Integrand size = 21, antiderivative size = 166 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^3 (a-b)^{5/2} f}-\frac {\text {arctanh}(\cos (e+f x))}{a^3 f}-\frac {b \sec (e+f x)}{4 a (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(7 a-4 b) b \sec (e+f x)}{8 a^2 (a-b)^2 f \left (a-b+b \sec ^2(e+f x)\right )} \]
-arctanh(cos(f*x+e))/a^3/f-1/4*b*sec(f*x+e)/a/(a-b)/f/(a-b+b*sec(f*x+e)^2) ^2-1/8*(7*a-4*b)*b*sec(f*x+e)/a^2/(a-b)^2/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a ^2-20*a*b+8*b^2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^3/(a-b)^ (5/2)/f
Time = 5.26 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.49 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{5/2}}+\frac {8 a^2 b^2 \cos (e+f x)}{(a-b)^2 (a+b+(a-b) \cos (2 (e+f x)))^2}-\frac {2 a (9 a-4 b) b \cos (e+f x)}{(a-b)^2 (a+b+(a-b) \cos (2 (e+f x)))}-8 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+8 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 a^3 f} \]
((Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(5/2) + (Sqrt[b]*(15*a^2 - 20*a*b + 8*b^2)*Arc Tan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(5/2) + (8* a^2*b^2*Cos[e + f*x])/((a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x)])^2) - ( 2*a*(9*a - 4*b)*b*Cos[e + f*x])/((a - b)^2*(a + b + (a - b)*Cos[2*(e + f*x )])) - 8*Log[Cos[(e + f*x)/2]] + 8*Log[Sin[(e + f*x)/2]])/(8*a^3*f)
Time = 0.42 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4147, 25, 316, 25, 402, 25, 397, 218, 219}
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.
| \(\displaystyle \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (e+f x) \left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {1}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int -\frac {-3 b \sec ^2(e+f x)+4 a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {-3 b \sec ^2(e+f x)+4 a-b}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {-\frac {\frac {b (7 a-4 b) \sec (e+f x)}{2 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {\int -\frac {8 a^2-9 b a+4 b^2-(7 a-4 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{2 a (a-b)}}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {8 a^2-9 b a+4 b^2-(7 a-4 b) b \sec ^2(e+f x)}{\left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a-b\right )}d\sec (e+f x)}{2 a (a-b)}+\frac {b (7 a-4 b) \sec (e+f x)}{2 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {-\frac {\frac {\frac {b \left (15 a^2-20 a b+8 b^2\right ) \int \frac {1}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{a}+\frac {8 (a-b)^2 \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}}{2 a (a-b)}+\frac {b (7 a-4 b) \sec (e+f x)}{2 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {\frac {\frac {8 (a-b)^2 \int \frac {1}{1-\sec ^2(e+f x)}d\sec (e+f x)}{a}+\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}}{2 a (a-b)}+\frac {b (7 a-4 b) \sec (e+f x)}{2 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\sqrt {b} \left (15 a^2-20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{a \sqrt {a-b}}+\frac {8 (a-b)^2 \text {arctanh}(\sec (e+f x))}{a}}{2 a (a-b)}+\frac {b (7 a-4 b) \sec (e+f x)}{2 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )}}{4 a (a-b)}-\frac {b \sec (e+f x)}{4 a (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
(-1/4*(b*Sec[e + f*x])/(a*(a - b)*(a - b + b*Sec[e + f*x]^2)^2) - (((Sqrt[ b]*(15*a^2 - 20*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/( a*Sqrt[a - b]) + (8*(a - b)^2*ArcTanh[Sec[e + f*x]])/a)/(2*a*(a - b)) + (( 7*a - 4*b)*b*Sec[e + f*x])/(2*a*(a - b)*(a - b + b*Sec[e + f*x]^2)))/(4*a* (a - b)))/f
3.1.83.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 0.65 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.10
| method | result | size |
| derivativedivides | \(\frac {\frac {b \left (\frac {-\frac {\left (9 a -4 b \right ) a \cos \left (f x +e \right )^{3}}{8 \left (a -b \right )}-\frac {a b \left (7 a -4 b \right ) \cos \left (f x +e \right )}{8 \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {\left (15 a^{2}-20 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {b \left (a -b \right )}}\right )}{a^{3}}-\frac {\ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{3}}+\frac {\ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{3}}}{f}\) | \(183\) |
