Integrand size = 29, antiderivative size = 314 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 a x}{b^4}+\frac {3 \sqrt {a^2-b^2} \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}-\frac {6 \left (2 a^6-a^4 b^2-b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 b^4 \sqrt {a^2-b^2} d}+\frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{b^3 d}-\frac {\cot (c+d x)}{a^3 d}-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac {2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))} \]
3*a*x/b^4+3*b*arctanh(cos(d*x+c))/a^4/d+cos(d*x+c)/b^3/d-cot(d*x+c)/a^3/d- 1/2*(a^2-b^2)^2*cos(d*x+c)/a^2/b^3/d/(a+b*sin(d*x+c))^2-3/2*(a^2-b^2)*cos( d*x+c)/a/b^3/d/(a+b*sin(d*x+c))+2*(a^2-b^2)*(2*a^2+b^2)*cos(d*x+c)/a^3/b^3 /d/(a+b*sin(d*x+c))-6*(2*a^6-a^4*b^2-b^6)*arctan((b+a*tan(1/2*d*x+1/2*c))/ (a^2-b^2)^(1/2))/a^4/b^4/d/(a^2-b^2)^(1/2)+3*(2*a^2+b^2)*arctan((b+a*tan(1 /2*d*x+1/2*c))/(a^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/a^2/b^4/d
Time = 5.58 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {6 a^5 c}{b^4}+\frac {6 a^5 d x}{b^4}+\frac {6 \left (-2 a^6+a^4 b^2-a^2 b^4+2 b^6\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2}}-a \cot \left (\frac {1}{2} (c+d x)\right )+6 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-6 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a \cos (c+d x) \left (6 a^5+a^3 b^2-5 a b^4-b \left (-9 a^4+a^2 b^2+4 b^4\right ) \sin (c+d x)+2 a^3 b^2 \sin ^2(c+d x)\right )}{b^3 (a+b \sin (c+d x))^2}+a \tan \left (\frac {1}{2} (c+d x)\right )}{2 a^4 d} \]
((6*a^5*c)/b^4 + (6*a^5*d*x)/b^4 + (6*(-2*a^6 + a^4*b^2 - a^2*b^4 + 2*b^6) *ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]) - a*Cot[(c + d*x)/2] + 6*b*Log[Cos[(c + d*x)/2]] - 6*b*Log[Sin[(c + d*x)/2] ] + (a*Cos[c + d*x]*(6*a^5 + a^3*b^2 - 5*a*b^4 - b*(-9*a^4 + a^2*b^2 + 4*b ^4)*Sin[c + d*x] + 2*a^3*b^2*Sin[c + d*x]^2))/(b^3*(a + b*Sin[c + d*x])^2) + a*Tan[(c + d*x)/2])/(2*a^4*d)
Time = 0.74 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, 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.
| \(\displaystyle \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^2 (a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle \int \left (-\frac {3 b \csc (c+d x)}{a^4}+\frac {\csc ^2(c+d x)}{a^3}-\frac {\left (a^2-b^2\right )^3}{a^2 b^4 (a+b \sin (c+d x))^3}-\frac {3 \left (2 a^6-a^4 b^2-b^6\right )}{a^4 b^4 (a+b \sin (c+d x))}+\frac {2 \left (2 a^6-3 a^4 b^2+b^6\right )}{a^3 b^4 (a+b \sin (c+d x))^2}+\frac {3 a}{b^4}-\frac {\sin (c+d x)}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 b \text {arctanh}(\cos (c+d x))}{a^4 d}-\frac {\cot (c+d x)}{a^3 d}+\frac {3 \left (2 a^2+b^2\right ) \sqrt {a^2-b^2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}-\frac {\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac {3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac {2 \left (2 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}-\frac {6 \left (2 a^6-a^4 b^2-b^6\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 b^4 d \sqrt {a^2-b^2}}+\frac {3 a x}{b^4}+\frac {\cos (c+d x)}{b^3 d}\) |
(3*a*x)/b^4 + (3*Sqrt[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x) /2])/Sqrt[a^2 - b^2]])/(a^2*b^4*d) - (6*(2*a^6 - a^4*b^2 - b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*b^4*Sqrt[a^2 - b^2]*d) + (3*b *ArcTanh[Cos[c + d*x]])/(a^4*d) + Cos[c + d*x]/(b^3*d) - Cot[c + d*x]/(a^3 *d) - ((a^2 - b^2)^2*Cos[c + d*x])/(2*a^2*b^3*d*(a + b*Sin[c + d*x])^2) - (3*(a^2 - b^2)*Cos[c + d*x])/(2*a*b^3*d*(a + b*Sin[c + d*x])) + (2*(a^2 - b^2)*(2*a^2 + b^2)*Cos[c + d*x])/(a^3*b^3*d*(a + b*Sin[c + d*x]))
3.13.70.3.1 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 = 1.95 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {3 a \,b^{2} \left (a^{4}+a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {b \left (4 a^{6}+9 a^{4} b^{2}-3 a^{2} b^{4}-10 b^{6}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {b^{2} a \left (13 a^{4}+a^{2} b^{2}-14 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2 a^{6} b -\frac {a^{4} b^{3}}{2}+\frac {5 a^{2} b^{5}}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (2 a^{6}-a^{4} b^{2}+a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} b^{4}}+\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+6 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(326\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}-\frac {2 \left (\frac {-\frac {3 a \,b^{2} \left (a^{4}+a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {b \left (4 a^{6}+9 a^{4} b^{2}-3 a^{2} b^{4}-10 b^{6}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {b^{2} a \left (13 a^{4}+a^{2} b^{2}-14 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-2 a^{6} b -\frac {a^{4} b^{3}}{2}+\frac {5 a^{2} b^{5}}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {3 \left (2 a^{6}-a^{4} b^{2}+a^{2} b^{4}-2 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} b^{4}}+\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+6 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}}{d}\) | \(326\) |
