Integrand size = 23, antiderivative size = 124 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{7/2} f}-\frac {15 \cot (e+f x)}{8 (a+b)^3 f}+\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
-15/8*cot(f*x+e)/(a+b)^3/f-15/8*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^( 1/2)/(a+b)^(7/2)/f+1/4*cot(f*x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2+5/8*cot(f *x+e)/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 6.88 (sec) , antiderivative size = 749, normalized size of antiderivative = 6.04 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (\frac {120 b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\csc (e) \csc (e+f x) \sec (2 e) \left (\left (-32 a^4-64 a^3 b+22 a^2 b^2+80 a b^3+16 b^4\right ) \sin (f x)+2 a \left (16 a^3+23 a^2 b-27 a b^2-4 b^3\right ) \sin (3 f x)-48 a^4 \sin (2 e-f x)-128 a^3 b \sin (2 e-f x)-106 a^2 b^2 \sin (2 e-f x)+80 a b^3 \sin (2 e-f x)+16 b^4 \sin (2 e-f x)+48 a^4 \sin (2 e+f x)+146 a^3 b \sin (2 e+f x)+182 a^2 b^2 \sin (2 e+f x)+80 a b^3 \sin (2 e+f x)+16 b^4 \sin (2 e+f x)-32 a^4 \sin (4 e+f x)-82 a^3 b \sin (4 e+f x)-54 a^2 b^2 \sin (4 e+f x)-80 a b^3 \sin (4 e+f x)-16 b^4 \sin (4 e+f x)-8 a^4 \sin (2 e+3 f x)+18 a^3 b \sin (2 e+3 f x)+54 a^2 b^2 \sin (2 e+3 f x)+8 a b^3 \sin (2 e+3 f x)+32 a^4 \sin (4 e+3 f x)+73 a^3 b \sin (4 e+3 f x)+24 a^2 b^2 \sin (4 e+3 f x)+8 a b^3 \sin (4 e+3 f x)-8 a^4 \sin (6 e+3 f x)-9 a^3 b \sin (6 e+3 f x)-24 a^2 b^2 \sin (6 e+3 f x)-8 a b^3 \sin (6 e+3 f x)+8 a^4 \sin (2 e+5 f x)-9 a^3 b \sin (2 e+5 f x)-2 a^2 b^2 \sin (2 e+5 f x)+9 a^3 b \sin (4 e+5 f x)+2 a^2 b^2 \sin (4 e+5 f x)+8 a^4 \sin (6 e+5 f x)\right )}{a^2}\right )}{512 (a+b)^3 f \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*((120*b*ArcTan[(Sec[f*x]*(C os[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])]*(a + 2*b + a*Cos[2*(e + f*x)])^2*(C os[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (Csc[ e]*Csc[e + f*x]*Sec[2*e]*((-32*a^4 - 64*a^3*b + 22*a^2*b^2 + 80*a*b^3 + 16 *b^4)*Sin[f*x] + 2*a*(16*a^3 + 23*a^2*b - 27*a*b^2 - 4*b^3)*Sin[3*f*x] - 4 8*a^4*Sin[2*e - f*x] - 128*a^3*b*Sin[2*e - f*x] - 106*a^2*b^2*Sin[2*e - f* x] + 80*a*b^3*Sin[2*e - f*x] + 16*b^4*Sin[2*e - f*x] + 48*a^4*Sin[2*e + f* x] + 146*a^3*b*Sin[2*e + f*x] + 182*a^2*b^2*Sin[2*e + f*x] + 80*a*b^3*Sin[ 2*e + f*x] + 16*b^4*Sin[2*e + f*x] - 32*a^4*Sin[4*e + f*x] - 82*a^3*b*Sin[ 4*e + f*x] - 54*a^2*b^2*Sin[4*e + f*x] - 80*a*b^3*Sin[4*e + f*x] - 16*b^4* Sin[4*e + f*x] - 8*a^4*Sin[2*e + 3*f*x] + 18*a^3*b*Sin[2*e + 3*f*x] + 54*a ^2*b^2*Sin[2*e + 3*f*x] + 8*a*b^3*Sin[2*e + 3*f*x] + 32*a^4*Sin[4*e + 3*f* x] + 73*a^3*b*Sin[4*e + 3*f*x] + 24*a^2*b^2*Sin[4*e + 3*f*x] + 8*a*b^3*Sin [4*e + 3*f*x] - 8*a^4*Sin[6*e + 3*f*x] - 9*a^3*b*Sin[6*e + 3*f*x] - 24*a^2 *b^2*Sin[6*e + 3*f*x] - 8*a*b^3*Sin[6*e + 3*f*x] + 8*a^4*Sin[2*e + 5*f*x] - 9*a^3*b*Sin[2*e + 5*f*x] - 2*a^2*b^2*Sin[2*e + 5*f*x] + 9*a^3*b*Sin[4*e + 5*f*x] + 2*a^2*b^2*Sin[4*e + 5*f*x] + 8*a^4*Sin[6*e + 5*f*x]))/a^2))/(51 2*(a + b)^3*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4620, 253, 253, 264, 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.
