Integrand size = 25, antiderivative size = 189 \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=-\frac {\left (18 b c d-9 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2} f}+\frac {(b c-3 d)^2 \cos (e+f x)}{2 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {(b c-3 d) \left (9 c d+b \left (c^2-4 d^2\right )\right ) \cos (e+f x)}{2 d \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))} \]
-(6*a*b*c*d-a^2*(2*c^2+d^2)-b^2*(c^2+2*d^2))*arctan((d+c*tan(1/2*f*x+1/2*e ))/(c^2-d^2)^(1/2))/(c^2-d^2)^(5/2)/f+1/2*(-a*d+b*c)^2*cos(f*x+e)/d/(c^2-d ^2)/f/(c+d*sin(f*x+e))^2-1/2*(-a*d+b*c)*(3*a*c*d+b*(c^2-4*d^2))*cos(f*x+e) /d/(c^2-d^2)^2/f/(c+d*sin(f*x+e))
Time = 1.01 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.01 \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {2 \left (\left (18+b^2\right ) c^2-18 b c d+\left (9+2 b^2\right ) d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}+\frac {(b c-3 d)^2 \cos (e+f x)}{(c-d) d (c+d) (c+d \sin (e+f x))^2}-\frac {\left (-27 c d^2+6 b d \left (c^2+2 d^2\right )+b^2 \left (c^3-4 c d^2\right )\right ) \cos (e+f x)}{(c-d)^2 d (c+d)^2 (c+d \sin (e+f x))}}{2 f} \]
((2*((18 + b^2)*c^2 - 18*b*c*d + (9 + 2*b^2)*d^2)*ArcTan[(d + c*Tan[(e + f *x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(5/2) + ((b*c - 3*d)^2*Cos[e + f*x]) /((c - d)*d*(c + d)*(c + d*Sin[e + f*x])^2) - ((-27*c*d^2 + 6*b*d*(c^2 + 2 *d^2) + b^2*(c^3 - 4*c*d^2))*Cos[e + f*x])/((c - d)^2*d*(c + d)^2*(c + d*S in[e + f*x])))/(2*f)
Time = 0.67 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3269, 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.
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3269 |
| \(\displaystyle \frac {\int \frac {2 d \left (\left (a^2+b^2\right ) c-2 a b d\right )+\left (b^2 c^2+2 a b d c-\left (a^2+2 b^2\right ) d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 d \left (\left (a^2+b^2\right ) c-2 a b d\right )+\left (b^2 c^2+2 a b d c-\left (a^2+2 b^2\right ) d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int \frac {d \left (-\left (\left (2 c^2+d^2\right ) a^2\right )+6 b c d a-b^2 \left (c^2+2 d^2\right )\right )}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {d \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {d \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {-\frac {2 d \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {4 d \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {-\frac {2 d \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {(b c-a d) \left (3 a c d+b c^2-4 b d^2\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\) |
((b*c - a*d)^2*Cos[e + f*x])/(2*d*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) + ((-2*d*(6*a*b*c*d - a^2*(2*c^2 + d^2) - b^2*(c^2 + 2*d^2))*ArcTan[(2*d + 2 *c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c^2 - d^2)^(3/2)*f) - ((b*c - a*d)*(b*c^2 + 3*a*c*d - 4*b*d^2)*Cos[e + f*x])/((c^2 - d^2)*f*(c + d*Sin[ e + f*x])))/(2*d*(c^2 - d^2))
3.7.84.3.1 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_)])^2, x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[ 1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1) *(2*b*c*d - a*(c^2 + d^2)) + (a^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1 ) + c^2*(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]
Leaf count of result is larger than twice the leaf count of optimal. \(466\) vs. \(2(187)=374\).
