Integrand size = 23, antiderivative size = 231 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{11/2} f}-\frac {\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac {(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )} \]
-1/5*(5*a^2-30*a*b+27*b^2)*cot(f*x+e)/a^5/f-1/15*(10*a-9*b)*cot(f*x+e)^3/a ^4/f-1/8*(15*a^2-70*a*b+63*b^2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2) /a^(11/2)/f-1/5*cot(f*x+e)^5/a/f/(a+b*tan(f*x+e)^2)^2-1/20*b*(5*a^2-10*a*b +9*b^2)*tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)^2-1/40*b*(35*a^2-110*a*b+99*b^ 2)*tan(f*x+e)/a^5/f/(a+b*tan(f*x+e)^2)
Time = 2.40 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.50 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {-960 \sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-\frac {2 \sqrt {a} \left (1600 a^4-165 a^3 b+637 a^2 b^2-28875 a b^3+33075 b^4+4 \left (416 a^4-447 a^3 b-1400 a^2 b^2+13125 a b^3-13230 b^4\right ) \cos (2 (e+f x))-4 \left (32 a^4-257 a^3 b-2821 a^2 b^2+8925 a b^3-6615 b^4\right ) \cos (4 (e+f x))-128 a^4 \cos (6 (e+f x))+1788 a^3 b \cos (6 (e+f x))-8800 a^2 b^2 \cos (6 (e+f x))+14700 a b^3 \cos (6 (e+f x))-7560 b^4 \cos (6 (e+f x))+64 a^4 \cos (8 (e+f x))-863 a^3 b \cos (8 (e+f x))+2479 a^2 b^2 \cos (8 (e+f x))-2625 a b^3 \cos (8 (e+f x))+945 b^4 \cos (8 (e+f x))\right ) \cot (e+f x) \csc ^4(e+f x)}{(a+b+(a-b) \cos (2 (e+f x)))^2}}{7680 a^{11/2} f} \]
(-960*Sqrt[b]*(15*a^2 - 70*a*b + 63*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqr t[a]] - (2*Sqrt[a]*(1600*a^4 - 165*a^3*b + 637*a^2*b^2 - 28875*a*b^3 + 330 75*b^4 + 4*(416*a^4 - 447*a^3*b - 1400*a^2*b^2 + 13125*a*b^3 - 13230*b^4)* Cos[2*(e + f*x)] - 4*(32*a^4 - 257*a^3*b - 2821*a^2*b^2 + 8925*a*b^3 - 661 5*b^4)*Cos[4*(e + f*x)] - 128*a^4*Cos[6*(e + f*x)] + 1788*a^3*b*Cos[6*(e + f*x)] - 8800*a^2*b^2*Cos[6*(e + f*x)] + 14700*a*b^3*Cos[6*(e + f*x)] - 75 60*b^4*Cos[6*(e + f*x)] + 64*a^4*Cos[8*(e + f*x)] - 863*a^3*b*Cos[8*(e + f *x)] + 2479*a^2*b^2*Cos[8*(e + f*x)] - 2625*a*b^3*Cos[8*(e + f*x)] + 945*b ^4*Cos[8*(e + f*x)])*Cot[e + f*x]*Csc[e + f*x]^4)/(a + b + (a - b)*Cos[2*( e + f*x)])^2)/(7680*a^(11/2)*f)
Time = 0.54 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4146, 365, 361, 1582, 25, 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.
| \(\displaystyle \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{\left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \frac {\cot ^4(e+f x) \left (5 a \tan ^2(e+f x)+10 a-9 b\right )}{\left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {\frac {-\frac {1}{4} b \int \frac {\cot ^4(e+f x) \left (\frac {3 \left (5 a^2-10 b a+9 b^2\right ) \tan ^4(e+f x)}{a^3}+4 \left (-\frac {9 b}{a^2}+\frac {10}{a}-\frac {5}{b}\right ) \tan ^2(e+f x)+4 \left (\frac {9}{a}-\frac {10}{b}\right )\right )}{\left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{4 a^3 \left (a+b \tan ^2(e+f x)\right )^2}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {-\frac {1}{4} b \left (\frac {\int -\frac {\cot ^4(e+f x) \left (-\frac {b^2 \left (35 a^2-110 b a+99 b^2\right ) \tan ^4(e+f x)}{a}+8 b \left (5 a^2-20 b a+18 b^2\right ) \tan ^2(e+f x)+8 a (10 a-9 b) b\right )}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{2 a^3 b^2}+\frac {\left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{2 a^4 \left (a+b \tan ^2(e+f x)\right )}\right )-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{4 a^3 \left (a+b \tan ^2(e+f x)\right )^2}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {1}{4} b \left (\frac {\left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{2 a^4 \left (a+b \tan ^2(e+f x)\right )}-\frac {\int \frac {\cot ^4(e+f x) \left (-\frac {b^2 \left (35 a^2-110 b a+99 b^2\right ) \tan ^4(e+f x)}{a}+8 b \left (5 a^2-20 b a+18 b^2\right ) \tan ^2(e+f x)+8 a (10 a-9 b) b\right )}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{2 a^3 b^2}\right )-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{4 a^3 \left (a+b \tan ^2(e+f x)\right )^2}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\frac {-\frac {1}{4} b \left (\frac {\left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{2 a^4 \left (a+b \tan ^2(e+f x)\right )}-\frac {\int \left (8 (10 a-9 b) b \cot ^4(e+f x)+\frac {8 b \left (5 a^2-30 b a+27 b^2\right ) \cot ^2(e+f x)}{a}-\frac {5 b^2 \left (15 a^2-70 b a+63 