Integrand size = 29, antiderivative size = 250 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{16 (a+b)^3 d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {b^7 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\sin (c+d x))^2}+\frac {7 a+9 b}{16 (a+b)^2 d (1-\sin (c+d x))}-\frac {1}{16 (a-b) d (1+\sin (c+d x))^2}-\frac {7 a-9 b}{16 (a-b)^2 d (1+\sin (c+d x))} \]
-csc(d*x+c)/a/d-1/16*(15*a^2+37*a*b+24*b^2)*ln(1-sin(d*x+c))/(a+b)^3/d-b*l n(sin(d*x+c))/a^2/d+1/16*(15*a^2-37*a*b+24*b^2)*ln(1+sin(d*x+c))/(a-b)^3/d -b^7*ln(a+b*sin(d*x+c))/a^2/(a^2-b^2)^3/d+1/16/(a+b)/d/(1-sin(d*x+c))^2+1/ 16*(7*a+9*b)/(a+b)^2/d/(1-sin(d*x+c))-1/16/(a-b)/d/(1+sin(d*x+c))^2+1/16*( -7*a+9*b)/(a-b)^2/d/(1+sin(d*x+c))
Time = 4.24 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.94 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^7 \left (-\frac {16 \csc (c+d x)}{a b^7}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (1-\sin (c+d x))}{b^7 (a+b)^3}-\frac {16 \log (\sin (c+d x))}{a^2 b^6}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (1+\sin (c+d x))}{(a-b)^3 b^7}-\frac {16 \log (a+b \sin (c+d x))}{a^2 (a-b)^3 (a+b)^3}+\frac {1}{b^7 (a+b) (-1+\sin (c+d x))^2}+\frac {-7 a-9 b}{b^7 (a+b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) b^7 (1+\sin (c+d x))^2}+\frac {-7 a+9 b}{(a-b)^2 b^7 (1+\sin (c+d x))}\right )}{16 d} \]
(b^7*((-16*Csc[c + d*x])/(a*b^7) - ((15*a^2 + 37*a*b + 24*b^2)*Log[1 - Sin [c + d*x]])/(b^7*(a + b)^3) - (16*Log[Sin[c + d*x]])/(a^2*b^6) + ((15*a^2 - 37*a*b + 24*b^2)*Log[1 + Sin[c + d*x]])/((a - b)^3*b^7) - (16*Log[a + b* Sin[c + d*x]])/(a^2*(a - b)^3*(a + b)^3) + 1/(b^7*(a + b)*(-1 + Sin[c + d* x])^2) + (-7*a - 9*b)/(b^7*(a + b)^2*(-1 + Sin[c + d*x])) - 1/((a - b)*b^7 *(1 + Sin[c + d*x])^2) + (-7*a + 9*b)/((a - b)^2*b^7*(1 + Sin[c + d*x])))) /(16*d)
Time = 0.57 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 27, 615, 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.
| \(\displaystyle \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (c+d x)^2 \cos (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {b^5 \int \frac {\csc ^2(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^7 \int \frac {\csc ^2(c+d x)}{b^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^7 \int \left (\frac {\csc ^2(c+d x)}{a b^8}-\frac {\csc (c+d x)}{a^2 b^7}+\frac {15 a^2+37 b a+24 b^2}{16 b^7 (a+b)^3 (b-b \sin (c+d x))}-\frac {1}{a^2 (a-b)^3 (a+b)^3 (a+b \sin (c+d x))}+\frac {15 a^2-37 b a+24 b^2}{16 (a-b)^3 b^7 (\sin (c+d x) b+b)}+\frac {7 a+9 b}{16 b^6 (a+b)^2 (b-b \sin (c+d x))^2}+\frac {7 a-9 b}{16 (a-b)^2 b^6 (\sin (c+d x) b+b)^2}+\frac {1}{8 b^5 (a+b) (b-b \sin (c+d x))^3}-\frac {1}{8 b^5 (b-a) (\sin (c+d x) b+b)^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^7 \left (-\frac {\log (b \sin (c+d x))}{a^2 b^6}-\frac {\log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )^3}-\frac {\left (15 a^2+37 a b+24 b^2\right ) \log (b-b \sin (c+d x))}{16 b^7 (a+b)^3}+\frac {\left (15 a^2-37 a b+24 b^2\right ) \log (b \sin (c+d x)+b)}{16 b^7 (a-b)^3}-\frac {\csc (c+d x)}{a b^7}-\frac {7 a-9 b}{16 b^6 (a-b)^2 (b \sin (c+d x)+b)}+\frac {7 a+9 b}{16 b^6 (a+b)^2 (b-b \sin (c+d x))}+\frac {1}{16 b^5 (a+b) (b-b \sin (c+d x))^2}-\frac {1}{16 b^5 (a-b) (b \sin (c+d x)+b)^2}\right )}{d}\) |
(b^7*(-(Csc[c + d*x]/(a*b^7)) - Log[b*Sin[c + d*x]]/(a^2*b^6) - ((15*a^2 + 37*a*b + 24*b^2)*Log[b - b*Sin[c + d*x]])/(16*b^7*(a + b)^3) - Log[a + b* Sin[c + d*x]]/(a^2*(a^2 - b^2)^3) + ((15*a^2 - 37*a*b + 24*b^2)*Log[b + b* Sin[c + d*x]])/(16*(a - b)^3*b^7) + 1/(16*b^5*(a + b)*(b - b*Sin[c + d*x]) ^2) + (7*a + 9*b)/(16*b^6*(a + b)^2*(b - b*Sin[c + d*x])) - 1/(16*(a - b)* b^5*(b + b*Sin[c + d*x])^2) - (7*a - 9*b)/(16*(a - b)^2*b^6*(b + b*Sin[c + d*x]))))/d