Time = 1.52 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.47
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (5 a^{2} c^{2} d^{2}-2 a^{2} d^{4}-6 a b \,c^{3} d +b^{2} c^{4}+2 b^{2} c^{2} d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c}+\frac {\left (4 a^{2} c^{4} d +7 a^{2} c^{2} d^{3}-2 a^{2} d^{5}-4 a b \,c^{5}-10 a b \,c^{3} d^{2}-4 a b c \,d^{4}+3 b^{2} c^{4} d +6 b^{2} c^{2} d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c^{2}}+\frac {\left (11 a^{2} c^{2} d^{2}-2 a^{2} d^{4}-10 a b \,c^{3} d -8 d^{3} a b c -b^{2} c^{4}+10 b^{2} c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {2 \left (4 a^{2} c^{2} d -a^{2} d^{3}-4 a b \,c^{3}-2 a b c \,d^{2}+3 b^{2} c^{2} d \right )}{2 c^{4}-4 c^{2} d^{2}+2 d^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {c^{2}-d^{2}}}}{f}\) | \(467\) |
| default | \(\frac {\frac {\frac {\left (5 a^{2} c^{2} d^{2}-2 a^{2} d^{4}-6 a b \,c^{3} d +b^{2} c^{4}+2 b^{2} c^{2} d^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c}+\frac {\left (4 a^{2} c^{4} d +7 a^{2} c^{2} d^{3}-2 a^{2} d^{5}-4 a b \,c^{5}-10 a b \,c^{3} d^{2}-4 a b c \,d^{4}+3 b^{2} c^{4} d +6 b^{2} c^{2} d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) c^{2}}+\frac {\left (11 a^{2} c^{2} d^{2}-2 a^{2} d^{4}-10 a b \,c^{3} d -8 d^{3} a b c -b^{2} c^{4}+10 b^{2} c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{4}-2 c^{2} d^{2}+d^{4}\right )}+\frac {2 \left (4 a^{2} c^{2} d -a^{2} d^{3}-4 a b \,c^{3}-2 a b c \,d^{2}+3 b^{2} c^{2} d \right )}{2 c^{4}-4 c^{2} d^{2}+2 d^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )}^{2}}+\frac {\left (2 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {c^{2}-d^{2}}}}{f}\) | \(467\) |
| risch | \(\text {Expression too large to display}\) | \(1294\) |
1/f*(2*(1/2*(5*a^2*c^2*d^2-2*a^2*d^4-6*a*b*c^3*d+b^2*c^4+2*b^2*c^2*d^2)/(c ^4-2*c^2*d^2+d^4)/c*tan(1/2*f*x+1/2*e)^3+1/2*(4*a^2*c^4*d+7*a^2*c^2*d^3-2* a^2*d^5-4*a*b*c^5-10*a*b*c^3*d^2-4*a*b*c*d^4+3*b^2*c^4*d+6*b^2*c^2*d^3)/(c ^4-2*c^2*d^2+d^4)/c^2*tan(1/2*f*x+1/2*e)^2+1/2*(11*a^2*c^2*d^2-2*a^2*d^4-1 0*a*b*c^3*d-8*a*b*c*d^3-b^2*c^4+10*b^2*c^2*d^2)/c/(c^4-2*c^2*d^2+d^4)*tan( 1/2*f*x+1/2*e)+1/2*(4*a^2*c^2*d-a^2*d^3-4*a*b*c^3-2*a*b*c*d^2+3*b^2*c^2*d) /(c^4-2*c^2*d^2+d^4))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)^2+ (2*a^2*c^2+a^2*d^2-6*a*b*c*d+b^2*c^2+2*b^2*d^2)/(c^4-2*c^2*d^2+d^4)/(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. 471 vs. \(2 (187) = 374\).
