Integrand size = 27, antiderivative size = 402 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {2 b^3 \left (3 a^2-b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{7/2} d}+\frac {b^3 \left (2 a^2+b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{7/2} d}+\frac {2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2} d}-\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {\cos (c+d x)}{2 (a+b)^3 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^3 d (1+\sin (c+d x))}+\frac {b^4 \cos (c+d x)}{2 a \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac {3 b^4 \cos (c+d x)}{2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac {b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))} \]
2*b^3*(3*a^2-b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a/(a^2- b^2)^(7/2)/d+b^3*(2*a^2+b^2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/ 2))/a/(a^2-b^2)^(7/2)/d+2*b^3*(6*a^4-3*a^2*b^2+b^4)*arctan((b+a*tan(1/2*d* x+1/2*c))/(a^2-b^2)^(1/2))/a^3/(a^2-b^2)^(7/2)/d-arctanh(cos(d*x+c))/a^3/d +1/2*cos(d*x+c)/(a+b)^3/d/(1-sin(d*x+c))+1/2*cos(d*x+c)/(a-b)^3/d/(1+sin(d *x+c))+1/2*b^4*cos(d*x+c)/a/(a^2-b^2)^2/d/(a+b*sin(d*x+c))^2+3/2*b^4*cos(d *x+c)/(a^2-b^2)^3/d/(a+b*sin(d*x+c))+b^4*(3*a^2-b^2)*cos(d*x+c)/a^2/(a^2-b ^2)^3/d/(a+b*sin(d*x+c))
Time = 5.85 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.69 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 b^3 \left (20 a^4-7 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{7/2}}-\frac {2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^3}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b^4 \cos (c+d x)}{a (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^2}+\frac {b^4 \left (9 a^2-2 b^2\right ) \cos (c+d x)}{a^2 (a-b)^3 (a+b)^3 (a+b \sin (c+d x))}}{2 d} \]
((2*b^3*(20*a^4 - 7*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[ a^2 - b^2]])/(a^3*(a^2 - b^2)^(7/2)) - (2*Log[Cos[(c + d*x)/2]])/a^3 + (2* Log[Sin[(c + d*x)/2]])/a^3 + (2*Sin[(c + d*x)/2])/((a + b)^3*(Cos[(c + d*x )/2] - Sin[(c + d*x)/2])) - (2*Sin[(c + d*x)/2])/((a - b)^3*(Cos[(c + d*x) /2] + Sin[(c + d*x)/2])) + (b^4*Cos[c + d*x])/(a*(a - b)^2*(a + b)^2*(a + b*Sin[c + d*x])^2) + (b^4*(9*a^2 - 2*b^2)*Cos[c + d*x])/(a^2*(a - b)^3*(a + b)^3*(a + b*Sin[c + d*x])))/(2*d)
Time = 0.85 (sec) , antiderivative size = 402, 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.
| \(\displaystyle \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x) \cos (c+d x)^2 (a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle \int \left (\frac {\csc (c+d x)}{a^3}+\frac {b^3 \left (3 a^2-b^2\right )}{a^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}-\frac {b^3}{a \left (b^2-a^2\right ) (a+b \sin (c+d x))^3}+\frac {6 a^4 b^3-3 a^2 b^5+b^7}{a^3 \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}-\frac {1}{2 (a+b)^3 (\sin (c+d x)-1)}-\frac {1}{2 (a-b)^3 (\sin (c+d x)+1)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {2 b^3 \left (3 a^2-b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac {b^3 \left (2 a^2+b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a d \left (a^2-b^2\right )^{7/2}}+\frac {b^4 \left (3 a^2-b^2\right ) \cos (c+d x)}{a^2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {3 b^4 \cos (c+d x)}{2 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac {b^4 \cos (c+d x)}{2 a d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac {2 b^3 \left (6 a^4-3 a^2 b^2+b^4\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{7/2}}+\frac {\cos (c+d x)}{2 d (a+b)^3 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^3 (\sin (c+d x)+1)}\) |
(2*b^3*(3*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a* (a^2 - b^2)^(7/2)*d) + (b^3*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/ Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(7/2)*d) + (2*b^3*(6*a^4 - 3*a^2*b^2 + b^ 4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(7/2 )*d) - ArcTanh[Cos[c + d*x]]/(a^3*d) + Cos[c + d*x]/(2*(a + b)^3*d*(1 - Si n[c + d*x])) + Cos[c + d*x]/(2*(a - b)^3*d*(1 + Sin[c + d*x])) + (b^4*Cos[ c + d*x])/(2*a*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (3*b^4*Cos[c + d* x])/(2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x])) + (b^4*(3*a^2 - b^2)*Cos[c + d*x])/(a^2*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))
3.15.75.3.1 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 = 2.59 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.69
| method | result | size |