3.14.68.3.1 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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.31 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (15 a^{2}-37 a b +24 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7 a +9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-15 a^{2}-37 a b -24 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {b^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2}}}{d}\) | \(219\) |
| default | \(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 a -9 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (15 a^{2}-37 a b +24 b^{2}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7 a +9 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-15 a^{2}-37 a b -24 b^{2}\right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {b^{7} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2}}}{d}\) | \(219\) |
| parallelrisch | \(\frac {-8 b^{7} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-15 \left (a^{2}+\frac {37}{15} a b +\frac {8}{5} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{3} a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \left (a +b \right ) \left (\left (a^{2}-\frac {37}{15} a b +\frac {8}{5} b^{2}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\left (\frac {8 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} \left (a -b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\left (\left (\left (a^{4}-\frac {9}{5} a^{2} b^{2}+\frac {4}{5} b^{4}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {3}{8} a^{4}-\frac {27}{40} a^{2} b^{2}+\frac {1}{5} b^{4}\right ) \cos \left (4 d x +4 c \right )+\frac {9 a^{4}}{40}-\frac {29 a^{2} b^{2}}{40}+\frac {3 b^{4}}{5}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2 b \left (\left (a^{2}-b^{2}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (3 a^{2}-5 b^{2}\right ) \cos \left (4 d x +4 c \right )}{4}-\frac {7 a^{2}}{4}+\frac {9 b^{2}}{4}\right ) a}{5}\right ) a \right ) \left (a -b \right )}{3}\right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} a^{2} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(429\) |
| norman | \(\frac {-\frac {1}{2 a d}-\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {2 \left (2 a^{2} b -3 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (2 a^{2} b -3 b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 \left (2 a^{2} b -4 b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (15 a^{4}-25 a^{2} b^{2}+6 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\left (15 a^{4}-25 a^{2} b^{2}+6 b^{4}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (5 a^{4}-13 a^{2} b^{2}+4 b^{4}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (5 a^{4}-13 a^{2} b^{2}+4 b^{4}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {\left (15 a^{2}-37 a b +24 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (15 a^{2}+37 a b +24 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {b^{7} \ln \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 )}{a^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(605\) |
| risch | \(\text {Expression too large to display}\) | \(1209\) |
1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(7*a-9*b)/(a-b)^2/(1+sin(d*x+c)) +1/16*(15*a^2-37*a*b+24*b^2)/(a-b)^3*ln(1+sin(d*x+c))-1/a/sin(d*x+c)-1/a^2 *b*ln(sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1/16*(7*a+9*b)/(a+b)^2/(s in(d*x+c)-1)+1/16/(a+b)^3*(-15*a^2-37*a*b-24*b^2)*ln(sin(d*x+c)-1)-b^7/(a+ b)^3/(a-b)^3/a^2*ln(a+b*sin(d*x+c)))