| default | \(\frac {\frac {b \left (\frac {-\frac {\left (9 a -4 b \right ) a \cos \left (f x +e \right )^{3}}{8 \left (a -b \right )}-\frac {a b \left (7 a -4 b \right ) \cos \left (f x +e \right )}{8 \left (a^{2}-2 a b +b^{2}\right )}}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {\left (15 a^{2}-20 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \left (a^{2}-2 a b +b^{2}\right ) \sqrt {b \left (a -b \right )}}\right )}{a^{3}}-\frac {\ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{3}}+\frac {\ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{3}}}{f}\) | \(183\) |
| risch | \(-\frac {b \left (9 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}-13 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+27 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}-11 a b \,{\mathrm e}^{5 i \left (f x +e \right )}-4 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}+27 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-11 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-4 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}+9 a^{2} {\mathrm e}^{i \left (f x +e \right )}-13 a b \,{\mathrm e}^{i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )^{2} a^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a^{3} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{a^{3} f}+\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{16 \left (a -b \right )^{3} f a}-\frac {5 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{4 \left (a -b \right )^{3} f \,a^{2}}+\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b^{2}}{2 \left (a -b \right )^{3} f \,a^{3}}-\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right )}{16 \left (a -b \right )^{3} f a}+\frac {5 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{4 \left (a -b \right )^{3} f \,a^{2}}-\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b^{2}}{2 \left (a -b \right )^{3} f \,a^{3}}\) | \(680\) |
1/f*(b/a^3*((-1/8*(9*a-4*b)*a/(a-b)*cos(f*x+e)^3-1/8*a*b*(7*a-4*b)/(a^2-2* a*b+b^2)*cos(f*x+e))/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)^2+1/8*(15*a^2-20*a* b+8*b^2)/(a^2-2*a*b+b^2)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b)) ^(1/2)))-1/2/a^3*ln(cos(f*x+e)+1)+1/2/a^3*ln(cos(f*x+e)-1))
Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (152) = 304\).
Time = 0.49 (sec) , antiderivative size = 1050, normalized size of antiderivative = 6.33 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
[-1/16*(2*(9*a^3*b - 13*a^2*b^2 + 4*a*b^3)*cos(f*x + e)^3 - ((15*a^4 - 50* a^3*b + 63*a^2*b^2 - 36*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 - 20*a* b^3 + 8*b^4 + 2*(15*a^3*b - 35*a^2*b^2 + 28*a*b^3 - 8*b^4)*cos(f*x + e)^2) *sqrt(-b/(a - b))*log(((a - b)*cos(f*x + e)^2 + 2*(a - b)*sqrt(-b/(a - b)) *cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 2*(7*a^2*b^2 - 4*a*b^3) *cos(f*x + e) + 8*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e )^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*cos( f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) - 8*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2 *b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 - 2*a^4*b^3 + a^3*b^4)*f), -1/8*((9*a^3*b - 13*a^2*b^2 + 4*a*b^3)*cos(f*x + e)^3 + ((1 5*a^4 - 50*a^3*b + 63*a^2*b^2 - 36*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 15*a^2* b^2 - 20*a*b^3 + 8*b^4 + 2*(15*a^3*b - 35*a^2*b^2 + 28*a*b^3 - 8*b^4)*cos( f*x + e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b ) + (7*a^2*b^2 - 4*a*b^3)*cos(f*x + e) + 4*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4 *a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*b^3 + b^4 + 2*(a^3*b - 3*a^2* b^2 + 3*a*b^3 - b^4)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) - 4*((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^4 + a^2*b^2 - 2*a*...
Timed out. \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (152) = 304\).
