Integrand size = 13, antiderivative size = 130 \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\frac {\left (3 a^4-12 a^2 b^2+8 b^4\right ) x}{8 a^5}+\frac {2 b \left (a^2-b^2\right )^{3/2} \text {arctanh}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^5}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{8 a^4} \]
1/8*(3*a^4-12*a^2*b^2+8*b^4)*x/a^5+2*b*(a^2-b^2)^(3/2)*arctanh((a+b*tan(1/ 2*x))/(a^2-b^2)^(1/2))/a^5-1/12*cos(x)^3*(4*b-3*a*sin(x))/a^2-1/8*cos(x)*( 8*b*(a^2-b^2)-a*(3*a^2-4*b^2)*sin(x))/a^4
Time = 0.91 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=-\frac {192 b \left (-a^2+b^2\right )^{3/2} \arctan \left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+24 a b \left (5 a^2-4 b^2\right ) \cos (x)+8 a^3 b \cos (3 x)-3 \left (4 \left (3 a^4-12 a^2 b^2+8 b^4\right ) x+8 a^2 \left (a^2-b^2\right ) \sin (2 x)+a^4 \sin (4 x)\right )}{96 a^5} \]
-1/96*(192*b*(-a^2 + b^2)^(3/2)*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + 24*a*b*(5*a^2 - 4*b^2)*Cos[x] + 8*a^3*b*Cos[3*x] - 3*(4*(3*a^4 - 12*a^2* b^2 + 8*b^4)*x + 8*a^2*(a^2 - b^2)*Sin[2*x] + a^4*Sin[4*x]))/a^5
Time = 0.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4360, 3042, 3344, 25, 3042, 3344, 3042, 3214, 3042, 3139, 1083, 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.
| \(\displaystyle \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^4}{a+b \csc (x)}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\sin (x) \cos ^4(x)}{a \sin (x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x) \cos (x)^4}{a \sin (x)+b}dx\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle \frac {\int -\frac {\cos ^2(x) \left (a b-\left (3 a^2-4 b^2\right ) \sin (x)\right )}{b+a \sin (x)}dx}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^2(x) \left (a b-\left (3 a^2-4 b^2\right ) \sin (x)\right )}{b+a \sin (x)}dx}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\cos (x)^2 \left (a b-\left (3 a^2-4 b^2\right ) \sin (x)\right )}{b+a \sin (x)}dx}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle -\frac {\frac {\int \frac {a b \left (5 a^2-4 b^2\right )-\left (3 a^4-12 b^2 a^2+8 b^4\right ) \sin (x)}{b+a \sin (x)}dx}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {a b \left (5 a^2-4 b^2\right )-\left (3 a^4-12 b^2 a^2+8 b^4\right ) \sin (x)}{b+a \sin (x)}dx}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle -\frac {\frac {\frac {8 b \left (a^2-b^2\right )^2 \int \frac {1}{b+a \sin (x)}dx}{a}-\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{a}}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {8 b \left (a^2-b^2\right )^2 \int \frac {1}{b+a \sin (x)}dx}{a}-\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{a}}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {\frac {\frac {16 b \left (a^2-b^2\right )^2 \int \frac {1}{b \tan ^2\left (\frac {x}{2}\right )+2 a \tan \left (\frac {x}{2}\right )+b}d\tan \left (\frac {x}{2}\right )}{a}-\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{a}}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {\frac {-\frac {32 b \left (a^2-b^2\right )^2 \int \frac {1}{4 \left (a^2-b^2\right )-\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )^2}d\left (2 a+2 b \tan \left (\frac {x}{2}\right )\right )}{a}-\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{a}}{2 a^2}+\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}}{4 a^2}-\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\cos ^3(x) (4 b-3 a \sin (x))}{12 a^2}-\frac {\frac {\cos (x) \left (8 b \left (a^2-b^2\right )-a \left (3 a^2-4 b^2\right ) \sin (x)\right )}{2 a^2}+\frac {-\frac {16 b \left (a^2-b^2\right )^{3/2} \text {arctanh}\left (\frac {2 a+2 b \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^2-b^2}}\right )}{a}-\frac {x \left (3 a^4-12 a^2 b^2+8 b^4\right )}{a}}{2 a^2}}{4 a^2}\) |
-1/12*(Cos[x]^3*(4*b - 3*a*Sin[x]))/a^2 - ((-(((3*a^4 - 12*a^2*b^2 + 8*b^4 )*x)/a) - (16*b*(a^2 - b^2)^(3/2)*ArcTanh[(2*a + 2*b*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/a)/(2*a^2) + (Cos[x]*(8*b*(a^2 - b^2) - a*(3*a^2 - 4*b^2)*Sin[x ]))/(2*a^2))/(4*a^2)
3.1.9.3.1 Defintions of rubi rules used
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_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 , 0] && 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]
Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(118)=236\).
