Integrand size = 21, antiderivative size = 140 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {2 a^3 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {\left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{3 \left (a^2-b^2\right )^2 d}+\frac {(b-a \cos (c+d x)) \csc ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \]
-2*a^3*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/( a+b)^(5/2)/d+1/3*(3*a^2*b-a*(2*a^2+b^2)*cos(d*x+c))*csc(d*x+c)/(a^2-b^2)^2 /d+1/3*(b-a*cos(d*x+c))*csc(d*x+c)^3/(a^2-b^2)/d
Time = 1.07 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {24 a^3 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} \left (10 a^2 b-4 b^3+\left (-6 a^3+3 a b^2\right ) \cos (c+d x)-6 a^2 b \cos (2 (c+d x))+2 a^3 \cos (3 (c+d x))+a b^2 \cos (3 (c+d x))\right ) \csc ^3(c+d x)}{12 (a-b)^2 (a+b)^2 \sqrt {a^2-b^2} d} \]
(24*a^3*b*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]*(10*a^2*b - 4*b^3 + (-6*a^3 + 3*a*b^2)*Cos[c + d*x] - 6*a^2*b*Cos[2 *(c + d*x)] + 2*a^3*Cos[3*(c + d*x)] + a*b^2*Cos[3*(c + d*x)])*Csc[c + d*x ]^3)/(12*(a - b)^2*(a + b)^2*Sqrt[a^2 - b^2]*d)
Time = 0.72 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.18, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4360, 25, 25, 3042, 25, 3345, 3042, 3345, 27, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cot (c+d x) \csc ^3(c+d x)}{-a \cos (c+d x)-b}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cot (c+d x) \csc ^3(c+d x)}{b+a \cos (c+d x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \left (b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\int \frac {\left (a b-2 a^2 \cos (c+d x)\right ) \csc ^2(c+d x)}{b+a \cos (c+d x)}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\int \frac {2 \sin \left (c+d x-\frac {\pi }{2}\right ) a^2+b a}{\cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (b-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3345 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\frac {\int \frac {3 a^3 b}{b+a \cos (c+d x)}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\frac {3 a^3 b \int \frac {1}{b+a \cos (c+d x)}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\frac {3 a^3 b \int \frac {1}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\frac {6 a^3 b \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\csc ^3(c+d x) (b-a \cos (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\frac {6 a^3 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\csc (c+d x) \left (3 a^2 b-a \left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )}}{3 \left (a^2-b^2\right )}\) |
((b - a*Cos[c + d*x])*Csc[c + d*x]^3)/(3*(a^2 - b^2)*d) - ((6*a^3*b*ArcTan h[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a ^2 - b^2)*d) - ((3*a^2*b - a*(2*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x])/((a ^2 - b^2)*d))/(3*(a^2 - b^2))
3.3.7.3.1 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[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt Q[p, -1] && IntegerQ[2*m]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.66 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.18
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a +b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(165\) |
| default | \(\frac {\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b}{3}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{8 \left (a -b \right )^{2}}-\frac {2 a^{3} b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {3 a +b}{8 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d}\) | \(165\) |
| risch | \(\frac {2 i \left (3 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-10 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-2 a^{3}-a \,b^{2}\right )}{3 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(299\) |
1/d*(1/8/(a-b)^2*(1/3*tan(1/2*d*x+1/2*c)^3*a-1/3*tan(1/2*d*x+1/2*c)^3*b+3* tan(1/2*d*x+1/2*c)*a-tan(1/2*d*x+1/2*c)*b)-2/(a-b)^2/(a+b)^2*a^3*b/((a-b)* (a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))-1/24/(a +b)/tan(1/2*d*x+1/2*c)^3-1/8*(3*a+b)/(a+b)^2/tan(1/2*d*x+1/2*c))
Time = 0.31 (sec) , antiderivative size = 558, normalized size of antiderivative = 3.99 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\left [-\frac {8 \, a^{4} b - 10 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{6 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5} + {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{3} b \cos \left (d x + c\right )^{2} - a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{3 \, {\left ({\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
[-1/6*(8*a^4*b - 10*a^2*b^3 + 2*b^5 + 2*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3 - 3*(a^3*b*cos(d*x + c)^2 - a^3*b)*sqrt(a^2 - b^2)*log((2*a*b*cos(d *x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 6*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 6*(a^5 - a^3* b^2)*cos(d*x + c))/(((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)*sin(d*x + c)), -1/3*(4*a^4*b - 5* a^2*b^3 + b^5 + (2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^3 + 3*(a^3*b*cos(d* x + c)^2 - a^3*b)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c ) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 3*(a^4*b - a^2*b^3)*cos( d*x + c)^2 - 3*(a^5 - a^3*b^2)*cos(d*x + c))/(((a^6 - 3*a^4*b^2 + 3*a^2*b^ 4 - b^6)*d*cos(d*x + c)^2 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)*sin(d*x + c))]
\[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\csc ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (128) = 256\).
Time = 0.33 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.92 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{3} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]
1/24*(48*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan( 1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*a^3*b/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + (a^2*tan(1/2*d*x + 1/2*c)^3 - 2*a*b *tan(1/2*d*x + 1/2*c)^3 + b^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*tan(1/2*d*x + 1/2*c) - 12*a*b*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (9*a*tan(1/2*d*x + 1/2*c)^2 + 3*b*tan(1/2*d*x + 1/2*c)^2 + a + b)/((a^2 + 2*a*b + b^2)*tan(1/2*d*x + 1/2*c)^3))/d
Time = 14.00 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.56 \[ \int \frac {\csc ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2}{8\,a-8\,b}+\frac {8\,a+8\,b}{{\left (8\,a-8\,b\right )}^2}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,d\,\left (8\,a-8\,b\right )}-\frac {\frac {a^2-2\,a\,b+b^2}{3\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3-5\,a^2\,b+a\,b^2+b^3\right )}{{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (8\,a^2-16\,a\,b+8\,b^2\right )}-\frac {2\,a^3\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
(tan(c/2 + (d*x)/2)*(2/(8*a - 8*b) + (8*a + 8*b)/(8*a - 8*b)^2))/d + tan(c /2 + (d*x)/2)^3/(3*d*(8*a - 8*b)) - ((a^2 - 2*a*b + b^2)/(3*(a + b)) + (ta n(c/2 + (d*x)/2)^2*(a*b^2 - 5*a^2*b + 3*a^3 + b^3))/(a + b)^2)/(d*tan(c/2 + (d*x)/2)^3*(8*a^2 - 16*a*b + 8*b^2)) - (2*a^3*b*atanh((tan(c/2 + (d*x)/2 )*(a^4 + b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a - b)^(3/2))))/(d*(a + b)^(5/2 )*(a - b)^(5/2))