Integrand size = 27, antiderivative size = 252 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 x}{8 b}-\frac {\left (a^2-3 b^2\right ) x}{2 b^3}-\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) x}{b^5}+\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^5 d}-\frac {\text {arctanh}(\cos (c+d x))}{a d}-\frac {a \cos (c+d x)}{b^2 d}-\frac {a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac {a \cos ^3(c+d x)}{3 b^2 d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 b d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 b d} \]
-3/8*x/b-1/2*(a^2-3*b^2)*x/b^3-(a^4-3*a^2*b^2+3*b^4)*x/b^5+2*(a^2-b^2)^(5/ 2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/b^5/d-arctanh(cos(d* x+c))/a/d-a*cos(d*x+c)/b^2/d-a*(a^2-3*b^2)*cos(d*x+c)/b^4/d+1/3*a*cos(d*x+ c)^3/b^2/d+3/8*cos(d*x+c)*sin(d*x+c)/b/d+1/2*(a^2-3*b^2)*cos(d*x+c)*sin(d* x+c)/b^3/d+1/4*cos(d*x+c)*sin(d*x+c)^3/b/d
Time = 0.60 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {96 a^5 c-240 a^3 b^2 c+180 a b^4 c+96 a^5 d x-240 a^3 b^2 d x+180 a b^4 d x-192 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )+24 a^2 b \left (4 a^2-9 b^2\right ) \cos (c+d x)-8 a^2 b^3 \cos (3 (c+d x))+96 b^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-96 b^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-24 a^3 b^2 \sin (2 (c+d x))+48 a b^4 \sin (2 (c+d x))+3 a b^4 \sin (4 (c+d x))}{96 a b^5 d} \]
-1/96*(96*a^5*c - 240*a^3*b^2*c + 180*a*b^4*c + 96*a^5*d*x - 240*a^3*b^2*d *x + 180*a*b^4*d*x - 192*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2]) /Sqrt[a^2 - b^2]] + 24*a^2*b*(4*a^2 - 9*b^2)*Cos[c + d*x] - 8*a^2*b^3*Cos[ 3*(c + d*x)] + 96*b^5*Log[Cos[(c + d*x)/2]] - 96*b^5*Log[Sin[(c + d*x)/2]] - 24*a^3*b^2*Sin[2*(c + d*x)] + 48*a*b^4*Sin[2*(c + d*x)] + 3*a*b^4*Sin[4 *(c + d*x)])/(a*b^5*d)
Time = 0.56 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3376, 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.
| \(\displaystyle \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x) (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle \int \left (\frac {\left (a^2-b^2\right )^3}{a b^5 (a+b \sin (c+d x))}+\frac {a \left (a^2-3 b^2\right ) \sin (c+d x)}{b^4}+\frac {\left (3 b^2-a^2\right ) \sin ^2(c+d x)}{b^3}+\frac {-a^4+3 a^2 b^2-3 b^4}{b^5}+\frac {a \sin ^3(c+d x)}{b^2}+\frac {\csc (c+d x)}{a}-\frac {\sin ^4(c+d x)}{b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a b^5 d}-\frac {a \left (a^2-3 b^2\right ) \cos (c+d x)}{b^4 d}+\frac {\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac {x \left (a^2-3 b^2\right )}{2 b^3}-\frac {x \left (a^4-3 a^2 b^2+3 b^4\right )}{b^5}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {a \cos ^3(c+d x)}{3 b^2 d}-\frac {a \cos (c+d x)}{b^2 d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 b d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 b d}-\frac {3 x}{8 b}\) |
(-3*x)/(8*b) - ((a^2 - 3*b^2)*x)/(2*b^3) - ((a^4 - 3*a^2*b^2 + 3*b^4)*x)/b ^5 + (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] )/(a*b^5*d) - ArcTanh[Cos[c + d*x]]/(a*d) - (a*Cos[c + d*x])/(b^2*d) - (a* (a^2 - 3*b^2)*Cos[c + d*x])/(b^4*d) + (a*Cos[c + d*x]^3)/(3*b^2*d) + (3*Co s[c + d*x]*Sin[c + d*x])/(8*b*d) + ((a^2 - 3*b^2)*Cos[c + d*x]*Sin[c + d*x ])/(2*b^3*d) + (Cos[c + d*x]*Sin[c + d*x]^3)/(4*b*d)
3.14.23.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Time = 0.86 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.33
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \left (\frac {\left (\frac {1}{2} a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b -3 a \,b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{3} b -7 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{3} b -\frac {19}{3} a \,b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{3} b -\frac {7 a \,b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{5}}+\frac {\left (2 a^{6}-6 a^{4} b^{2}+6 a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{5} \sqrt {a^{2}-b^{2}}}}{d}\) | \(336\) |