| derivativedivides | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 b^{3} \left (\frac {\left (\frac {11}{2} a^{3} b^{2}-2 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (10 a^{4}+17 a^{2} b^{2}-6 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \,b^{2} \left (29 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (10 a^{2}-3 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\left (20 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{3}}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(278\) |
| default | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3}}+\frac {2 b^{3} \left (\frac {\left (\frac {11}{2} a^{3} b^{2}-2 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (10 a^{4}+17 a^{2} b^{2}-6 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {a \,b^{2} \left (29 a^{2}-8 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\frac {a^{2} b \left (10 a^{2}-3 b^{2}\right )}{2}}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}+\frac {\left (20 a^{4}-7 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{3} \left (a +b \right )^{3} a^{3}}+\frac {1}{\left (a -b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{\left (a +b \right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(278\) |
| risch | \(\frac {6 i a^{4} b^{3}-16 i a^{6} b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i b^{7} {\mathrm e}^{4 i \left (d x +c \right )}+11 i a^{2} b^{5}-2 a^{5} {\mathrm e}^{5 i \left (d x +c \right )} b^{2}-14 a^{3} b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+a \,b^{6} {\mathrm e}^{5 i \left (d x +c \right )}-36 i a^{4} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-8 i a^{6} {\mathrm e}^{4 i \left (d x +c \right )} b +8 a^{7} {\mathrm e}^{3 i \left (d x +c \right )}+4 a^{5} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+24 a^{3} b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-6 a \,b^{6} {\mathrm e}^{3 i \left (d x +c \right )}-2 i b^{7}-14 i a^{4} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{2} b^{5} {\mathrm e}^{4 i \left (d x +c \right )}+22 a^{5} b^{2} {\mathrm e}^{i \left (d x +c \right )}+30 a^{3} b^{4} {\mathrm e}^{i \left (d x +c \right )}-7 a \,b^{6} {\mathrm e}^{i \left (d x +c \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} a^{2} d \left (a^{2}-b^{2}\right )^{3}}+\frac {10 i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {7 i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d a}+\frac {i b^{7} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d \,a^{3}}-\frac {10 i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {7 i b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d a}-\frac {i b^{7} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\) | \(896\) |
1/d*(1/a^3*ln(tan(1/2*d*x+1/2*c))+2*b^3/(a-b)^3/(a+b)^3/a^3*(((11/2*a^3*b^ 2-2*a*b^4)*tan(1/2*d*x+1/2*c)^3+1/2*b*(10*a^4+17*a^2*b^2-6*b^4)*tan(1/2*d* x+1/2*c)^2+1/2*a*b^2*(29*a^2-8*b^2)*tan(1/2*d*x+1/2*c)+1/2*a^2*b*(10*a^2-3 *b^2))/(tan(1/2*d*x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*(20*a^4-7*a ^2*b^2+2*b^4)/(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)))+1/(a-b)^3/(tan(1/2*d*x+1/2*c)+1)-1/(a+b)^3/(tan(1/2*d*x+1/2* c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 768 vs. \(2 (377) = 754\).
Time = 1.80 (sec) , antiderivative size = 1623, normalized size of antiderivative = 4.04 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
[-1/4*(4*a^10 - 12*a^8*b^2 + 12*a^6*b^4 - 4*a^4*b^6 + 2*(10*a^8*b^2 - 2*a^ 6*b^4 - 11*a^4*b^6 + 3*a^2*b^8)*cos(d*x + c)^2 - ((20*a^4*b^5 - 7*a^2*b^7 + 2*b^9)*cos(d*x + c)^3 - 2*(20*a^5*b^4 - 7*a^3*b^6 + 2*a*b^8)*cos(d*x + c )*sin(d*x + c) - (20*a^6*b^3 + 13*a^4*b^5 - 5*a^2*b^7 + 2*b^9)*cos(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)) + 2*((a^8* b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2*b^8 + b^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x + c) - ( a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^4*b^6 - 3*a^2*b^8 + b^10)*cos(d*x + c)) *log(1/2*cos(d*x + c) + 1/2) - 2*((a^8*b^2 - 4*a^6*b^4 + 6*a^4*b^6 - 4*a^2 *b^8 + b^10)*cos(d*x + c)^3 - 2*(a^9*b - 4*a^7*b^3 + 6*a^5*b^5 - 4*a^3*b^7 + a*b^9)*cos(d*x + c)*sin(d*x + c) - (a^10 - 3*a^8*b^2 + 2*a^6*b^4 + 2*a^ 4*b^6 - 3*a^2*b^8 + b^10)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*( 2*a^9*b - 6*a^7*b^3 + 6*a^5*b^5 - 2*a^3*b^7 - (6*a^7*b^3 + 5*a^5*b^5 - 13* a^3*b^7 + 2*a*b^9)*cos(d*x + c)^2)*sin(d*x + c))/((a^11*b^2 - 4*a^9*b^4 + 6*a^7*b^6 - 4*a^5*b^8 + a^3*b^10)*d*cos(d*x + c)^3 - 2*(a^12*b - 4*a^10*b^ 3 + 6*a^8*b^5 - 4*a^6*b^7 + a^4*b^9)*d*cos(d*x + c)*sin(d*x + c) - (a^13 - 3*a^11*b^2 + 2*a^9*b^4 + 2*a^7*b^6 - 3*a^5*b^8 + a^3*b^10)*d*cos(d*x + c) ), -1/2*(2*a^10 - 6*a^8*b^2 + 6*a^6*b^4 - 2*a^4*b^6 + (10*a^8*b^2 - 2*a...