| risch | \(\frac {3 a x}{b^{4}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{3} d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{3} d}-\frac {i \left (3 i a^{3} b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-24 i a \,b^{5} {\mathrm e}^{3 i \left (d x +c \right )}-4 i a^{3} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+20 i a^{5} b \,{\mathrm e}^{3 i \left (d x +c \right )}-14 i a^{5} b \,{\mathrm e}^{i \left (d x +c \right )}+i a^{3} b^{3} {\mathrm e}^{i \left (d x +c \right )}+10 a^{6} {\mathrm e}^{4 i \left (d x +c \right )}+3 a^{4} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 a^{2} b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-6 b^{6} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+21 i a \,b^{5} {\mathrm e}^{i \left (d x +c \right )}-6 i a^{5} b \,{\mathrm e}^{5 i \left (d x +c \right )}-10 a^{6} {\mathrm e}^{2 i \left (d x +c \right )}-8 a^{4} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+18 a^{2} b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{6} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{4} b^{2}-a^{2} b^{4}-6 b^{6}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} b^{4} a^{3} d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{4} d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{4} d}-\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{2 d \,b^{2} a^{2}}-\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}+\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}+\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{2 d \,b^{2} a^{2}}+\frac {3 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {-i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{4}}\) | \(747\) |
1/d*(1/2*tan(1/2*d*x+1/2*c)/a^3-1/2/a^3/tan(1/2*d*x+1/2*c)-3/a^4*b*ln(tan( 1/2*d*x+1/2*c))-2/a^4/b^4*((-3/2*a*b^2*(a^4+a^2*b^2-2*b^4)*tan(1/2*d*x+1/2 *c)^3-1/2*b*(4*a^6+9*a^4*b^2-3*a^2*b^4-10*b^6)*tan(1/2*d*x+1/2*c)^2-1/2*b^ 2*a*(13*a^4+a^2*b^2-14*b^4)*tan(1/2*d*x+1/2*c)-2*a^6*b-1/2*a^4*b^3+5/2*a^2 *b^5)/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+3/2*(2*a^6-a^4*b ^2+a^2*b^4-2*b^6)/(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)))+2/b^4*(b/(1+tan(1/2*d*x+1/2*c)^2)+3*a*arctan(tan(1/2*d*x +1/2*c))))
Time = 0.62 (sec) , antiderivative size = 1171, normalized size of antiderivative = 3.73 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
[1/4*(24*a^6*b*d*x*cos(d*x + c)^2 - 24*a^6*b*d*x + 2*(9*a^5*b^2 - a^3*b^4 - 6*a*b^6)*cos(d*x + c)^3 - 3*(4*a^5*b + 2*a^3*b^3 + 4*a*b^5 - 2*(2*a^5*b + a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 + (2*a^6 + 3*a^4*b^2 + 3*a^2*b^4 + 2*b ^6 - (2*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2)*sin(d*x + c))*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)) - 6*(3*a^5*b^2 - a^3*b^ 4 - 2*a*b^6)*cos(d*x + c) + 6*(2*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos (d*x + c)^2 - a^2*b^5 - b^7)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 6 *(2*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos(d*x + c)^2 - a^2*b^5 - b^7)* sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*(6*a^5*b^2*d*x*cos(d*x + c) ^2 + 2*a^4*b^3*cos(d*x + c)^3 - 6*(a^7 + a^5*b^2)*d*x - 3*(2*a^6*b + a^4*b ^3 - 3*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/(2*a^5*b^5*d*cos(d*x + c)^2 - 2*a^5*b^5*d + (a^4*b^6*d*cos(d*x + c)^2 - (a^6*b^4 + a^4*b^6)*d)*sin(d*x + c)), 1/2*(12*a^6*b*d*x*cos(d*x + c)^2 - 12*a^6*b*d*x + (9*a^5*b^2 - a^3*b ^4 - 6*a*b^6)*cos(d*x + c)^3 - 3*(4*a^5*b + 2*a^3*b^3 + 4*a*b^5 - 2*(2*a^5 *b + a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 + (2*a^6 + 3*a^4*b^2 + 3*a^2*b^4 + 2*b^6 - (2*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a ^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3 *(3*a^5*b^2 - a^3*b^4 - 2*a*b^6)*cos(d*x + c) + 3*(2*a*b^6*cos(d*x + c)...
