Integrand size = 23, antiderivative size = 242 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{11/2} f}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 (a+b)^5 f}-\frac {(10 a+b) \cot ^3(e+f x)}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )} \]
-1/5*(5*a^2-20*a*b+2*b^2)*cot(f*x+e)/(a+b)^5/f-1/15*(10*a+b)*cot(f*x+e)^3/ (a+b)^4/f-1/8*(15*a^2-40*a*b+8*b^2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2)) *b^(1/2)/(a+b)^(11/2)/f-1/5*cot(f*x+e)^5/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2-1/ 20*b*(5*a^2+4*b^2)*tan(f*x+e)/(a+b)^4/f/(a+b+b*tan(f*x+e)^2)^2-1/40*b*(35* a^2-40*a*b+24*b^2)*tan(f*x+e)/(a+b)^5/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 5.57 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.98 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (-8 (4 a-11 b) (a+b) (a+2 b+a \cos (2 (e+f x)))^2 \cot (e) \csc ^2(e+f x)-24 (a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2 \cot (e) \csc ^4(e+f x)+\frac {15 b \left (15 a^2-40 a b+8 b^2\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+8 \left (8 a^2-59 a b+23 b^2\right ) (a+2 b+a \cos (2 (e+f x)))^2 \csc (e) \csc (e+f x) \sin (f x)+8 (4 a-11 b) (a+b) (a+2 b+a \cos (2 (e+f x)))^2 \csc (e) \csc ^3(e+f x) \sin (f x)+24 (a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2 \csc (e) \csc ^5(e+f x) \sin (f x)-60 b^2 (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))+15 b (a+2 b+a \cos (2 (e+f x))) \sec (2 e) \left (\left (9 a^2+16 a b-8 b^2\right ) \sin (2 e)+3 a (-3 a+2 b) \sin (2 f x)\right )\right )}{960 (a+b)^5 f \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-8*(4*a - 11*b)*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Cot[e]*Csc[e + f*x]^2 - 24*(a + b)^2*(a + 2 *b + a*Cos[2*(e + f*x)])^2*Cot[e]*Csc[e + f*x]^4 + (15*b*(15*a^2 - 40*a*b + 8*b^2)*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])]*(a + 2* b + a*Cos[2*(e + f*x)])^2*(Cos[2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Co s[e] - I*Sin[e])^4]) + 8*(8*a^2 - 59*a*b + 23*b^2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]*Sin[f*x] + 8*(4*a - 11*b)*(a + b)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]^3*Sin[f*x] + 24*(a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2*Csc[e]*Csc[e + f*x]^5*Sin[f*x] - 60*b^2*(a + b )*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]) + 15*b*(a + 2*b + a*Cos[2*( e + f*x)])*Sec[2*e]*((9*a^2 + 16*a*b - 8*b^2)*Sin[2*e] + 3*a*(-3*a + 2*b)* Sin[2*f*x])))/(960*(a + b)^5*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.56 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4620, 365, 361, 25, 1582, 1584, 2009}
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.
| \(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (5 (a+b) \tan ^2(e+f x)+10 a+b\right )}{\left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {\frac {-\frac {1}{4} b \int -\frac {\cot ^4(e+f x) \left (-\frac {3 \left (5 a^2+4 b^2\right ) \tan ^4(e+f x)}{(a+b)^3}+\frac {4 \left (5 a^2+4 b^2\right ) \tan ^2(e+f x)}{b (a+b)^2}+\frac {4 (10 a+b)}{b (a+b)}\right )}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{4 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {1}{4} b \int \frac {\cot ^4(e+f x) \left (-\frac {3 \left (5 a^2+4 b^2\right ) \tan ^4(e+f x)}{(a+b)^3}+\frac {4 \left (5 a^2+4 b^2\right ) \tan ^2(e+f x)}{b (a+b)^2}+\frac {4 (10 a+b)}{b (a+b)}\right )}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{4 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {\frac {1}{4} b \left (\frac {\int \frac {\cot ^4(e+f x) \left (-\frac {b^2 \left (35 a^2-40 b a+24 b^2\right ) \tan ^4(e+f x)}{a+b}+8 b \left (5 a^2-10 b a+3 b^2\right ) \tan ^2(e+f x)+8 b (a+b) (10 a+b)\right )}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{2 b^2 (a+b)^3}-\frac {\left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{2 (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}\right )-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{4 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\frac {\frac {1}{4} b \left (\frac {\int \left (8 b (10 a+b) \cot ^4(e+f x)+\frac {8 b \left (5 a^2-20 b a+2 b^2\right ) \cot ^2(e+f x)}{a+b}-\frac {5 b^2 \left (15 a^2-40 b a+8 b^2\right )}{(a+b) \left (b \tan ^2(e+f x)+a+b\right )}\right )d\tan (e+f x)}{2 b^2 (a+b)^3}-\frac {\left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{2 (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}\right )-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{4 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {1}{4} b \left (\frac {-\frac {5 b^{3/2} \left (15 a^2-40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {8 b \left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{a+b}-\frac {8}{3} b (10 a+b) \cot ^3(e+f x)}{2 b^2 (a+b)^3}-\frac {\left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{2 (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )}\right )-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{4 (a+b)^3 \left (a+b \tan ^2(e+f x)+b\right )^2}}{5 (a+b)}-\frac {\cot ^5(e+f x)}{5 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
(-1/5*Cot[e + f*x]^5/((a + b)*(a + b + b*Tan[e + f*x]^2)^2) + (-1/4*(b*(5* a^2 + 4*b^2)*Tan[e + f*x])/((a + b)^3*(a + b + b*Tan[e + f*x]^2)^2) + (b*( ((-5*b^(3/2)*(15*a^2 - 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[ a + b]])/(a + b)^(3/2) - (8*b*(5*a^2 - 20*a*b + 2*b^2)*Cot[e + f*x])/(a + b) - (8*b*(10*a + b)*Cot[e + f*x]^3)/3)/(2*b^2*(a + b)^3) - ((35*a^2 - 40* a*b + 24*b^2)*Tan[e + f*x])/(2*(a + b)^4*(a + b + b*Tan[e + f*x]^2))))/4)/ (5*(a + b)))/f
3.1.66.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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 = 2.55 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.74
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -a \,b^{2}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}+a b -8 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-40 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{5}}-\frac {1}{5 \left (a +b \right )^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a -b}{3 \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-4 a b +b^{2}}{\left (a +b \right )^{5} \tan \left (f x +e \right )}}{f}\) | \(180\) |
| default | \(\frac {-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -a \,b^{2}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}+a b -8 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-40 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{5}}-\frac {1}{5 \left (a +b \right )^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a -b}{3 \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-4 a b +b^{2}}{\left (a +b \right )^{5} \tan \left (f x +e \right )}}{f}\) | \(180\) |
| risch | \(\text {Expression too large to display}\) | \(948\) |
1/f*(-b/(a+b)^5*(((7/8*a^2*b-a*b^2)*tan(f*x+e)^3+1/8*a*(9*a^2+a*b-8*b^2)*t an(f*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(15*a^2-40*a*b+8*b^2)/((a+b)*b)^(1/2 )*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))-1/5/(a+b)^3/tan(f*x+e)^5-1/3*(2*a- b)/(a+b)^4/tan(f*x+e)^3-(a^2-4*a*b+b^2)/(a+b)^5/tan(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 668 vs. \(2 (222) = 444\).
Time = 0.36 (sec) , antiderivative size = 1423, normalized size of antiderivative = 5.88 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
[-1/480*(4*(64*a^4 - 607*a^3*b + 274*a^2*b^2)*cos(f*x + e)^9 - 4*(160*a^4 - 1533*a^3*b + 1599*a^2*b^2 - 488*a*b^3)*cos(f*x + e)^7 + 4*(120*a^4 - 120 5*a^3*b + 2769*a^2*b^2 - 1392*a*b^3 + 184*b^4)*cos(f*x + e)^5 + 20*(75*a^3 *b - 305*a^2*b^2 + 320*a*b^3 - 56*b^4)*cos(f*x + e)^3 - 15*((15*a^4 - 40*a ^3*b + 8*a^2*b^2)*cos(f*x + e)^8 - 2*(15*a^4 - 55*a^3*b + 48*a^2*b^2 - 8*a *b^3)*cos(f*x + e)^6 + (15*a^4 - 100*a^3*b + 183*a^2*b^2 - 72*a*b^3 + 8*b^ 4)*cos(f*x + e)^4 + 15*a^2*b^2 - 40*a*b^3 + 8*b^4 + 2*(15*a^3*b - 55*a^2*b ^2 + 48*a*b^3 - 8*b^4)*cos(f*x + e)^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))*si n(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*(15*a^2*b^2 - 40*a*b^3 + 8*b^4)*cos(f*x + e))/(((a^7 + 5*a^6* b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^8 - 2*(a ^7 + 4*a^6*b + 5*a^5*b^2 - 5*a^3*b^4 - 4*a^2*b^5 - a*b^6)*f*cos(f*x + e)^6 + (a^7 + a^6*b - 9*a^5*b^2 - 25*a^4*b^3 - 25*a^3*b^4 - 9*a^2*b^5 + a*b^6 + b^7)*f*cos(f*x + e)^4 + 2*(a^6*b + 4*a^5*b^2 + 5*a^4*b^3 - 5*a^2*b^5 - 4 *a*b^6 - b^7)*f*cos(f*x + e)^2 + (a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^ 2*b^5 + 5*a*b^6 + b^7)*f)*sin(f*x + e)), -1/240*(2*(64*a^4 - 607*a^3*b + 2 74*a^2*b^2)*cos(f*x + e)^9 - 2*(160*a^4 - 1533*a^3*b + 1599*a^2*b^2 - 488* a*b^3)*cos(f*x + e)^7 + 2*(120*a^4 - 1205*a^3*b + 2769*a^2*b^2 - 1392*a...
Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.79 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (15 \, a^{2} b^{2} - 40 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{8} + 25 \, {\left (15 \, a^{3} b - 25 \, a^{2} b^{2} - 32 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{6} + 8 \, {\left (15 \, a^{4} - 10 \, a^{3} b - 57 \, a^{2} b^{2} - 24 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + 24 \, a^{4} + 96 \, a^{3} b + 144 \, a^{2} b^{2} + 96 \, a b^{3} + 24 \, b^{4} + 8 \, {\left (10 \, a^{4} + 31 \, a^{3} b + 33 \, a^{2} b^{2} + 13 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{9} + 2 \, {\left (a^{6} b + 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} + 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} + 6 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{7} + {\left (a^{7} + 7 \, a^{6} b + 21 \, a^{5} b^{2} + 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 21 \, a^{2} b^{5} + 7 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]
-1/120*(15*(15*a^2*b - 40*a*b^2 + 8*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b )*b))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sqrt((a + b)*b)) + (15*(15*a^2*b^2 - 40*a*b^3 + 8*b^4)*tan(f*x + e)^8 + 25*(15*a^3* b - 25*a^2*b^2 - 32*a*b^3 + 8*b^4)*tan(f*x + e)^6 + 8*(15*a^4 - 10*a^3*b - 57*a^2*b^2 - 24*a*b^3 + 8*b^4)*tan(f*x + e)^4 + 24*a^4 + 96*a^3*b + 144*a ^2*b^2 + 96*a*b^3 + 24*b^4 + 8*(10*a^4 + 31*a^3*b + 33*a^2*b^2 + 13*a*b^3 + b^4)*tan(f*x + e)^2)/((a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5 *a*b^6 + b^7)*tan(f*x + e)^9 + 2*(a^6*b + 6*a^5*b^2 + 15*a^4*b^3 + 20*a^3* b^4 + 15*a^2*b^5 + 6*a*b^6 + b^7)*tan(f*x + e)^7 + (a^7 + 7*a^6*b + 21*a^5 *b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2*b^5 + 7*a*b^6 + b^7)*tan(f*x + e)^ 5))/f
Time = 0.46 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.52 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 8 \, a b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) + a^{2} b^{2} \tan \left (f x + e\right ) - 8 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 60 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 5 \, a b \tan \left (f x + e\right )^{2} - 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]
-1/120*(15*(15*a^2*b - 40*a*b^2 + 8*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn (b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sqrt(a*b + b^2)) + 15*(7*a^2*b^2*tan(f*x + e)^3 - 8*a*b^3*tan(f*x + e)^3 + 9*a^3*b*tan(f*x + e) + a^2*b^2*tan(f*x + e ) - 8*a*b^3*tan(f*x + e))/((a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a* b^4 + b^5)*(b*tan(f*x + e)^2 + a + b)^2) + 8*(15*a^2*tan(f*x + e)^4 - 60*a *b*tan(f*x + e)^4 + 15*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 5*a*b* tan(f*x + e)^2 - 5*b^2*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((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)^5))/f
Time = 21.39 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a+b\right )}{15\,{\left (a+b\right )}^2}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (15\,a^2\,b-40\,a\,b^2+8\,b^3\right )}{24\,{\left (a+b\right )}^4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{15\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^8\,\left (15\,a^2\,b^2-40\,a\,b^3+8\,b^4\right )}{8\,{\left (a+b\right )}^5}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )}{{\left (a+b\right )}^{11/2}}\right )\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{8\,f\,{\left (a+b\right )}^{11/2}} \]
- (1/(5*(a + b)) + (tan(e + f*x)^2*(10*a + b))/(15*(a + b)^2) + (5*tan(e + f*x)^6*(15*a^2*b - 40*a*b^2 + 8*b^3))/(24*(a + b)^4) + (tan(e + f*x)^4*(1 5*a^2 - 40*a*b + 8*b^2))/(15*(a + b)^3) + (tan(e + f*x)^8*(8*b^4 - 40*a*b^ 3 + 15*a^2*b^2))/(8*(a + b)^5))/(f*(tan(e + f*x)^5*(2*a*b + a^2 + b^2) + t an(e + f*x)^7*(2*a*b + 2*b^2) + b^2*tan(e + f*x)^9)) - (b^(1/2)*atan((b^(1 /2)*tan(e + f*x)*(5*a*b^4 + 5*a^4*b + a^5 + b^5 + 10*a^2*b^3 + 10*a^3*b^2) )/(a + b)^(11/2))*(15*a^2 - 40*a*b + 8*b^2))/(8*f*(a + b)^(11/2))