b^2\right )}{a \left (b \tan ^2(e+f x)+a\right )}\right )d\tan (e+f x)}{2 a^3 b^2}\right )-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{4 a^3 \left (a+b \tan ^2(e+f x)\right )^2}}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{4 a^3 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {1}{4} b \left (\frac {\left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{2 a^4 \left (a+b \tan ^2(e+f x)\right )}-\frac {-\frac {8 b \left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{a}-\frac {5 b^{3/2} \left (15 a^2-70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2}}-\frac {8}{3} b (10 a-9 b) \cot ^3(e+f x)}{2 a^3 b^2}\right )}{5 a}-\frac {\cot ^5(e+f x)}{5 a \left (a+b \tan ^2(e+f x)\right )^2}}{f}\) |
(-1/5*Cot[e + f*x]^5/(a*(a + b*Tan[e + f*x]^2)^2) + (-1/4*(b*(5*a^2 - 10*a *b + 9*b^2)*Tan[e + f*x])/(a^3*(a + b*Tan[e + f*x]^2)^2) - (b*(-1/2*((-5*b ^(3/2)*(15*a^2 - 70*a*b + 63*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/ a^(3/2) - (8*b*(5*a^2 - 30*a*b + 27*b^2)*Cot[e + f*x])/a - (8*(10*a - 9*b) *b*Cot[e + f*x]^3)/3)/(a^3*b^2) + ((35*a^2 - 110*a*b + 99*b^2)*Tan[e + f*x ])/(2*a^4*(a + b*Tan[e + f*x]^2))))/4)/(5*a))/f
3.1.91.3.1 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[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 4.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{5 a^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a -3 b}{3 a^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-6 a b +6 b^{2}}{a^{5} \tan \left (f x +e \right )}-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {11}{4} a \,b^{2}+\frac {15}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}-26 a b +17 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-70 a b +63 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\) | \(175\) |
| default | \(\frac {-\frac {1}{5 a^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a -3 b}{3 a^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}-6 a b +6 b^{2}}{a^{5} \tan \left (f x +e \right )}-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {11}{4} a \,b^{2}+\frac {15}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {a \left (9 a^{2}-26 a b +17 b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-70 a b +63 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\) | \(175\) |
| risch | \(\text {Expression too large to display}\) | \(974\) |
1/f*(-1/5/a^3/tan(f*x+e)^5-1/3*(2*a-3*b)/a^4/tan(f*x+e)^3-(a^2-6*a*b+6*b^2 )/a^5/tan(f*x+e)-b/a^5*(((7/8*a^2*b-11/4*a*b^2+15/8*b^3)*tan(f*x+e)^3+1/8* a*(9*a^2-26*a*b+17*b^2)*tan(f*x+e))/(a+b*tan(f*x+e)^2)^2+1/8*(15*a^2-70*a* b+63*b^2)/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (211) = 422\).
Time = 0.37 (sec) , antiderivative size = 1199, normalized size of antiderivative = 5.19 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
[-1/480*(4*(64*a^4 - 863*a^3*b + 2479*a^2*b^2 - 2625*a*b^3 + 945*b^4)*cos( f*x + e)^9 - 4*(160*a^4 - 2173*a^3*b + 7158*a^2*b^2 - 8925*a*b^3 + 3780*b^ 4)*cos(f*x + e)^7 + 4*(120*a^4 - 1685*a^3*b + 7104*a^2*b^2 - 11025*a*b^3 + 5670*b^4)*cos(f*x + e)^5 + 20*(75*a^3*b - 530*a^2*b^2 + 1155*a*b^3 - 756* b^4)*cos(f*x + e)^3 - 15*((15*a^4 - 100*a^3*b + 218*a^2*b^2 - 196*a*b^3 + 63*b^4)*cos(f*x + e)^8 - 2*(15*a^4 - 115*a^3*b + 303*a^2*b^2 - 329*a*b^3 + 126*b^4)*cos(f*x + e)^6 + (15*a^4 - 160*a^3*b + 573*a^2*b^2 - 798*a*b^3 + 378*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 - 70*a*b^3 + 63*b^4 + 2*(15*a^3*b - 100*a^2*b^2 + 203*a*b^3 - 126*b^4)*cos(f*x + e)^2)*sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 + 4*((a^2 + a *b)*cos(f*x + e)^3 - a*b*cos(f*x + e))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^ 2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2))*sin (f*x + e) + 60*(15*a^2*b^2 - 70*a*b^3 + 63*b^4)*cos(f*x + e))/(((a^7 - 2*a ^6*b + a^5*b^2)*f*cos(f*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^ 2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e)^4 + 2*(a^ 6*b - 2*a^5*b^2)*f*cos(f*x + e)^2)*sin(f*x + e)), -1/240*(2*(64*a^4 - 863* a^3*b + 2479*a^2*b^2 - 2625*a*b^3 + 945*b^4)*cos(f*x + e)^9 - 2*(160*a^4 - 2173*a^3*b + 7158*a^2*b^2 - 8925*a*b^3 + 3780*b^4)*cos(f*x + e)^7 + 2*(12 0*a^4 - 1685*a^3*b + 7104*a^2*b^2 - 11025*a*b^3 + 5670*b^4)*cos(f*x + e)^5 + 10*(75*a^3*b - 530*a^2*b^2 + 1155*a*b^3 - 756*b^4)*cos(f*x + e)^3 - ...