Time = 2.42 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.70 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, b^{7} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) - 4 \, a^{7} + 8 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + 16 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - {\left (15 \, a^{7} + 8 \, a^{6} b - 42 \, a^{5} b^{2} - 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (15 \, a^{7} - 8 \, a^{6} b - 42 \, a^{5} b^{2} + 24 \, a^{4} b^{3} + 35 \, a^{3} b^{4} - 24 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, {\left (15 \, a^{7} - 42 \, a^{5} b^{2} + 35 \, a^{3} b^{4} - 8 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, a^{7} - 14 \, a^{5} b^{2} + 9 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} + 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} + 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]
-1/16*(16*b^7*cos(d*x + c)^4*log(b*sin(d*x + c) + a)*sin(d*x + c) - 4*a^7 + 8*a^5*b^2 - 4*a^3*b^4 + 16*(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*cos(d*x + c)^4*log(1/2*sin(d*x + c))*sin(d*x + c) - (15*a^7 + 8*a^6*b - 42*a^5*b^ 2 - 24*a^4*b^3 + 35*a^3*b^4 + 24*a^2*b^5)*cos(d*x + c)^4*log(sin(d*x + c) + 1)*sin(d*x + c) + (15*a^7 - 8*a^6*b - 42*a^5*b^2 + 24*a^4*b^3 + 35*a^3*b ^4 - 24*a^2*b^5)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)*sin(d*x + c) + 2*(1 5*a^7 - 42*a^5*b^2 + 35*a^3*b^4 - 8*a*b^6)*cos(d*x + c)^4 - 2*(5*a^7 - 14* a^5*b^2 + 9*a^3*b^4)*cos(d*x + c)^2 + 4*(a^6*b - 2*a^4*b^3 + a^2*b^5 + 2*( a^6*b - 3*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/((a^8 - 3*a^6 *b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^4*sin(d*x + c))
Timed out. \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, b^{7} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} - \frac {{\left (15 \, a^{2} - 37 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 37 \, a b + 24 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left ({\left (15 \, a^{4} - 27 \, a^{2} b^{2} + 8 \, b^{4}\right )} \sin \left (d x + c\right )^{4} + 8 \, a^{4} - 16 \, a^{2} b^{2} + 8 \, b^{4} - 4 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{3} - {\left (25 \, a^{4} - 45 \, a^{2} b^{2} + 16 \, b^{4}\right )} \sin \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{3} b - 5 \, a b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{5} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{3} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )} + \frac {16 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{16 \, d} \]
-1/16*(16*b^7*log(b*sin(d*x + c) + a)/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b ^6) - (15*a^2 - 37*a*b + 24*b^2)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3* a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(sin(d*x + c) - 1)/(a^3 + 3*a ^2*b + 3*a*b^2 + b^3) + 2*((15*a^4 - 27*a^2*b^2 + 8*b^4)*sin(d*x + c)^4 + 8*a^4 - 16*a^2*b^2 + 8*b^4 - 4*(a^3*b - 2*a*b^3)*sin(d*x + c)^3 - (25*a^4 - 45*a^2*b^2 + 16*b^4)*sin(d*x + c)^2 + 2*(3*a^3*b - 5*a*b^3)*sin(d*x + c) )/((a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)^5 - 2*(a^5 - 2*a^3*b^2 + a*b^4)* sin(d*x + c)^3 + (a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c)) + 16*b*log(sin(d* x + c))/a^2)/d
