Integrand size = 25, antiderivative size = 119 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\frac {d^2 x}{b^2}+\frac {2 (b c-3 d) \left (3 b c+9 d-2 b^2 d\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{b^2 \left (9-b^2\right )^{3/2} f}+\frac {(b c-3 d)^2 \cos (e+f x)}{b \left (9-b^2\right ) f (3+b \sin (e+f x))} \]
d^2*x/b^2+2*(-a*d+b*c)*(a^2*d+a*b*c-2*b^2*d)*arctan((b+a*tan(1/2*f*x+1/2*e ))/(a^2-b^2)^(1/2))/b^2/(a^2-b^2)^(3/2)/f+(-a*d+b*c)^2*cos(f*x+e)/b/(a^2-b ^2)/f/(a+b*sin(f*x+e))
Time = 0.73 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\frac {d^2 (e+f x)-\frac {2 \left (2 b^3 c d+27 d^2-3 b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{3/2}}-\frac {b (b c-3 d)^2 \cos (e+f x)}{(-3+b) (3+b) (3+b \sin (e+f x))}}{b^2 f} \]
(d^2*(e + f*x) - (2*(2*b^3*c*d + 27*d^2 - 3*b^2*(c^2 + 2*d^2))*ArcTan[(b + 3*Tan[(e + f*x)/2])/Sqrt[9 - b^2]])/(9 - b^2)^(3/2) - (b*(b*c - 3*d)^2*Co s[e + f*x])/((-3 + b)*(3 + b)*(3 + b*Sin[e + f*x])))/(b^2*f)
Time = 0.57 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.36, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3269, 25, 3042, 3214, 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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3269 |
| \(\displaystyle \frac {\int -\frac {b \left (2 b c d-a \left (c^2+d^2\right )\right )-\left (a^2-b^2\right ) d^2 \sin (e+f x)}{a+b \sin (e+f x)}dx}{b \left (a^2-b^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {\int \frac {b \left (2 b c d-a \left (c^2+d^2\right )\right )-\left (a^2-b^2\right ) d^2 \sin (e+f x)}{a+b \sin (e+f x)}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {\int \frac {b \left (2 b c d-a \left (c^2+d^2\right )\right )-\left (a^2-b^2\right ) d^2 \sin (e+f x)}{a+b \sin (e+f x)}dx}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {-\frac {(b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \int \frac {1}{a+b \sin (e+f x)}dx}{b}-\frac {d^2 x \left (a^2-b^2\right )}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {-\frac {(b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \int \frac {1}{a+b \sin (e+f x)}dx}{b}-\frac {d^2 x \left (a^2-b^2\right )}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {-\frac {2 (b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 b \tan \left (\frac {1}{2} (e+f x)\right )+a}d\tan \left (\frac {1}{2} (e+f x)\right )}{b f}-\frac {d^2 x \left (a^2-b^2\right )}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {\frac {4 (b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )}{b f}-\frac {d^2 x \left (a^2-b^2\right )}{b}}{b \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}-\frac {-\frac {2 (b c-a d) \left (a^2 d+a b c-2 b^2 d\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (e+f x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{b f \sqrt {a^2-b^2}}-\frac {d^2 x \left (a^2-b^2\right )}{b}}{b \left (a^2-b^2\right )}\) |
-((-(((a^2 - b^2)*d^2*x)/b) - (2*(b*c - a*d)*(a*b*c + a^2*d - 2*b^2*d)*Arc Tan[(2*b + 2*a*Tan[(e + f*x)/2])/(2*Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]* f))/(b*(a^2 - b^2))) + ((b*c - a*d)^2*Cos[e + f*x])/(b*(a^2 - b^2)*f*(a + b*Sin[e + f*x]))
3.8.8.3.1 Defintions of rubi rules used
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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
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]
Time = 1.01 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.85
| method | result | size |