Time = 0.31 (sec) , antiderivative size = 1027, normalized size of antiderivative = 5.43 \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \]
[1/4*(2*(b^2*c^5 + 2*a*b*c^4*d + 2*a*b*c^2*d^3 - 4*a*b*d^5 - (3*a^2 + 5*b^ 2)*c^3*d^2 + (3*a^2 + 4*b^2)*c*d^4)*cos(f*x + e)*sin(f*x + e) - (6*a*b*c^3 *d + 6*a*b*c*d^3 - (2*a^2 + b^2)*c^4 - 3*(a^2 + b^2)*c^2*d^2 - (a^2 + 2*b^ 2)*d^4 - (6*a*b*c*d^3 - (2*a^2 + b^2)*c^2*d^2 - (a^2 + 2*b^2)*d^4)*cos(f*x + e)^2 + 2*(6*a*b*c^2*d^2 - (2*a^2 + b^2)*c^3*d - (a^2 + 2*b^2)*c*d^3)*si n(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))*s qrt(-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*b*c^5 - 2*a*b*c^3*d^2 - 2*a*b*c*d^4 - a^2*d^5 - (4*a^2 + 3*b^2)*c^4 *d + (5*a^2 + 3*b^2)*c^2*d^3)*cos(f*x + e))/((c^6*d^2 - 3*c^4*d^4 + 3*c^2* d^6 - d^8)*f*cos(f*x + e)^2 - 2*(c^7*d - 3*c^5*d^3 + 3*c^3*d^5 - c*d^7)*f* sin(f*x + e) - (c^8 - 2*c^6*d^2 + 2*c^2*d^6 - d^8)*f), 1/2*((b^2*c^5 + 2*a *b*c^4*d + 2*a*b*c^2*d^3 - 4*a*b*d^5 - (3*a^2 + 5*b^2)*c^3*d^2 + (3*a^2 + 4*b^2)*c*d^4)*cos(f*x + e)*sin(f*x + e) - (6*a*b*c^3*d + 6*a*b*c*d^3 - (2* a^2 + b^2)*c^4 - 3*(a^2 + b^2)*c^2*d^2 - (a^2 + 2*b^2)*d^4 - (6*a*b*c*d^3 - (2*a^2 + b^2)*c^2*d^2 - (a^2 + 2*b^2)*d^4)*cos(f*x + e)^2 + 2*(6*a*b*c^2 *d^2 - (2*a^2 + b^2)*c^3*d - (a^2 + 2*b^2)*c*d^3)*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (4*a* b*c^5 - 2*a*b*c^3*d^2 - 2*a*b*c*d^4 - a^2*d^5 - (4*a^2 + 3*b^2)*c^4*d + (5 *a^2 + 3*b^2)*c^2*d^3)*cos(f*x + e))/((c^6*d^2 - 3*c^4*d^4 + 3*c^2*d^6 ...
Timed out. \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]
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. 586 vs. \(2 (187) = 374\).
Time = 0.34 (sec) , antiderivative size = 586, normalized size of antiderivative = 3.10 \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {\frac {{\left (2 \, a^{2} c^{2} + b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2} + 2 \, b^{2} d^{2}\right )} {\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 )}}{{\left (c^{4} - 2 \, c^{2} d^{2} + d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 5 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a b c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, a^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, b^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, a b c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, b^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a b c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 10 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 11 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a b c^{5} + 4 \, a^{2} c^{4} d + 3 \, b^{2} c^{4} d - 2 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}}{{\left (c^{6} - 2 \, c^{4} d^{2} + c^{2} d^{4}\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} \]
