}
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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c+d \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (2 c-5 d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (2 c-5 d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (2 c-5 d)+2 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {-\frac {\int -\frac {\left (2 c^2-7 d c+15 d^2\right ) a^2+(2 c-7 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 c^2-7 d c+15 d^2\right ) a^2+(2 c-7 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 c^2-7 d c+15 d^2\right ) a^2+(2 c-7 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {15 a^3 d^3}{c+d \sin (e+f x)}dx}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {15 a d^3 \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {15 a d^3 \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {-\frac {30 a d^3 \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (c-d)}-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {\frac {60 a d^3 \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f (c-d)}-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {-\frac {a^2 \left (2 c^2-9 c d+22 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}-\frac {30 a d^3 \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f (c-d) \sqrt {c^2-d^2}}}{3 a^2 (c-d)}-\frac {a (2 c-7 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\) |
-1/5*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^3) + (-1/3*(a*(2*c - 7*d )*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^2) + ((-30*a*d^3*ArcTan[(2 *d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c - d)*Sqrt[c^2 - d^2]* f) - (a^2*(2*c^2 - 9*c*d + 22*d^2)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])))/(3*a^2*(c - d)))/(5*a^2*(c - d))
3.5.77.3.1 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, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && 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^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + 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] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], 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[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Time = 1.35 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12
| method | result | size |
| derivativedivides | \(\frac {-\frac {-4 c +6 d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 c -10 d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (c^{2}-3 c d +3 d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}}{a^{3} f}\) | \(200\) |
| default | \(\frac {-\frac {-4 c +6 d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 c -10 d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (c^{2}-3 c d +3 d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 d^{3} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}}{a^{3} f}\) | \(200\) |
| risch | \(-\frac {2 \left (2 c^{2}+22 d^{2}-9 c d -15 i c d \,{\mathrm e}^{3 i \left (f x +e \right )}+15 d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-20 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-145 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+45 i d \,{\mathrm e}^{i \left (f x +e \right )} c +75 c d \,{\mathrm e}^{2 i \left (f x +e \right )}+75 i d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-10 i c^{2} {\mathrm e}^{i \left (f x +e \right )}-95 i d^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left (c -d \right )^{3} f \,a^{3}}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}+\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}\) | \(331\) |
2/f/a^3*(-1/2*(-4*c+6*d)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(8*c-10*d)/( c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3-(c^2-3*c*d+3*d^2)/(c-d)^3/(tan(1/2*f*x+1/2 *e)+1)-4/5/(c-d)/(tan(1/2*f*x+1/2*e)+1)^5+2/(c-d)/(tan(1/2*f*x+1/2*e)+1)^4 -d^3/(c-d)^3/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2- d^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 830 vs. \(2 (175) = 350\).
Time = 0.33 (sec) , antiderivative size = 1744, normalized size of antiderivative = 9.80 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]
[1/30*(6*c^4 - 12*c^3*d + 12*c*d^3 - 6*d^4 - 2*(2*c^4 - 9*c^3*d + 20*c^2*d ^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^3 + 2*(4*c^4 - 18*c^3*d + 25*c^2*d^2 + 18*c*d^3 - 29*d^4)*cos(f*x + e)^2 + 15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f* x + e)^2 - 2*d^3*cos(f*x + e) - 4*d^3 + (d^3*cos(f*x + e)^2 - 2*d^3*cos(f* x + e) - 4*d^3)*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e ) - c^2 - d^2)) + 6*(3*c^4 - 11*c^3*d + 15*c^2*d^2 + 11*c*d^3 - 18*d^4)*co s(f*x + e) - 2*(3*c^4 - 6*c^3*d + 6*c*d^3 - 3*d^4 - (2*c^4 - 9*c^3*d + 20* c^2*d^2 + 9*c*d^3 - 22*d^4)*cos(f*x + e)^2 - 3*(2*c^4 - 9*c^3*d + 15*c^2*d ^2 + 9*c*d^3 - 17*d^4)*cos(f*x + e))*sin(f*x + e))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2 *a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3* c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f + ((a^3*c ^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)* f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^ 3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a ^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f)*sin(f*x + e)), 1...
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f or more de
Time = 0.38 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=-\frac {2 \, {\left (\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d^{3}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} \sqrt {c^{2} - d^{2}}} + \frac {15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 105 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 135 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 135 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 185 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 75 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 115 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} - 24 \, c d + 32 \, d^{2}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \]
-2/15*(15*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))*d^3/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) + (15*c^2*tan(1/2*f*x + 1/2*e)^4 - 45*c*d*tan (1/2*f*x + 1/2*e)^4 + 45*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1/2*f*x + 1/2*e)^3 - 105*c*d*tan(1/2*f*x + 1/2*e)^3 + 135*d^2*tan(1/2*f*x + 1/2*e)^ 3 + 40*c^2*tan(1/2*f*x + 1/2*e)^2 - 135*c*d*tan(1/2*f*x + 1/2*e)^2 + 185*d ^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x + 1/2*e) - 75*c*d*tan(1/2*f *x + 1/2*e) + 115*d^2*tan(1/2*f*x + 1/2*e) + 7*c^2 - 24*c*d + 32*d^2)/((a^ 3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^5) )/f
Time = 10.21 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {2\,d^3\,\mathrm {atan}\left (\frac {\frac {d^3\,\left (-2\,a^3\,c^3\,d+6\,a^3\,c^2\,d^2-6\,a^3\,c\,d^3+2\,a^3\,d^4\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^3\,c^3-3\,a^3\,c^2\,d+3\,a^3\,c\,d^2-a^3\,d^3\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}}{2\,d^3}\right )}{a^3\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (7\,c^2-24\,c\,d+32\,d^2\right )}{15\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c^2-15\,c\,d+23\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (c^2-3\,c\,d+3\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^2-7\,c\,d+9\,d^2\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,c^2-27\,c\,d+37\,d^2\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]
(2*d^3*atan(((d^3*(2*a^3*d^4 - 6*a^3*c*d^3 - 2*a^3*c^3*d + 6*a^3*c^2*d^2)) /(a^3*(c + d)^(1/2)*(c - d)^(7/2)) - (2*c*d^3*tan(e/2 + (f*x)/2)*(a^3*c^3 - a^3*d^3 + 3*a^3*c*d^2 - 3*a^3*c^2*d))/(a^3*(c + d)^(1/2)*(c - d)^(7/2))) /(2*d^3)))/(a^3*f*(c + d)^(1/2)*(c - d)^(7/2)) - ((2*(7*c^2 - 24*c*d + 32* d^2))/(15*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)*(4*c^2 - 15 *c*d + 23*d^2))/(3*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^4* (c^2 - 3*c*d + 3*d^2))/((c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/ 2)^3*(2*c^2 - 7*c*d + 9*d^2))/((c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^2*(8*c^2 - 27*c*d + 37*d^2))/(3*(c - d)*(c^2 - 2*c*d + d^2)))/(f *(10*a^3*tan(e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/ 2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x)/2) ))