| default | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \left (\frac {\left (\frac {1}{2} a^{2} b^{2}-\frac {9}{8} b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{3} b -3 a \,b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{2} b^{2}-\frac {1}{8} b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{3} b -7 a \,b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{2} b^{2}+\frac {1}{8} b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{3} b -\frac {19}{3} a \,b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {1}{2} a^{2} b^{2}+\frac {9}{8} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a^{3} b -\frac {7 a \,b^{3}}{3}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {\left (8 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}\right )}{b^{5}}+\frac {\left (2 a^{6}-6 a^{4} b^{2}+6 a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{5} \sqrt {a^{2}-b^{2}}}}{d}\) | \(336\) |
| risch | \(-\frac {x \,a^{4}}{b^{5}}+\frac {5 x \,a^{2}}{2 b^{3}}-\frac {15 x}{8 b}-\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{4} d}+\frac {9 a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{4} d}+\frac {9 a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} d}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d b a}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d b a}+\frac {i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{5}}-\frac {2 i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}+a \right )}{b}\right )}{d \,b^{3}}+\frac {2 i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{3}}-\frac {i \sqrt {a^{2}-b^{2}}\, a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (\sqrt {a^{2}-b^{2}}-a \right )}{b}\right )}{d \,b^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d a}-\frac {\sin \left (4 d x +4 c \right )}{32 b d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d \,b^{2}}+\frac {\sin \left (2 d x +2 c \right ) a^{2}}{4 b^{3} d}-\frac {\sin \left (2 d x +2 c \right )}{2 b d}\) | \(533\) |
1/d*(1/a*ln(tan(1/2*d*x+1/2*c))-2/b^5*(((1/2*a^2*b^2-9/8*b^4)*tan(1/2*d*x+ 1/2*c)^7+(a^3*b-3*a*b^3)*tan(1/2*d*x+1/2*c)^6+(1/2*a^2*b^2-1/8*b^4)*tan(1/ 2*d*x+1/2*c)^5+(3*a^3*b-7*a*b^3)*tan(1/2*d*x+1/2*c)^4+(-1/2*a^2*b^2+1/8*b^ 4)*tan(1/2*d*x+1/2*c)^3+(3*a^3*b-19/3*a*b^3)*tan(1/2*d*x+1/2*c)^2+(-1/2*a^ 2*b^2+9/8*b^4)*tan(1/2*d*x+1/2*c)+a^3*b-7/3*a*b^3)/(1+tan(1/2*d*x+1/2*c)^2 )^4+1/8*(8*a^4-20*a^2*b^2+15*b^4)*arctan(tan(1/2*d*x+1/2*c)))+(2*a^6-6*a^4 *b^2+6*a^2*b^4-2*b^6)/a/b^5/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/ 2*c)+2*b)/(a^2-b^2)^(1/2)))
Time = 0.77 (sec) , antiderivative size = 508, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {8 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 12 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 24 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a b^{5} d}, \frac {8 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 12 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 12 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x - 24 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 24 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, a b^{4} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, a b^{5} d}\right ] \]