Timed out. \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.92 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (15 \, a^{2} b^{2} - 70 \, a b^{3} + 63 \, b^{4}\right )} \tan \left (f x + e\right )^{8} + 25 \, {\left (15 \, a^{3} b - 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \tan \left (f x + e\right )^{6} + 8 \, {\left (15 \, a^{4} - 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \tan \left (f x + e\right )^{4} + 24 \, a^{4} + 8 \, {\left (10 \, a^{4} - 9 \, a^{3} b\right )} \tan \left (f x + e\right )^{2}}{a^{5} b^{2} \tan \left (f x + e\right )^{9} + 2 \, a^{6} b \tan \left (f x + e\right )^{7} + a^{7} \tan \left (f x + e\right )^{5}} + \frac {15 \, {\left (15 \, a^{2} b - 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}}}{120 \, f} \]
-1/120*((15*(15*a^2*b^2 - 70*a*b^3 + 63*b^4)*tan(f*x + e)^8 + 25*(15*a^3*b - 70*a^2*b^2 + 63*a*b^3)*tan(f*x + e)^6 + 8*(15*a^4 - 70*a^3*b + 63*a^2*b ^2)*tan(f*x + e)^4 + 24*a^4 + 8*(10*a^4 - 9*a^3*b)*tan(f*x + e)^2)/(a^5*b^ 2*tan(f*x + e)^9 + 2*a^6*b*tan(f*x + e)^7 + a^7*tan(f*x + e)^5) + 15*(15*a ^2*b - 70*a*b^2 + 63*b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^5) )/f
Time = 0.90 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left (15 \, a^{2} b - 70 \, a b^{2} + 63 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a^{5}} + \frac {15 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 22 \, a b^{3} \tan \left (f x + e\right )^{3} + 15 \, b^{4} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) - 26 \, a^{2} b^{2} \tan \left (f x + e\right ) + 17 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2} a^{5}} + \frac {8 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 90 \, a b \tan \left (f x + e\right )^{4} + 90 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 15 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )}}{a^{5} \tan \left (f x + e\right )^{5}}}{120 \, f} \]
-1/120*(15*(15*a^2*b - 70*a*b^2 + 63*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sg n(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/(sqrt(a*b)*a^5) + 15*(7*a^2*b^2*t an(f*x + e)^3 - 22*a*b^3*tan(f*x + e)^3 + 15*b^4*tan(f*x + e)^3 + 9*a^3*b* tan(f*x + e) - 26*a^2*b^2*tan(f*x + e) + 17*a*b^3*tan(f*x + e))/((b*tan(f* x + e)^2 + a)^2*a^5) + 8*(15*a^2*tan(f*x + e)^4 - 90*a*b*tan(f*x + e)^4 + 90*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 - 15*a*b*tan(f*x + e)^2 + 3* a^2)/(a^5*tan(f*x + e)^5))/f
Time = 12.41 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{15\,a^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a-9\,b\right )}{15\,a^2}+\frac {5\,b\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{24\,a^4}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{8\,a^5}}{f\,\left (a^2\,{\mathrm {tan}\left (e+f\,x\right )}^5+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^7+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{8\,a^{11/2}\,f} \]
- (1/(5*a) + (tan(e + f*x)^4*(15*a^2 - 70*a*b + 63*b^2))/(15*a^3) + (tan(e + f*x)^2*(10*a - 9*b))/(15*a^2) + (5*b*tan(e + f*x)^6*(15*a^2 - 70*a*b + 63*b^2))/(24*a^4) + (b^2*tan(e + f*x)^8*(15*a^2 - 70*a*b + 63*b^2))/(8*a^5 ))/(f*(a^2*tan(e + f*x)^5 + b^2*tan(e + f*x)^9 + 2*a*b*tan(e + f*x)^7)) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x))/a^(1/2))*(15*a^2 - 70*a*b + 63*b^2))/ (8*a^(11/2)*f)