| derivativedivides | \(\frac {\frac {2 d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {-\frac {b^{2} \left (d^{2} a^{2}-2 a b c d +b^{2} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {b \left (d^{2} a^{2}-2 a b c d +b^{2} c^{2}\right )}{a^{2}-b^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (a^{3} d^{2}-a \,b^{2} c^{2}-2 a \,b^{2} d^{2}+2 b^{3} c d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{2}}}{f}\) | \(220\) |
| default | \(\frac {\frac {2 d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b^{2}}-\frac {2 \left (\frac {-\frac {b^{2} \left (d^{2} a^{2}-2 a b c d +b^{2} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {b \left (d^{2} a^{2}-2 a b c d +b^{2} c^{2}\right )}{a^{2}-b^{2}}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (a^{3} d^{2}-a \,b^{2} c^{2}-2 a \,b^{2} d^{2}+2 b^{3} c d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{2}}}{f}\) | \(220\) |
| risch | \(\frac {d^{2} x}{b^{2}}-\frac {2 i \left (d^{2} a^{2}-2 a b c d +b^{2} c^{2}\right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{b^{2} \left (a^{2}-b^{2}\right ) f \left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+2 a \,{\mathrm e}^{i \left (f x +e \right )}+i b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{3} d^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a \,c^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}-\frac {2 b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) c d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{3} d^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a \,c^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {2 b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) c d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}\) | \(761\) |
1/f*(2*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))-2/b^2*((-b^2*(a^2*d^2-2*a*b*c*d+ b^2*c^2)/(a^2-b^2)/a*tan(1/2*f*x+1/2*e)-b*(a^2*d^2-2*a*b*c*d+b^2*c^2)/(a^2 -b^2))/(tan(1/2*f*x+1/2*e)^2*a+2*b*tan(1/2*f*x+1/2*e)+a)+(a^3*d^2-a*b^2*c^ 2-2*a*b^2*d^2+2*b^3*c*d)/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e )+2*b)/(a^2-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (124) = 248\).
Time = 0.31 (sec) , antiderivative size = 665, normalized size of antiderivative = 5.59 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\left [\frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d^{2} f x \sin \left (f x + e\right ) + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d^{2} f x + {\left (a^{2} b^{2} c^{2} - 2 \, a b^{3} c d - {\left (a^{4} - 2 \, a^{2} b^{2}\right )} d^{2} + {\left (a b^{3} c^{2} - 2 \, b^{4} c d - {\left (a^{3} b - 2 \, a b^{3}\right )} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} c^{2} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} c d + {\left (a^{4} b - a^{2} b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} f\right )}}, \frac {{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d^{2} f x \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d^{2} f x - {\left (a^{2} b^{2} c^{2} - 2 \, a b^{3} c d - {\left (a^{4} - 2 \, a^{2} b^{2}\right )} d^{2} + {\left (a b^{3} c^{2} - 2 \, b^{4} c d - {\left (a^{3} b - 2 \, a b^{3}\right )} d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + {\left ({\left (a^{2} b^{3} - b^{5}\right )} c^{2} - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} c d + {\left (a^{4} b - a^{2} b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} f}\right ] \]