\[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.40 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.02 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {{\left (20 \, a^{4} b^{3} - 7 \, a^{2} b^{5} + 2 \, b^{7}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3} - 3 \, a b^{2}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}} + \frac {11 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, a^{4} b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 17 \, a^{2} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 29 \, a^{3} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10 \, a^{4} b^{4} - 3 \, a^{2} b^{6}}{{\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}}}{d} \]
((20*a^4*b^3 - 7*a^2*b^5 + 2*b^7)*(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)))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*sqrt(a^2 - b^2)) + 2*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) - a^3 - 3*a*b^2)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(tan(1/2*d*x + 1/2*c)^2 - 1)) + (11*a^3*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*a*b^7*tan(1/2*d*x + 1/2*c)^3 + 10*a^4*b^4*tan(1/2*d*x + 1/2*c)^2 + 17 *a^2*b^6*tan(1/2*d*x + 1/2*c)^2 - 6*b^8*tan(1/2*d*x + 1/2*c)^2 + 29*a^3*b^ 5*tan(1/2*d*x + 1/2*c) - 8*a*b^7*tan(1/2*d*x + 1/2*c) + 10*a^4*b^4 - 3*a^2 *b^6)/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) + log(abs(tan(1/2*d*x + 1/2*c)))/a^3)/d
Time = 17.19 (sec) , antiderivative size = 2999, normalized size of antiderivative = 7.46 \[ \int \frac {\csc (c+d x) \sec ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]
log(tan(c/2 + (d*x)/2))/(a^3*d) - ((2*a^6 - 3*b^6 + 10*a^2*b^4 + 6*a^4*b^2 )/(a*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) + (tan(c/2 + (d*x)/2)*(2*a^6*b - 8*b^7 + 29*a^2*b^5 + 22*a^4*b^3))/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2 )) - (2*tan(c/2 + (d*x)/2)^3*(2*a^6*b - 2*b^7 + 13*a^2*b^5 + 2*a^4*b^3))/( a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)^5*(6*a^6*b - 4*b^7 + 11*a^2*b^5 + 2*a^4*b^3))/(a^2*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2 )) + (2*tan(c/2 + (d*x)/2)^2*(2*a^8 - 3*b^8 + 10*a^2*b^6 + 8*a^4*b^4 - 2*a ^6*b^2))/(a^3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)^4 *(17*a^2*b^6 - 6*b^8 - 2*a^8 + 18*a^4*b^4 + 18*a^6*b^2))/(a^3*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^6 - a^2 - tan(c/2 + ( d*x)/2)^2*(a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 + 4*b^2) + 4*a*b*tan(c /2 + (d*x)/2)^5 - 4*a*b*tan(c/2 + (d*x)/2))) + (b^3*atan(((b^3*(-(a + b)^7 *(a - b)^7)^(1/2)*(10*a^4 + b^4 - (7*a^2*b^2)/2)*(tan(c/2 + (d*x)/2)*(2*a^ 24 + 8*a^4*b^20 - 76*a^6*b^18 + 346*a^8*b^16 - 938*a^10*b^14 + 1612*a^12*b ^12 - 1790*a^14*b^10 + 1276*a^16*b^8 - 566*a^18*b^6 + 148*a^20*b^4 - 22*a^ 22*b^2) - 2*a^23*b + 4*a^5*b^19 - 37*a^7*b^17 + 164*a^9*b^15 - 433*a^11*b^ 13 + 722*a^13*b^11 - 769*a^15*b^9 + 512*a^17*b^7 - 199*a^19*b^5 + 38*a^21* b^3 + (b^3*(-(a + b)^7*(a - b)^7)^(1/2)*(10*a^4 + b^4 - (7*a^2*b^2)/2)*(ta n(c/2 + (d*x)/2)*(6*a^27 + 8*a^7*b^20 - 78*a^9*b^18 + 342*a^11*b^16 - 888* a^13*b^14 + 1512*a^15*b^12 - 1764*a^17*b^10 + 1428*a^19*b^8 - 792*a^21*...