Time = 0.97 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.96
| method | result | size |
| default | \(\frac {\frac {2 \left (\left (-\frac {5}{8} a^{4}+\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {x}{2}\right )^{7}+\left (-2 a^{3} b +a \,b^{3}\right ) \tan \left (\frac {x}{2}\right )^{6}+\left (\frac {3}{8} a^{4}+\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {x}{2}\right )^{5}+\left (-4 a^{3} b +3 a \,b^{3}\right ) \tan \left (\frac {x}{2}\right )^{4}+\left (-\frac {3}{8} a^{4}-\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}+\left (-\frac {10}{3} a^{3} b +3 a \,b^{3}\right ) \tan \left (\frac {x}{2}\right )^{2}+\left (\frac {5}{8} a^{4}-\frac {1}{2} a^{2} b^{2}\right ) \tan \left (\frac {x}{2}\right )-\frac {4 a^{3} b}{3}+a \,b^{3}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{4}}+\frac {\left (3 a^{4}-12 a^{2} b^{2}+8 b^{4}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{4}}{a^{5}}-\frac {2 b \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \arctan \left (\frac {2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{5} \sqrt {-a^{2}+b^{2}}}\) | \(255\) |
| risch | \(\frac {3 x}{8 a}-\frac {3 x \,b^{2}}{2 a^{3}}+\frac {x \,b^{4}}{a^{5}}-\frac {5 b \,{\mathrm e}^{i x}}{8 a^{2}}+\frac {b^{3} {\mathrm e}^{i x}}{2 a^{4}}-\frac {5 b \,{\mathrm e}^{-i x}}{8 a^{2}}+\frac {b^{3} {\mathrm e}^{-i x}}{2 a^{4}}+\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{a^{3}}-\frac {\sqrt {a^{2}-b^{2}}\, b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{a^{5}}-\frac {\sqrt {a^{2}-b^{2}}\, b \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{a^{3}}+\frac {\sqrt {a^{2}-b^{2}}\, b^{3} \ln \left ({\mathrm e}^{i x}+\frac {i b \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{a^{5}}+\frac {\sin \left (4 x \right )}{32 a}-\frac {b \cos \left (3 x \right )}{12 a^{2}}+\frac {\sin \left (2 x \right )}{4 a}-\frac {\sin \left (2 x \right ) b^{2}}{4 a^{3}}\) | \(367\) |
2/a^5*(((-5/8*a^4+1/2*a^2*b^2)*tan(1/2*x)^7+(-2*a^3*b+a*b^3)*tan(1/2*x)^6+ (3/8*a^4+1/2*a^2*b^2)*tan(1/2*x)^5+(-4*a^3*b+3*a*b^3)*tan(1/2*x)^4+(-3/8*a ^4-1/2*a^2*b^2)*tan(1/2*x)^3+(-10/3*a^3*b+3*a*b^3)*tan(1/2*x)^2+(5/8*a^4-1 /2*a^2*b^2)*tan(1/2*x)-4/3*a^3*b+a*b^3)/(1+tan(1/2*x)^2)^4+1/8*(3*a^4-12*a ^2*b^2+8*b^4)*arctan(tan(1/2*x)))-2*b*(a^4-2*a^2*b^2+b^4)/a^5/(-a^2+b^2)^( 1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))
Time = 0.28 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.58 \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\left [-\frac {8 \, a^{3} b \cos \left (x\right )^{3} + 12 \, {\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} - 2 \, {\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x + 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) - 3 \, {\left (2 \, a^{4} \cos \left (x\right )^{3} + {\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{24 \, a^{5}}, -\frac {8 \, a^{3} b \cos \left (x\right )^{3} - 24 \, {\left (a^{2} b - b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) - 3 \, {\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x + 24 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (x\right ) - 3 \, {\left (2 \, a^{4} \cos \left (x\right )^{3} + {\left (3 \, a^{4} - 4 \, a^{2} b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{24 \, a^{5}}\right ] \]
[-1/24*(8*a^3*b*cos(x)^3 + 12*(a^2*b - b^3)*sqrt(a^2 - b^2)*log(-((a^2 - 2 *b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 - 2*(b*cos(x)*sin(x) + a*cos(x)) *sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 24*(a^3*b - a*b^3)*cos(x) - 3*(2*a^4*cos(x)^3 + (3 *a^4 - 4*a^2*b^2)*cos(x))*sin(x))/a^5, -1/24*(8*a^3*b*cos(x)^3 - 24*(a^2*b - b^3)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b ^2)*cos(x))) - 3*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x + 24*(a^3*b - a*b^3)*cos(x ) - 3*(2*a^4*cos(x)^3 + (3*a^4 - 4*a^2*b^2)*cos(x))*sin(x))/a^5]
\[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\int \frac {\cos ^{4}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\text {Exception raised: ValueError} \]
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. 278 vs. \(2 (117) = 234\).
