Integrand size = 25, antiderivative size = 127 \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2 d^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{9 (c-d)^2 \sqrt {c^2-d^2} f}-\frac {(c-4 d) \cos (e+f x)}{27 (c-d)^2 f (1+\sin (e+f x))}-\frac {\cos (e+f x)}{3 (c-d) f (3+3 \sin (e+f x))^2} \]
-1/3*(c-4*d)*cos(f*x+e)/a^2/(c-d)^2/f/(1+sin(f*x+e))-1/3*cos(f*x+e)/(c-d)/ f/(a+a*sin(f*x+e))^2+2*d^2*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2) )/a^2/(c-d)^2/f/(c^2-d^2)^(1/2)
Time = 1.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )-(c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (c-4 d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+\frac {6 d^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{\sqrt {c^2-d^2}}\right )}{27 (c-d)^2 f (1+\sin (e+f x))^2} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2] - (c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - 4*d)*Sin[(e + f*x)/2]*(C os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (6*d^2*ArcTan[(d + c*Tan[(e + f*x) /2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/Sqrt[c^2 - d^2]))/(27*(c - d)^2*f*(1 + Sin[e + f*x])^2)
Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3245, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {a (c-3 d)+a d \sin (e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (c-3 d)+a d \sin (e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (c-3 d)+a d \sin (e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 a^2 d^2}{c+d \sin (e+f x)}dx}{a^2 (c-d)}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 d^2 \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 d^2 \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {6 d^2 \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (c-d)}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {-\frac {12 d^2 \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f (c-d)}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {6 d^2 \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f (c-d) \sqrt {c^2-d^2}}-\frac {(c-4 d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\) |
-1/3*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^2) + ((6*d^2*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c - d)*Sqrt[c^2 - d^2]*f) - ((c - 4*d)*Cos[e + f*x])/((c - d)*f*(1 + Sin[e + f*x])))/(3*a^2*(c - d) )
3.5.67.3.1 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 = 1.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}-\frac {2 \left (c -2 d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{2} f}\) | \(132\) |
| default | \(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}-\frac {2 \left (c -2 d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{2} f}\) | \(132\) |
| risch | \(\frac {2 d \,{\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 c}{3}-\frac {8 d}{3}-2 i c \,{\mathrm e}^{i \left (f x +e \right )}+6 i d \,{\mathrm e}^{i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} \left (c -d \right )^{2} f \,a^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}\) | \(231\) |
2/f/a^2*(d^2/(c-d)^2/(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))-(c-2*d)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)-2/3/(c-d)/(tan( 1/2*f*x+1/2*e)+1)^3+1/(c-d)/(tan(1/2*f*x+1/2*e)+1)^2)
Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (122) = 244\).
Time = 0.30 (sec) , antiderivative size = 989, normalized size of antiderivative = 7.79 \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\left [\frac {2 \, c^{3} - 2 \, c^{2} d - 2 \, c d^{2} + 2 \, d^{3} + 2 \, {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2} \cos \left (f x + e\right ) - 2 \, d^{2} - {\left (d^{2} \cos \left (f x + e\right ) + 2 \, d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right ) - 2 \, {\left (c^{3} - c^{2} d - c d^{2} + d^{3} - {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f - {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}}, \frac {c^{3} - c^{2} d - c d^{2} + d^{3} + {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2} \cos \left (f x + e\right ) - 2 \, d^{2} - {\left (d^{2} \cos \left (f x + e\right ) + 2 \, d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{3} - c^{2} d - c d^{2} + d^{3} - {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f - {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \]
[1/6*(2*c^3 - 2*c^2*d - 2*c*d^2 + 2*d^3 + 2*(c^3 - 4*c^2*d - c*d^2 + 4*d^3 )*cos(f*x + e)^2 - 3*(d^2*cos(f*x + e)^2 - d^2*cos(f*x + e) - 2*d^2 - (d^2 *cos(f*x + e) + 2*d^2)*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*c os(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)) + 2*(2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x + e ) - 2*(c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*co s(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2* a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3 *d + 2*a^2*c*d^3 - a^2*d^4)*f)*sin(f*x + e)), 1/3*(c^3 - c^2*d - c*d^2 + d ^3 + (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e)^2 - 3*(d^2*cos(f*x + e)^ 2 - d^2*cos(f*x + e) - 2*d^2 - (d^2*cos(f*x + e) + 2*d^2)*sin(f*x + e))*sq rt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2 *c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c ^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3...
Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} - \frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, c - 5 \, d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \]
2/3*(3*(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^2/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sqrt(c^ 2 - d^2)) - (3*c*tan(1/2*f*x + 1/2*e)^2 - 6*d*tan(1/2*f*x + 1/2*e)^2 + 3*c *tan(1/2*f*x + 1/2*e) - 9*d*tan(1/2*f*x + 1/2*e) + 2*c - 5*d)/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*(tan(1/2*f*x + 1/2*e) + 1)^3))/f
Time = 8.14 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.97 \[ \int \frac {1}{(3+3 \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2\,d^2\,\mathrm {atan}\left (\frac {\frac {d^2\,\left (2\,a^2\,c^2\,d-4\,a^2\,c\,d^2+2\,a^2\,d^3\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^2-2\,a^2\,c\,d+a^2\,d^2\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}}{2\,d^2}\right )}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (2\,c-5\,d\right )}{3\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (c-3\,d\right )}{{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c-2\,d\right )}{{\left (c-d\right )}^2}}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )} \]
(2*d^2*atan(((d^2*(2*a^2*d^3 - 4*a^2*c*d^2 + 2*a^2*c^2*d))/(a^2*(c + d)^(1 /2)*(c - d)^(5/2)) + (2*c*d^2*tan(e/2 + (f*x)/2)*(a^2*c^2 + a^2*d^2 - 2*a^ 2*c*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)))/(2*d^2)))/(a^2*f*(c + d)^(1/2)* (c - d)^(5/2)) - ((2*(2*c - 5*d))/(3*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(c - 3*d))/(c - d)^2 + (2*tan(e/2 + (f*x)/2)^2*(c - 2*d))/(c - d)^2)/(f*(3*a ^2*tan(e/2 + (f*x)/2)^2 + a^2*tan(e/2 + (f*x)/2)^3 + a^2 + 3*a^2*tan(e/2 + (f*x)/2)))