| \(\displaystyle \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^2 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {5 \int \frac {\cot ^2(e+f x)}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 (a+b)}+\frac {\cot (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {\cot ^2(e+f x)}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{2 (a+b)}+\frac {\cot (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{4 (a+b)}+\frac {\cot (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {b \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a+b}-\frac {\cot (e+f x)}{a+b}\right )}{2 (a+b)}+\frac {\cot (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{4 (a+b)}+\frac {\cot (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\cot (e+f x)}{a+b}\right )}{2 (a+b)}+\frac {\cot (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{4 (a+b)}+\frac {\cot (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
(Cot[e + f*x]/(4*(a + b)*(a + b + b*Tan[e + f*x]^2)^2) + (5*((3*(-((Sqrt[b ]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2)) - Cot[e + f*x ]/(a + b)))/(2*(a + b)) + Cot[e + f*x]/(2*(a + b)*(a + b + b*Tan[e + f*x]^ 2))))/(4*(a + b)))/f
3.1.64.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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 = 1.37 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {7 b \tan \left (f x +e \right )^{3}}{8}+\left (\frac {9 a}{8}+\frac {9 b}{8}\right ) \tan \left (f x +e \right )}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \tan \left (f x +e \right )}}{f}\) | \(97\) |
| default | \(\frac {-\frac {b \left (\frac {\frac {7 b \tan \left (f x +e \right )^{3}}{8}+\left (\frac {9 a}{8}+\frac {9 b}{8}\right ) \tan \left (f x +e \right )}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3}}-\frac {1}{\left (a +b \right )^{3} \tan \left (f x +e \right )}}{f}\) | \(97\) |
| risch | \(-\frac {i \left (8 a^{4} {\mathrm e}^{8 i \left (f x +e \right )}+9 a^{3} b \,{\mathrm e}^{8 i \left (f x +e \right )}+24 a^{2} b^{2} {\mathrm e}^{8 i \left (f x +e \right )}+8 a \,b^{3} {\mathrm e}^{8 i \left (f x +e \right )}+32 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}+82 a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+54 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+80 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}+16 b^{4} {\mathrm e}^{6 i \left (f x +e \right )}+48 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}+128 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+106 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-80 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}-16 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}+32 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}+46 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}-54 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-8 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}+8 a^{4}-9 a^{3} b -2 a^{2} b^{2}\right )}{4 a^{2} \left (a +b \right )^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}-\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 \left (a +b \right )^{4} f}+\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 \left (a +b \right )^{4} f}\) | \(461\) |
1/f*(-b/(a+b)^3*((7/8*b*tan(f*x+e)^3+(9/8*a+9/8*b)*tan(f*x+e))/(a+b+b*tan( f*x+e)^2)^2+15/8/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))-1/( a+b)^3/tan(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (108) = 216\).
Time = 0.31 (sec) , antiderivative size = 615, normalized size of antiderivative = 4.96 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (5 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + 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 ) \sin \left (f x + e\right ) + 60 \, b^{2} \cos \left (f x + e\right )}{32 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (5 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + 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 ) \sin \left (f x + e\right ) + 30 \, b^{2} \cos \left (f x + e\right )}{16 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/32*(4*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 + 20*(5*a*b - b^2)*cos(f* x + e)^3 - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + 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))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos (f*x + e)^2 + b^2))*sin(f*x + e) + 60*b^2*cos(f*x + e))/(((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f)*sin(f *x + e)), -1/16*(2*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 + 10*(5*a*b - b^ 2)*cos(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*s qrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/( b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*b^2*cos(f*x + e))/(((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*f*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*f*cos(f*x + e)^2 + (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 )*f)*sin(f*x + e))]
Timed out. \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (108) = 216\).
Time = 0.27 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.77 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, b^{2} \tan \left (f x + e\right )^{4} + 25 \, {\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )}}{8 \, f} \]
-1/8*(15*b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 3*a^2*b + 3*a*b^ 2 + b^3)*sqrt((a + b)*b)) + (15*b^2*tan(f*x + e)^4 + 25*(a*b + b^2)*tan(f* x + e)^2 + 8*a^2 + 16*a*b + 8*b^2)/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)* tan(f*x + e)^5 + 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*tan(f*x + e)^3 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*tan(f* x + e)))/f
Time = 0.45 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.43 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {7 \, b^{2} \tan \left (f x + e\right )^{3} + 9 \, a b \tan \left (f x + e\right ) + 9 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{8 \, f} \]
-1/8*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt (a*b + b^2)))*b/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b + b^2)) + (7*b^2 *tan(f*x + e)^3 + 9*a*b*tan(f*x + e) + 9*b^2*tan(f*x + e))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)^2) + 8/((a^3 + 3*a^2*b + 3*a* b^2 + b^3)*tan(f*x + e)))/f
Time = 19.49 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.18 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {1}{a+b}+\frac {25\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,{\left (a+b\right )}^2}+\frac {15\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{8\,{\left (a+b\right )}^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3\,\left (2\,b^2+2\,a\,b\right )+\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+2\,a\,b+b^2\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{8\,f\,{\left (a+b\right )}^{7/2}} \]
- (1/(a + b) + (25*b*tan(e + f*x)^2)/(8*(a + b)^2) + (15*b^2*tan(e + f*x)^ 4)/(8*(a + b)^3))/(f*(tan(e + f*x)^3*(2*a*b + 2*b^2) + tan(e + f*x)*(2*a*b + a^2 + b^2) + b^2*tan(e + f*x)^5)) - (15*b^(1/2)*atan((b^(1/2)*tan(e + f *x)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(a + b)^(7/2)))/(8*f*(a + b)^(7/2))