Integrand size = 23, antiderivative size = 130 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+4 b) x}{2 a^3}-\frac {\sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 \sqrt {a+b} f}-\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {b \tan (e+f x)}{a^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
1/2*(a+4*b)*x/a^3-1/2*(3*a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^( 1/2)/a^3/f/(a+b)^(1/2)-1/2*cos(f*x+e)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)- b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 8.73 (sec) , antiderivative size = 699, normalized size of antiderivative = 5.38 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^2 \sec ^4(e+f x) \left (-\frac {2 \left (16 x+\frac {\left (-a^3+6 a^2 b+24 a b^2+16 b^3\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{a^2}-\frac {-64 (a+2 b) x+\frac {\left (a^4-16 a^3 b-144 a^2 b^2-256 a b^3-128 b^4\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {16 a \cos (2 f x) \sin (2 e)}{f}+\frac {16 a \cos (2 e) \sin (2 f x)}{f}-\frac {\left (a^3+18 a^2 b+48 a b^2+32 b^3\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}}{a^3}+\frac {2 \left (\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}\right )}{b^{3/2} f}+\frac {-\frac {a \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\frac {\sqrt {b} (a+2 b) \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}}{b^{3/2} f}\right )}{256 \left (a+b \sec ^2(e+f x)\right )^2} \]
((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*((-2*(16*x + ((-a^3 + 6*a ^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin [e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*S in[e])^4]) + ((a^2 + 8*a*b + 8*b^2)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/( b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin [e]))))/a^2 - (-64*(a + 2*b)*x + ((a^4 - 16*a^3*b - 144*a^2*b^2 - 256*a*b^ 3 - 128*b^4)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x ]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Co s[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (16*a*Cos[2*f*x]*Sin[2*e])/f + (16*a*Cos[2*e]*Sin[2*f*x])/f - ((a^3 + 18*a ^2*b + 48*a*b^2 + 32*b^3)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)* f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + Sin[e])))/a^3 + (2*(((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2 ) - (a*Sqrt[b]*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))) )/(b^(3/2)*f) + (-((a*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^ (3/2)) + (Sqrt[b]*(a + 2*b)*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2* (e + f*x)])))/(b^(3/2)*f)))/(256*(a + b*Sec[e + f*x]^2)^2)
Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4620, 373, 402, 27, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 373 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \tan ^2(e+f x)+a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 (a+b) \left (-2 b \tan ^2(e+f x)+a+2 b\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {2 b \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {-2 b \tan ^2(e+f x)+a+2 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{a}-\frac {2 b \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a+4 b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b (3 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {2 b \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a+4 b) \arctan (\tan (e+f x))}{a}-\frac {b (3 a+4 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{a}-\frac {2 b \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {(a+4 b) \arctan (\tan (e+f x))}{a}-\frac {\sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{a}-\frac {2 b \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)+b\right )}}{2 a}-\frac {\tan (e+f x)}{2 a \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
(-1/2*Tan[e + f*x]/(a*(1 + Tan[e + f*x]^2)*(a + b + b*Tan[e + f*x]^2)) + ( (((a + 4*b)*ArcTan[Tan[e + f*x]])/a - (Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b] *Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/a - (2*b*Tan[e + f*x])/(a*(a + b + b*Tan[e + f*x]^2)))/(2*a))/f
3.1.49.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 1.68 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}}{f}\) | \(109\) |
| default | \(\frac {-\frac {b \left (\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}+\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (1+\tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +4 b \right ) \arctan \left (\tan \left (f x +e \right )\right )}{2}}{a^{3}}}{f}\) | \(109\) |
| risch | \(\frac {x}{2 a^{2}}+\frac {2 x b}{a^{3}}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{2} f}-\frac {i b \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right ) f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{\left (a +b \right ) f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right ) f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{\left (a +b \right ) f \,a^{3}}\) | \(340\) |
1/f*(-1/a^3*b*(1/2*a*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(3*a+4*b)/((a+b)* b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+1/a^3*(-1/2*a*tan(f*x+e)/(1 +tan(f*x+e)^2)+1/2*(a+4*b)*arctan(tan(f*x+e))))