Time = 0.62 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.67 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, b^{8} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}} - \frac {{\left (15 \, a^{2} - 37 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (15 \, a^{2} + 37 \, a b + 24 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {16 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (6 \, a^{4} b \sin \left (d x + c\right )^{4} - 18 \, a^{2} b^{3} \sin \left (d x + c\right )^{4} + 18 \, b^{5} \sin \left (d x + c\right )^{4} + 7 \, a^{5} \sin \left (d x + c\right )^{3} - 18 \, a^{3} b^{2} \sin \left (d x + c\right )^{3} + 11 \, a b^{4} \sin \left (d x + c\right )^{3} - 16 \, a^{4} b \sin \left (d x + c\right )^{2} + 48 \, a^{2} b^{3} \sin \left (d x + c\right )^{2} - 44 \, b^{5} \sin \left (d x + c\right )^{2} - 9 \, a^{5} \sin \left (d x + c\right ) + 22 \, a^{3} b^{2} \sin \left (d x + c\right ) - 13 \, a b^{4} \sin \left (d x + c\right ) + 12 \, a^{4} b - 34 \, a^{2} b^{3} + 28 \, b^{5}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}} - \frac {16 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )}}{16 \, d} \]
-1/16*(16*b^8*log(abs(b*sin(d*x + c) + a))/(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7) - (15*a^2 - 37*a*b + 24*b^2)*log(abs(sin(d*x + c) + 1))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (15*a^2 + 37*a*b + 24*b^2)*log(abs(sin(d*x + c) - 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 16*b*log(abs(sin(d*x + c)))/a^2 + 2*(6*a^4*b*sin(d*x + c)^4 - 18*a^2*b^3*sin(d*x + c)^4 + 18*b^5*sin(d*x + c)^4 + 7*a^5*sin(d*x + c)^3 - 18*a^3*b^2*sin(d*x + c)^3 + 11*a*b^4*sin(d*x + c)^3 - 16*a^4*b*sin(d*x + c)^2 + 48*a^2*b^3*sin(d*x + c)^2 - 44*b^5*sin (d*x + c)^2 - 9*a^5*sin(d*x + c) + 22*a^3*b^2*sin(d*x + c) - 13*a*b^4*sin( d*x + c) + 12*a^4*b - 34*a^2*b^3 + 28*b^5)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(sin(d*x + c)^2 - 1)^2) - 16*(b*sin(d*x + c) - a)/(a^2*sin(d*x + c)) )/d
Time = 12.31 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.49 \[ \int \frac {\csc ^2(c+d x) \sec ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b^2}{8\,{\left (a-b\right )}^3}-\frac {7\,b}{16\,{\left (a-b\right )}^2}+\frac {15}{16\,\left (a-b\right )}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {7\,b}{16\,{\left (a+b\right )}^2}+\frac {15}{16\,\left (a+b\right )}+\frac {b^2}{8\,{\left (a+b\right )}^3}\right )}{d}-\frac {\frac {1}{a}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (a^2\,b-2\,b^3\right )}{2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {\sin \left (c+d\,x\right )\,\left (3\,a^2\,b-5\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\sin \left (c+d\,x\right )}^4\,\left (15\,a^4-27\,a^2\,b^2+8\,b^4\right )}{8\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (25\,a^4-45\,a^2\,b^2+16\,b^4\right )}{8\,a\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}-\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2\,d}-\frac {b^7\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^8-3\,a^6\,b^2+3\,a^4\,b^4-a^2\,b^6\right )} \]
(log(sin(c + d*x) + 1)*(b^2/(8*(a - b)^3) - (7*b)/(16*(a - b)^2) + 15/(16* (a - b))))/d - (log(sin(c + d*x) - 1)*((7*b)/(16*(a + b)^2) + 15/(16*(a + b)) + b^2/(8*(a + b)^3)))/d - (1/a - (sin(c + d*x)^3*(a^2*b - 2*b^3))/(2*( a^4 + b^4 - 2*a^2*b^2)) + (sin(c + d*x)*(3*a^2*b - 5*b^3))/(4*(a^4 + b^4 - 2*a^2*b^2)) + (sin(c + d*x)^4*(15*a^4 + 8*b^4 - 27*a^2*b^2))/(8*a*(a^4 + b^4 - 2*a^2*b^2)) - (sin(c + d*x)^2*(25*a^4 + 16*b^4 - 45*a^2*b^2))/(8*a*( a^4 + b^4 - 2*a^2*b^2)))/(d*(sin(c + d*x) - 2*sin(c + d*x)^3 + sin(c + d*x )^5)) - (b*log(sin(c + d*x)))/(a^2*d) - (b^7*log(a + b*sin(c + d*x)))/(d*( a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))