Integrand size = 25, antiderivative size = 129 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {27 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d)^2 \sqrt {c^2-d^2} f}+\frac {9 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {9 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))} \]
1/2*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^2-1/2*a^2*(c+4*d)*cos( f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+e))+3*a^2*arctan((d+c*tan(1/2*f*x+1/2*e))/ (c^2-d^2)^(1/2))/(c+d)^2/f/(c^2-d^2)^(1/2)
Time = 0.41 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.06 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {9 \cos (e+f x) \left (-\frac {6 \text {arctanh}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right )}{(-c-d)^{3/2} \sqrt {c-d} \sqrt {\cos ^2(e+f x)}}-\frac {4 c+d+(c+4 d) \sin (e+f x)}{(c+d) (c+d \sin (e+f x))^2}\right )}{2 (c+d) f} \]
(9*Cos[e + f*x]*((-6*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]*Sqrt[1 + Sin[e + f*x]])])/((-c - d)^(3/2)*Sqrt[c - d]*Sqrt[Cos[e + f *x]^2]) - (4*c + d + (c + 4*d)*Sin[e + f*x])/((c + d)*(c + d*Sin[e + f*x]) ^2)))/(2*(c + d)*f)
Time = 0.57 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3241, 25, 3042, 3233, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}-\frac {a \int -\frac {4 a d+a (c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 d (c+d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a \int \frac {4 a d+a (c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \frac {4 a d+a (c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {a \left (-\frac {\int -\frac {3 a (c-d) d}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \left (\frac {3 a d (c-d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \left (\frac {3 a d (c-d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {a \left (\frac {6 a d (c-d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {a \left (-\frac {12 a d (c-d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2}+\frac {a \left (\frac {6 a d (c-d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a (c+4 d) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}\right )}{2 d (c+d)}\) |
(a^2*(c - d)*Cos[e + f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) + (a*((6 *a*(c - d)*d*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c ^2 - d^2)^(3/2)*f) - (a*(c + 4*d)*Cos[e + f*x])/((c + d)*f*(c + d*Sin[e + f*x]))))/(2*d*(c + d))
3.5.41.3.1 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_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*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[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Time = 1.03 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.94
| method | result | size |
| derivativedivides | \(\frac {2 a^{2} \left (\frac {\frac {\left (c^{2}-4 c d -2 d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {\left (4 c^{3}+c^{2} d +8 c \,d^{2}+2 d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}-\frac {\left (c^{2}+12 c d +2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}-\frac {4 c +d}{2 \left (c^{2}+2 c d +d^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {3 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\) | \(250\) |
| default | \(\frac {2 a^{2} \left (\frac {\frac {\left (c^{2}-4 c d -2 d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {\left (4 c^{3}+c^{2} d +8 c \,d^{2}+2 d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}-\frac {\left (c^{2}+12 c d +2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}-\frac {4 c +d}{2 \left (c^{2}+2 c d +d^{2}\right )}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {3 \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\) | \(250\) |
| risch | \(\frac {i a^{2} \left (-2 i c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}-4 i c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+i d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+2 i c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}+12 i c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}+i d^{3} {\mathrm e}^{i \left (f x +e \right )}+2 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+8 d \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+d^{2} {\mathrm e}^{2 i \left (f x +e \right )} c +4 d^{3} {\mathrm e}^{2 i \left (f x +e \right )}-c \,d^{2}-4 d^{3}\right )}{\left (c +d \right )^{2} \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )^{2} f \,d^{2}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} f}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{2} f}\) | \(362\) |
2/f*a^2*((1/2*(c^2-4*c*d-2*d^2)/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)^3-1/2 *(4*c^3+c^2*d+8*c*d^2+2*d^3)/(c^2+2*c*d+d^2)/c^2*tan(1/2*f*x+1/2*e)^2-1/2* (c^2+12*c*d+2*d^2)/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)-1/2*(4*c+d)/(c^2+2 *c*d+d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)^2+3/2/(c^2+2* c*d+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)))
Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (129) = 258\).