[1/24*(8*a^2*b^3*cos(d*x + c)^3 - 12*b^5*log(1/2*cos(d*x + c) + 1/2) + 12* b^5*log(-1/2*cos(d*x + c) + 1/2) - 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*d*x + 12*(a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 24*(a^4*b - 2*a^2*b^3)*cos(d*x + c) - 3*(2*a*b^4*cos(d*x + c)^3 - (4*a^3*b^2 - 7*a*b^4)*cos(d*x + c))*sin(d*x + c))/(a*b^5*d), 1/24 *(8*a^2*b^3*cos(d*x + c)^3 - 12*b^5*log(1/2*cos(d*x + c) + 1/2) + 12*b^5*l og(-1/2*cos(d*x + c) + 1/2) - 3*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*d*x - 24*( a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt( a^2 - b^2)*cos(d*x + c))) - 24*(a^4*b - 2*a^2*b^3)*cos(d*x + c) - 3*(2*a*b ^4*cos(d*x + c)^3 - (4*a^3*b^2 - 7*a*b^4)*cos(d*x + c))*sin(d*x + c))/(a*b ^5*d)]
\[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cos ^{6}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.35 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.58 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} - \frac {3 \, {\left (8 \, a^{4} - 20 \, a^{2} b^{2} + 15 \, b^{4}\right )} {\left (d x + c\right )}}{b^{5}} + \frac {48 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b^{5}} - \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 72 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 168 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 152 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{3} - 56 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} b^{4}}}{24 \, d} \]
1/24*(24*log(abs(tan(1/2*d*x + 1/2*c)))/a - 3*(8*a^4 - 20*a^2*b^2 + 15*b^4 )*(d*x + c)/b^5 + 48*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1/2*(d* x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b ^2)))/(sqrt(a^2 - b^2)*a*b^5) - 2*(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 27*b^ 3*tan(1/2*d*x + 1/2*c)^7 + 24*a^3*tan(1/2*d*x + 1/2*c)^6 - 72*a*b^2*tan(1/ 2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 3*b^3*tan(1/2*d*x + 1 /2*c)^5 + 72*a^3*tan(1/2*d*x + 1/2*c)^4 - 168*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 3*b^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^3 *tan(1/2*d*x + 1/2*c)^2 - 152*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan( 1/2*d*x + 1/2*c) + 27*b^3*tan(1/2*d*x + 1/2*c) + 24*a^3 - 56*a*b^2)/((tan( 1/2*d*x + 1/2*c)^2 + 1)^4*b^4))/d
Time = 13.46 (sec) , antiderivative size = 4866, normalized size of antiderivative = 19.31 \[ \int \frac {\cos ^5(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
log(tan(c/2 + (d*x)/2))/(a*d) + ((2*(7*a*b^2 - 3*a^3))/(3*b^4) + (2*tan(c/ 2 + (d*x)/2)^6*(3*a*b^2 - a^3))/b^4 + (2*tan(c/2 + (d*x)/2)^4*(7*a*b^2 - 3 *a^3))/b^4 + (2*tan(c/2 + (d*x)/2)^2*(19*a*b^2 - 9*a^3))/(3*b^4) + (tan(c/ 2 + (d*x)/2)*(4*a^2 - 9*b^2))/(4*b^3) + (tan(c/2 + (d*x)/2)^3*(4*a^2 - b^2 ))/(4*b^3) - (tan(c/2 + (d*x)/2)^5*(4*a^2 - b^2))/(4*b^3) - (tan(c/2 + (d* x)/2)^7*(4*a^2 - 9*b^2))/(4*b^3))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (atan( (((a^4*8i + b^4*15i - a^2*b^2*20i)*((840*a*b^14 - 112*a^15 - 3760*a^3*b^12 + (15271*a^5*b^10)/2 - (18191*a^7*b^8)/2 + (13697*a^9*b^6)/2 - 3252*a^11* b^4 + 900*a^13*b^2)/b^11 + (tan(c/2 + (d*x)/2)*(3664*b^20 - 18084*a^2*b^18 + 41125*a^4*b^16 - 56140*a^6*b^14 + 50084*a^8*b^12 - 29728*a^10*b^10 + 11 392*a^12*b^8 - 2560*a^14*b^6 + 256*a^16*b^4))/(2*b^16) - ((a^4*8i + b^4*15 i - a^2*b^2*20i)*((128*b^16 + 94*a^2*b^14 - (2395*a^4*b^12)/2 + 2068*a^6*b ^10 - 1600*a^8*b^8 + 608*a^10*b^6 - 96*a^12*b^4)/b^11 + (tan(c/2 + (d*x)/2 )*(4944*a*b^20 - 18960*a^3*b^18 + 29985*a^5*b^16 - 24664*a^7*b^14 + 10816* a^9*b^12 - 2240*a^11*b^10 + 128*a^13*b^8))/(2*b^16) - ((a^4*8i + b^4*15i - a^2*b^2*20i)*((256*a*b^16 - 452*a^3*b^14 + 340*a^5*b^12 - 112*a^7*b^10)/b ^11 - (((128*a^2*b^16 - 96*a^4*b^14)/b^11 + (tan(c/2 + (d*x)/2)*(1024*a*b^ 22 - 1088*a^3*b^20 + 128*a^5*b^18))/(2*b^16))*(a^4*8i + b^4*15i - a^2*b^2* 20i))/(8*b^5) + (tan(c/2 + (d*x)/2)*(1024*b^22 - 2368*a^2*b^20 + 2432*a...