Time = 0.28 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.14 \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\frac {{\left (3 \, a^{4} - 12 \, a^{2} b^{2} + 8 \, b^{4}\right )} x}{8 \, a^{5}} - \frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{5}} - \frac {15 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{7} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{7} + 48 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{6} - 24 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{6} - 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 96 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{4} - 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} + 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 80 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 72 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 15 \, a^{3} \tan \left (\frac {1}{2} \, x\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) + 32 \, a^{2} b - 24 \, b^{3}}{12 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{4} a^{4}} \]
1/8*(3*a^4 - 12*a^2*b^2 + 8*b^4)*x/a^5 - 2*(a^4*b - 2*a^2*b^3 + b^5)*(pi*f loor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2))) /(sqrt(-a^2 + b^2)*a^5) - 1/12*(15*a^3*tan(1/2*x)^7 - 12*a*b^2*tan(1/2*x)^ 7 + 48*a^2*b*tan(1/2*x)^6 - 24*b^3*tan(1/2*x)^6 - 9*a^3*tan(1/2*x)^5 - 12* a*b^2*tan(1/2*x)^5 + 96*a^2*b*tan(1/2*x)^4 - 72*b^3*tan(1/2*x)^4 + 9*a^3*t an(1/2*x)^3 + 12*a*b^2*tan(1/2*x)^3 + 80*a^2*b*tan(1/2*x)^2 - 72*b^3*tan(1 /2*x)^2 - 15*a^3*tan(1/2*x) + 12*a*b^2*tan(1/2*x) + 32*a^2*b - 24*b^3)/((t an(1/2*x)^2 + 1)^4*a^4)
Time = 19.21 (sec) , antiderivative size = 2055, normalized size of antiderivative = 15.81 \[ \int \frac {\cos ^4(x)}{a+b \csc (x)} \, dx=\text {Too large to display} \]
(atan(((((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10*b^4 + (9*a^12*b^ 2)/2)/a^11 - ((a^4*3i + b^4*8i - a^2*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 - ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/ (2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (tan(x/2)*(128*a^10*b ^6 - 256*a^12*b^4 + 128*a^14*b^2))/(2*a^12)))/(8*a^5) + (tan(x/2)*(18*a^14 *b - 128*a^4*b^11 + 576*a^6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^12*b^ 3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i)*1i)/(8*a^5) + (((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10*b^4 + (9*a^12*b^2)/2)/a^11 + ((a^4*3i + b^4*8i - a^2*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 + ( (32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12))*(a^4*3i + b ^4*8i - a^2*b^2*12i))/(8*a^5) + (tan(x/2)*(128*a^10*b^6 - 256*a^12*b^4 + 1 28*a^14*b^2))/(2*a^12)))/(8*a^5) + (tan(x/2)*(18*a^14*b - 128*a^4*b^11 + 5 76*a^6*b^9 - 960*a^8*b^7 + 712*a^10*b^5 - 217*a^12*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i)*1i)/(8*a^5))/((32*b^13 - 152*a^2*b^11 + 280*a^4*b ^9 - 247*a^6*b^7 + 102*a^8*b^5 - 15*a^10*b^3)/a^11 + (tan(x/2)*(128*b^14 - 640*a^2*b^12 + 1280*a^4*b^10 - 1296*a^6*b^8 + 690*a^8*b^6 - 180*a^10*b^4 + 18*a^12*b^2))/a^12 - (((32*a^4*b^10 - 96*a^6*b^8 + 96*a^8*b^6 - 36*a^10* b^4 + (9*a^12*b^2)/2)/a^11 - ((a^4*3i + b^4*8i - a^2*b^2*12i)*((12*a^14*b + 16*a^10*b^5 - 28*a^12*b^3)/a^11 - ((32*a^3*b^2 + (tan(x/2)*(192*a^16*b - 128*a^14*b^3))/(2*a^12))*(a^4*3i + b^4*8i - a^2*b^2*12i))/(8*a^5) + (t...