((2*a^2*c^2 + b^2*c^2 - 6*a*b*c*d + a^2*d^2 + 2*b^2*d^2)*(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)))/((c^4 - 2*c^2*d^2 + d^4)*sqrt(c^2 - d^2)) + (b^2*c^5*tan(1/2*f*x + 1 /2*e)^3 - 6*a*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 + 5*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 2*b^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 2*a^2*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 4*a*b*c^5*tan(1/2*f*x + 1/2*e)^2 + 4*a^2*c^4*d*tan(1/2*f*x + 1/2*e)^2 + 3*b^2*c^4*d*tan(1/2*f*x + 1/2*e)^2 - 10*a*b*c^3*d^2*tan(1/2*f* x + 1/2*e)^2 + 7*a^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 + 6*b^2*c^2*d^3*tan(1/ 2*f*x + 1/2*e)^2 - 4*a*b*c*d^4*tan(1/2*f*x + 1/2*e)^2 - 2*a^2*d^5*tan(1/2* f*x + 1/2*e)^2 - b^2*c^5*tan(1/2*f*x + 1/2*e) - 10*a*b*c^4*d*tan(1/2*f*x + 1/2*e) + 11*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e) + 10*b^2*c^3*d^2*tan(1/2*f*x + 1/2*e) - 8*a*b*c^2*d^3*tan(1/2*f*x + 1/2*e) - 2*a^2*c*d^4*tan(1/2*f*x + 1/2*e) - 4*a*b*c^5 + 4*a^2*c^4*d + 3*b^2*c^4*d - 2*a*b*c^3*d^2 - a^2*c^2* d^3)/((c^6 - 2*c^4*d^2 + c^2*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2* f*x + 1/2*e) + c)^2))/f
Time = 11.00 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.39 \[ \int \frac {(3+b \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,c^4\,d-4\,c^2\,d^3+2\,d^5\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{2\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}\right )\,\left (c^4-2\,c^2\,d^2+d^4\right )}{2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{f\,{\left (c+d\right )}^{5/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {-4\,a^2\,c^2\,d+a^2\,d^3+4\,a\,b\,c^3+2\,a\,b\,c\,d^2-3\,b^2\,c^2\,d}{c^4-2\,c^2\,d^2+d^4}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (-11\,a^2\,c^2\,d^2+2\,a^2\,d^4+10\,a\,b\,c^3\,d+8\,a\,b\,c\,d^3+b^2\,c^4-10\,b^2\,c^2\,d^2\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (5\,a^2\,c^2\,d^2-2\,a^2\,d^4-6\,a\,b\,c^3\,d+b^2\,c^4+2\,b^2\,c^2\,d^2\right )}{c\,\left (c^4-2\,c^2\,d^2+d^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (-4\,a^2\,c^2\,d+a^2\,d^3+4\,a\,b\,c^3+2\,a\,b\,c\,d^2-3\,b^2\,c^2\,d\right )}{c^2\,\left (c^4-2\,c^2\,d^2+d^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]
(atan(((((2*c^4*d + 2*d^5 - 4*c^2*d^3)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2* b^2*d^2 - 6*a*b*c*d))/(2*(c + d)^(5/2)*(c - d)^(5/2)*(c^4 + d^4 - 2*c^2*d^ 2)) + (c*tan(e/2 + (f*x)/2)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6 *a*b*c*d))/((c + d)^(5/2)*(c - d)^(5/2)))*(c^4 + d^4 - 2*c^2*d^2))/(2*a^2* c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))*(2*a^2*c^2 + a^2*d^2 + b ^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))/(f*(c + d)^(5/2)*(c - d)^(5/2)) - ((a^2*d ^3 - 4*a^2*c^2*d - 3*b^2*c^2*d + 4*a*b*c^3 + 2*a*b*c*d^2)/(c^4 + d^4 - 2*c ^2*d^2) + (tan(e/2 + (f*x)/2)*(2*a^2*d^4 + b^2*c^4 - 11*a^2*c^2*d^2 - 10*b ^2*c^2*d^2 + 8*a*b*c*d^3 + 10*a*b*c^3*d))/(c*(c^4 + d^4 - 2*c^2*d^2)) - (t an(e/2 + (f*x)/2)^3*(b^2*c^4 - 2*a^2*d^4 + 5*a^2*c^2*d^2 + 2*b^2*c^2*d^2 - 6*a*b*c^3*d))/(c*(c^4 + d^4 - 2*c^2*d^2)) + (tan(e/2 + (f*x)/2)^2*(c^2 + 2*d^2)*(a^2*d^3 - 4*a^2*c^2*d - 3*b^2*c^2*d + 4*a*b*c^3 + 2*a*b*c*d^2))/(c ^2*(c^4 + d^4 - 2*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*tan(e/2 + (f*x)/2)^3 + 4*c*d*tan(e/2 + (f*x)/2)))