[1/2*(2*(a^4*b - 2*a^2*b^3 + b^5)*d^2*f*x*sin(f*x + e) + 2*(a^5 - 2*a^3*b^ 2 + a*b^4)*d^2*f*x + (a^2*b^2*c^2 - 2*a*b^3*c*d - (a^4 - 2*a^2*b^2)*d^2 + (a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)*d^2)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 - 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*c os(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*((a^2*b^3 - b^5)*c^2 - 2*(a^3*b^2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^4*b^3 - 2*a^2*b^5 + b^7)*f*sin(f*x + e) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*f), ((a ^4*b - 2*a^2*b^3 + b^5)*d^2*f*x*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*d ^2*f*x - (a^2*b^2*c^2 - 2*a*b^3*c*d - (a^4 - 2*a^2*b^2)*d^2 + (a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)*d^2)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(- (a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + ((a^2*b^3 - b^5)*c^ 2 - 2*(a^3*b^2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^4*b ^3 - 2*a^2*b^5 + b^7)*f*sin(f*x + e) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*f)]
Timed out. \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.30 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.03 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\frac {\frac {{\left (f x + e\right )} d^{2}}{b^{2}} + \frac {2 \, {\left (a b^{2} c^{2} - 2 \, b^{3} c d - a^{3} d^{2} + 2 \, a b^{2} d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (b^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b^{2} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{2} b d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )}}{{\left (a^{3} b - a b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}}{f} \]
((f*x + e)*d^2/b^2 + 2*(a*b^2*c^2 - 2*b^3*c*d - a^3*d^2 + 2*a*b^2*d^2)*(pi *floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b )/sqrt(a^2 - b^2)))/((a^2*b^2 - b^4)*sqrt(a^2 - b^2)) + 2*(b^3*c^2*tan(1/2 *f*x + 1/2*e) - 2*a*b^2*c*d*tan(1/2*f*x + 1/2*e) + a^2*b*d^2*tan(1/2*f*x + 1/2*e) + a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)/((a^3*b - a*b^3)*(a*tan(1/2*f *x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)))/f
Time = 16.67 (sec) , antiderivative size = 5776, normalized size of antiderivative = 48.54 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^2} \, dx=\text {Too large to display} \]
((2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(b*(a^2 - b^2)) + (2*tan(e/2 + (f*x)/ 2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*(a^2 - b^2)))/(f*(a + 2*b*tan(e/2 + (f*x)/2) + a*tan(e/2 + (f*x)/2)^2)) - (2*d^2*atan(((d^2*((32*tan(e/2 + (f *x)/2)*(2*a^7*b*d^4 - 2*a*b^7*d^4 + a^3*b^5*c^4 + 9*a^3*b^5*d^4 - 8*a^5*b^ 3*d^4 + 4*a*b^7*c^2*d^2 - 8*a^2*b^6*c*d^3 - 4*a^2*b^6*c^3*d + 4*a^4*b^4*c* d^3 + 4*a^3*b^5*c^2*d^2 - 2*a^5*b^3*c^2*d^2))/(b^7 - 2*a^2*b^5 + a^4*b^3) - (32*(a^6*b*d^4 + a^2*b^5*d^4 - 2*a^4*b^3*d^4))/(b^6 - 2*a^2*b^4 + a^4*b^ 2) + (d^2*((32*(a*b^8*d^2 + a^3*b^6*c^2 - a^5*b^4*c^2 - a^3*b^6*d^2 - 2*a^ 2*b^7*c*d + 2*a^4*b^5*c*d))/(b^6 - 2*a^2*b^4 + a^4*b^2) - (d^2*((32*(a^2*b ^9 - 2*a^4*b^7 + a^6*b^5))/(b^6 - 2*a^2*b^4 + a^4*b^2) + (32*tan(e/2 + (f* x)/2)*(3*a*b^11 - 8*a^3*b^9 + 7*a^5*b^7 - 2*a^7*b^5))/(b^7 - 2*a^2*b^5 + a ^4*b^3))*1i)/b^2 + (32*tan(e/2 + (f*x)/2)*(2*a^2*b^8*c^2 - 2*a^4*b^6*c^2 + 4*a^2*b^8*d^2 - 6*a^4*b^6*d^2 + 2*a^6*b^4*d^2 - 4*a*b^9*c*d + 4*a^3*b^7*c *d))/(b^7 - 2*a^2*b^5 + a^4*b^3))*1i)/b^2))/b^2 - (d^2*((32*(a^6*b*d^4 + a ^2*b^5*d^4 - 2*a^4*b^3*d^4))/(b^6 - 2*a^2*b^4 + a^4*b^2) - (32*tan(e/2 + ( f*x)/2)*(2*a^7*b*d^4 - 2*a*b^7*d^4 + a^3*b^5*c^4 + 9*a^3*b^5*d^4 - 8*a^5*b ^3*d^4 + 4*a*b^7*c^2*d^2 - 8*a^2*b^6*c*d^3 - 4*a^2*b^6*c^3*d + 4*a^4*b^4*c *d^3 + 4*a^3*b^5*c^2*d^2 - 2*a^5*b^3*c^2*d^2))/(b^7 - 2*a^2*b^5 + a^4*b^3) + (d^2*((32*(a*b^8*d^2 + a^3*b^6*c^2 - a^5*b^4*c^2 - a^3*b^6*d^2 - 2*a^2* b^7*c*d + 2*a^4*b^5*c*d))/(b^6 - 2*a^2*b^4 + a^4*b^2) + (d^2*((32*(a^2*...