Time = 0.83 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.03 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 20 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right ) - b}{\sqrt {a b - b^{2}} \cos \left (f x + e\right ) + \sqrt {a b - b^{2}}}\right )}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {a b - b^{2}}} + \frac {2 \, {\left (9 \, a^{3} b - 6 \, a^{2} b^{2} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {68 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {32 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {27 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {120 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {48 \, b^{4} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {9 \, a^{3} b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {28 \, a^{2} b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {16 \, a b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{2}} - \frac {4 \, \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{3}}}{8 \, f} \]
-1/8*((15*a^2*b - 20*a*b^2 + 8*b^3)*arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2)))/((a^5 - 2*a^4*b + a^3*b^2)*sqrt(a*b - b^2)) + 2*(9*a^3*b - 6*a^2*b^2 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 68*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 32*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 90*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f* x + e) + 1)^2 + 120*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 48*b ^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^ 3/(cos(f*x + e) + 1)^3 - 28*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1 )^3 + 16*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 - 2*a^4*b + a^3*b^2)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2) ^2) - 4*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/a^3)/f
Time = 15.57 (sec) , antiderivative size = 1844, normalized size of antiderivative = 11.11 \[ \int \frac {\csc (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
log(tan(e/2 + (f*x)/2))/(a^3*f) - ((3*(3*a*b - 2*b^2))/(4*a*(a^2 - 2*a*b + b^2)) + (3*tan(e/2 + (f*x)/2)^4*(40*a*b^3 + 9*a^3*b - 16*b^4 - 30*a^2*b^2 ))/(4*a^3*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f*x)/2)^6*(9*a^2*b - 28*a*b^2 + 16*b^3))/(4*a^2*(a^2 - 2*a*b + b^2)) - (tan(e/2 + (f*x)/2)^2*(27*a^2*b - 68*a*b^2 + 32*b^3))/(4*a^2*(a^2 - 2*a*b + b^2)))/(f*(tan(e/2 + (f*x)/2)^ 2*(8*a*b - 4*a^2) + tan(e/2 + (f*x)/2)^6*(8*a*b - 4*a^2) + a^2*tan(e/2 + ( f*x)/2)^8 + tan(e/2 + (f*x)/2)^4*(6*a^2 - 16*a*b + 16*b^2) + a^2)) + (b^(1 /2)*atan(((tan(e/2 + (f*x)/2)^2*((((b^(3/2)*(15*a^2 - 20*a*b + 8*b^2)^3*(4 096*a^15*b - 128*a^16 + 6144*a^7*b^9 - 46080*a^8*b^8 + 150784*a^9*b^7 - 28 1216*a^10*b^6 + 327168*a^11*b^5 - 243584*a^12*b^4 + 113920*a^13*b^3 - 3110 4*a^14*b^2))/(32768*a^9*(a - b)^(15/2)*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6* b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)) - (b^(1/2)*(15*a^2 - 20*a*b + 8*b^2)*(1536*a*b^9 + 720*a^9*b - 11520*a^2*b^8 + 37760*a^3*b^7 - 70400*a^ 4*b^6 + 81384*a^5*b^5 - 59564*a^6*b^4 + 26864*a^7*b^3 - 6780*a^8*b^2))/(12 8*a^3*(a - b)^(5/2)*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6*b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)))*(3072*a*b^4 - 1090*a^4*b + 111*a^5 - 768*b^5 - 4752*a^2*b^3 + 3424*a^3*b^2))/(2*a^5*(a - b)^(13/2)*(960*a*b^4 - 1055*a^4* b + 256*a^5 - 192*b^5 - 1920*a^2*b^3 + 1960*a^3*b^2)) + (((576*a*b^6 - 64* b^7 - 1920*a^2*b^5 + 3160*a^3*b^4 - 2625*a^4*b^3 + 900*a^5*b^2)/(8*(a^11 - 6*a^10*b + a^5*b^6 - 6*a^6*b^5 + 15*a^7*b^4 - 20*a^8*b^3 + 15*a^9*b^2)...