Time = 0.30 (sec) , antiderivative size = 441, normalized size of antiderivative = 3.39 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a b + 4 \, b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac {2 \, {\left (a^{2} + 4 \, a b\right )} f x \cos \left (f x + e\right )^{2} + 2 \, {\left (a b + 4 \, b^{2}\right )} f x + {\left ({\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 4 \, b^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left (a^{2} \cos \left (f x + e\right )^{3} + 2 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \]
[1/8*(4*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 4*(a*b + 4*b^2)*f*x + ((3*a^2 + 4*a*b)*cos(f*x + e)^2 + 3*a*b + 4*b^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a^2 + 3* a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(-b/(a + b))*s in(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) - 4* (a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a^4*f*cos(f*x + e )^2 + a^3*b*f), 1/4*(2*(a^2 + 4*a*b)*f*x*cos(f*x + e)^2 + 2*(a*b + 4*b^2)* f*x + ((3*a^2 + 4*a*b)*cos(f*x + e)^2 + 3*a*b + 4*b^2)*sqrt(b/(a + b))*arc tan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin (f*x + e))) - 2*(a^2*cos(f*x + e)^3 + 2*a*b*cos(f*x + e))*sin(f*x + e))/(a ^4*f*cos(f*x + e)^2 + a^3*b*f)]
\[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\sin ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {2 \, b \tan \left (f x + e\right )^{3} + {\left (a + 2 \, b\right )} \tan \left (f x + e\right )}{a^{2} b \tan \left (f x + e\right )^{4} + a^{3} + a^{2} b + {\left (a^{3} + 2 \, a^{2} b\right )} \tan \left (f x + e\right )^{2}} - \frac {{\left (f x + e\right )} {\left (a + 4 \, b\right )}}{a^{3}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3}}}{2 \, f} \]
-1/2*((2*b*tan(f*x + e)^3 + (a + 2*b)*tan(f*x + e))/(a^2*b*tan(f*x + e)^4 + a^3 + a^2*b + (a^3 + 2*a^2*b)*tan(f*x + e)^2) - (f*x + e)*(a + 4*b)/a^3 + (3*a*b + 4*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)* a^3))/f
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (f x + e\right )} {\left (a + 4 \, b\right )}}{a^{3}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a b + 4 \, b^{2}\right )}}{\sqrt {a b + b^{2}} a^{3}} - \frac {2 \, b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right ) + 2 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )} a^{2}}}{2 \, f} \]
1/2*((f*x + e)*(a + 4*b)/a^3 - (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arct an(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3*a*b + 4*b^2)/(sqrt(a*b + b^2)*a^3) - (2*b*tan(f*x + e)^3 + a*tan(f*x + e) + 2*b*tan(f*x + e))/((b*tan(f*x + e )^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x + e)^2 + a + b)*a^2))/f
Time = 18.50 (sec) , antiderivative size = 816, normalized size of antiderivative = 6.28 \[ \int \frac {\sin ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+2\,b\right )}{2\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^3}{a^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (a+2\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atan}\left (\frac {b^2\,\mathrm {tan}\left (e+f\,x\right )}{4\,\left (\frac {b^2}{4}+\frac {b^3}{a}\right )}+\frac {b^3\,\mathrm {tan}\left (e+f\,x\right )}{b^3+\frac {a\,b^2}{4}}\right )\,\left (a\,1{}\mathrm {i}+b\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a^3\,f}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^4+b\,a^3}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^4+b\,a^3}}{\frac {\frac {3\,a^2\,b^3}{2}+8\,a\,b^4+8\,b^5}{a^6}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}+\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}-\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (5\,a^2\,b^3+16\,a\,b^4+16\,b^5\right )}{a^4}-\frac {\left (\frac {2\,a^7\,b^2+4\,a^6\,b^3}{a^6}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^7\,b^2+16\,a^6\,b^3\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4\,\left (a^4+b\,a^3\right )}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}}{a^4+b\,a^3}}\right )\,\left (\frac {3\,a}{4}+b\right )\,\sqrt {-b\,\left (a+b\right )}\,2{}\mathrm {i}}{f\,\left (a^4+b\,a^3\right )} \]
(atan(((((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 + (((4*a^6*b^3 + 2*a^7*b^2)/a^6 + (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*( -b*(a + b))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/ (a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2)*1i)/(a^3*b + a^4) + (((tan (e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 - (((4*a^6*b^3 + 2*a^7*b^2) /a^6 - (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b*(a + b))^( 1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4) )*((3*a)/4 + b)*(-b*(a + b))^(1/2)*1i)/(a^3*b + a^4))/((8*a*b^4 + 8*b^5 + (3*a^2*b^3)/2)/a^6 + (((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 + (((4*a^6*b^3 + 2*a^7*b^2)/a^6 + (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*( (3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^ 4) - (((tan(e + f*x)*(16*a*b^4 + 16*b^5 + 5*a^2*b^3))/a^4 - (((4*a^6*b^3 + 2*a^7*b^2)/a^6 - (tan(e + f*x)*(16*a^6*b^3 + 8*a^7*b^2)*((3*a)/4 + b)*(-b *(a + b))^(1/2))/(a^4*(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a ^3*b + a^4))*((3*a)/4 + b)*(-b*(a + b))^(1/2))/(a^3*b + a^4)))*((3*a)/4 + b)*(-b*(a + b))^(1/2)*2i)/(f*(a^3*b + a^4)) - (atan((b^2*tan(e + f*x))/(4* (b^2/4 + b^3/a)) + (b^3*tan(e + f*x))/((a*b^2)/4 + b^3))*(a*1i + b*4i)*1i) /(2*a^3*f) - ((tan(e + f*x)*(a + 2*b))/(2*a^2) + (b*tan(e + f*x)^3)/a^2)/( f*(a + b + b*tan(e + f*x)^4 + tan(e + f*x)^2*(a + 2*b)))