Time = 0.31 (sec) , antiderivative size = 679, normalized size of antiderivative = 5.26 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\left [\frac {2 \, {\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}, \frac {{\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}\right ] \]
[1/4*(2*(a^2*c^3 + 4*a^2*c^2*d - a^2*c*d^2 - 4*a^2*d^3)*cos(f*x + e)*sin(f *x + e) - 3*(a^2*d^2*cos(f*x + e)^2 - 2*a^2*c*d*sin(f*x + e) - a^2*c^2 - a ^2*d^2)*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)) + 2*(4 *a^2*c^3 + a^2*c^2*d - 4*a^2*c*d^2 - a^2*d^3)*cos(f*x + e))/((c^4*d^2 + 2* c^3*d^3 - 2*c*d^5 - d^6)*f*cos(f*x + e)^2 - 2*(c^5*d + 2*c^4*d^2 - 2*c^2*d ^4 - c*d^5)*f*sin(f*x + e) - (c^6 + 2*c^5*d + c^4*d^2 - c^2*d^4 - 2*c*d^5 - d^6)*f), 1/2*((a^2*c^3 + 4*a^2*c^2*d - a^2*c*d^2 - 4*a^2*d^3)*cos(f*x + e)*sin(f*x + e) - 3*(a^2*d^2*cos(f*x + e)^2 - 2*a^2*c*d*sin(f*x + e) - a^2 *c^2 - a^2*d^2)*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d ^2)*cos(f*x + e))) + (4*a^2*c^3 + a^2*c^2*d - 4*a^2*c*d^2 - a^2*d^3)*cos(f *x + e))/((c^4*d^2 + 2*c^3*d^3 - 2*c*d^5 - d^6)*f*cos(f*x + e)^2 - 2*(c^5* d + 2*c^4*d^2 - 2*c^2*d^4 - c*d^5)*f*sin(f*x + e) - (c^6 + 2*c^5*d + c^4*d ^2 - c^2*d^4 - 2*c*d^5 - d^6)*f)]
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (129) = 258\).
Time = 0.35 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.59 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} a^{2}}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt {c^{2} - d^{2}}} + \frac {a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 8 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a^{2} c^{3} - a^{2} c^{2} d}{{\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \]
(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)))*a^2/((c^2 + 2*c*d + d^2)*sqrt(c^2 - d^2)) + (a^2 *c^3*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 2*a^2*c *d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^3*tan(1/2*f*x + 1/2*e)^2 - a^2*c^2*d *tan(1/2*f*x + 1/2*e)^2 - 8*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 2*a^2*d^3*t an(1/2*f*x + 1/2*e)^2 - a^2*c^3*tan(1/2*f*x + 1/2*e) - 12*a^2*c^2*d*tan(1/ 2*f*x + 1/2*e) - 2*a^2*c*d^2*tan(1/2*f*x + 1/2*e) - 4*a^2*c^3 - a^2*c^2*d) /((c^4 + 2*c^3*d + c^2*d^2)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f
Time = 9.75 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.81 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {3\,a^2\,\mathrm {atan}\left (\frac {\left (\frac {3\,a^2\,\left (2\,c^2\,d+4\,c\,d^2+2\,d^3\right )}{2\,{\left (c+d\right )}^{5/2}\,\sqrt {c-d}\,\left (c^2+2\,c\,d+d^2\right )}+\frac {3\,a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}\right )\,\left (c^2+2\,c\,d+d^2\right )}{3\,a^2}\right )}{f\,{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}-\frac {\frac {4\,a^2\,c+a^2\,d}{c^2+2\,c\,d+d^2}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^2+12\,a^2\,c\,d+2\,a^2\,d^2\right )}{c\,\left (c^2+2\,c\,d+d^2\right )}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (-c^2+4\,c\,d+2\,d^2\right )}{c\,\left (c^2+2\,c\,d+d^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (4\,a^2\,c+a^2\,d\right )}{c^2\,\left (c^2+2\,c\,d+d^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]
(3*a^2*atan((((3*a^2*(4*c*d^2 + 2*c^2*d + 2*d^3))/(2*(c + d)^(5/2)*(c - d) ^(1/2)*(2*c*d + c^2 + d^2)) + (3*a^2*c*tan(e/2 + (f*x)/2))/((c + d)^(5/2)* (c - d)^(1/2)))*(2*c*d + c^2 + d^2))/(3*a^2)))/(f*(c + d)^(5/2)*(c - d)^(1 /2)) - ((4*a^2*c + a^2*d)/(2*c*d + c^2 + d^2) + (tan(e/2 + (f*x)/2)*(a^2*c ^2 + 2*a^2*d^2 + 12*a^2*c*d))/(c*(2*c*d + c^2 + d^2)) + (a^2*tan(e/2 + (f* x)/2)^3*(4*c*d - c^2 + 2*d^2))/(c*(2*c*d + c^2 + d^2)) + (tan(e/2 + (f*x)/ 2)^2*(c^2 + 2*d^2)*(4*a^2*c + a^2*d))/(c^2*(2*c*d + c^2 + d^2)))/(f*(tan(e /2 + (f*x)/2)^2*(2*c^2 + 4*d^2) + c^2*tan(e/2 + (f*x)/2)^4 + c^2 + 4*c*d*t an(e/2 + (f*x)/2)^3 + 4*c*d*tan(e/2 + (f*x)/2)))