Timed out. \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.37 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {6 \, {\left (d x + c\right )} a}{b^{4}} - \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {6 \, {\left (2 \, a^{6} - a^{4} b^{2} + a^{2} b^{4} - 2 \, 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^{4} b^{4}} + \frac {2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a b^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} a^{4} b^{3}} + \frac {2 \, {\left (3 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{4} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 10 \, b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{5} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 14 \, a b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{4} b^{3}}}{2 \, d} \]
1/2*(6*(d*x + c)*a/b^4 - 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + tan(1/2* d*x + 1/2*c)/a^3 - 6*(2*a^6 - a^4*b^2 + a^2*b^4 - 2*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^4*b^4) + (2*b^4*tan(1/2*d*x + 1/2*c)^3 - a*b^3*ta n(1/2*d*x + 1/2*c)^2 + 4*a^4*tan(1/2*d*x + 1/2*c) + 2*b^4*tan(1/2*d*x + 1/ 2*c) - a*b^3)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*a^4*b^3) + 2*(3*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a *b^5*tan(1/2*d*x + 1/2*c)^3 + 4*a^6*tan(1/2*d*x + 1/2*c)^2 + 9*a^4*b^2*tan (1/2*d*x + 1/2*c)^2 - 3*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 - 10*b^6*tan(1/2*d* x + 1/2*c)^2 + 13*a^5*b*tan(1/2*d*x + 1/2*c) + a^3*b^3*tan(1/2*d*x + 1/2*c ) - 14*a*b^5*tan(1/2*d*x + 1/2*c) + 4*a^6 + a^4*b^2 - 5*a^2*b^4)/((a*tan(1 /2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^4*b^3))/d
Time = 13.75 (sec) , antiderivative size = 4223, normalized size of antiderivative = 13.45 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
tan(c/2 + (d*x)/2)/(2*a^3*d) + ((tan(c/2 + (d*x)/2)^6*(6*a^4 - 12*b^4 + 5* a^2*b^2))/b^2 - a^2 - (tan(c/2 + (d*x)/2)^2*(32*b^4 - 42*a^4 + a^2*b^2))/b ^2 + (tan(c/2 + (d*x)/2)^4*(48*a^4 - 44*b^4 + 5*a^2*b^2))/b^2 + (2*tan(c/2 + (d*x)/2)*(6*a^5 - 7*a*b^4 + a^3*b^2))/b^3 + (4*tan(c/2 + (d*x)/2)^3*(6* a^6 - 5*b^6 - 6*a^2*b^4 + 9*a^4*b^2))/(a*b^3) + (2*tan(c/2 + (d*x)/2)^5*(6 *a^6 - 10*b^6 - 5*a^2*b^4 + 9*a^4*b^2))/(a*b^3))/(d*(2*a^5*tan(c/2 + (d*x) /2)^7 + tan(c/2 + (d*x)/2)^3*(6*a^5 + 8*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(6 *a^5 + 8*a^3*b^2) + 2*a^5*tan(c/2 + (d*x)/2) + 8*a^4*b*tan(c/2 + (d*x)/2)^ 2 + 16*a^4*b*tan(c/2 + (d*x)/2)^4 + 8*a^4*b*tan(c/2 + (d*x)/2)^6)) + (6*a* atan((6480*tan(c/2 + (d*x)/2))/((6480*b^4)/a^4 - (12960*b^2)/a^2 - (5184*b ^6)/a^6 + (5184*b^8)/a^8 + (6480*a*tan(c/2 + (d*x)/2))/b - (5184*b*tan(c/2 + (d*x)/2))/a + (5184*b^3*tan(c/2 + (d*x)/2))/a^3 - (12960*a^3*tan(c/2 + (d*x)/2))/b^3 + (6480*a^5*tan(c/2 + (d*x)/2))/b^5 + 6480) - (12960*tan(c/2 + (d*x)/2))/((6480*b^2)/a^2 + (6480*a^2)/b^2 - (5184*b^4)/a^4 + (5184*b^6 )/a^6 - (5184*a*tan(c/2 + (d*x)/2))/b + (5184*b*tan(c/2 + (d*x)/2))/a + (6 480*a^3*tan(c/2 + (d*x)/2))/b^3 - (12960*a^5*tan(c/2 + (d*x)/2))/b^5 + (64 80*a^7*tan(c/2 + (d*x)/2))/b^7 - 12960) - (6480*a)/(6480*b + 6480*a*tan(c/ 2 + (d*x)/2) - (12960*b^3)/a^2 + (6480*b^5)/a^4 - (5184*b^7)/a^6 + (5184*b ^9)/a^8 - (5184*b^2*tan(c/2 + (d*x)/2))/a - (12960*a^3*tan(c/2 + (d*x)/2)) /b^2 + (5184*b^4*tan(c/2 + (d*x)/2))/a^3 + (6480*a^5*tan(c/2 + (d*x)/2)...