Integrand size = 19, antiderivative size = 153 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^5 d}-\frac {\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac {34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}-\frac {173 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}-\frac {488 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )} \]
arctanh(sin(d*x+c))/a^5/d-1/9*sin(d*x+c)/d/(a+a*cos(d*x+c))^5-13/63*sin(d* x+c)/a/d/(a+a*cos(d*x+c))^4-34/105*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^3-173 /315*sin(d*x+c)/a^3/d/(a+a*cos(d*x+c))^2-488/315*sin(d*x+c)/d/(a^5+a^5*cos (d*x+c))
Time = 1.62 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.38 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (80640 \cos ^9\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \left (35973 \sin \left (\frac {d x}{2}\right )-25515 \sin \left (c+\frac {d x}{2}\right )+29757 \sin \left (c+\frac {3 d x}{2}\right )-11235 \sin \left (2 c+\frac {3 d x}{2}\right )+14733 \sin \left (2 c+\frac {5 d x}{2}\right )-2835 \sin \left (3 c+\frac {5 d x}{2}\right )+4077 \sin \left (3 c+\frac {7 d x}{2}\right )-315 \sin \left (4 c+\frac {7 d x}{2}\right )+488 \sin \left (4 c+\frac {9 d x}{2}\right )\right )\right )}{2520 a^5 d (1+\cos (c+d x))^5} \]
-1/2520*(Cos[(c + d*x)/2]*(80640*Cos[(c + d*x)/2]^9*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2] *(35973*Sin[(d*x)/2] - 25515*Sin[c + (d*x)/2] + 29757*Sin[c + (3*d*x)/2] - 11235*Sin[2*c + (3*d*x)/2] + 14733*Sin[2*c + (5*d*x)/2] - 2835*Sin[3*c + (5*d*x)/2] + 4077*Sin[3*c + (7*d*x)/2] - 315*Sin[4*c + (7*d*x)/2] + 488*Si n[4*c + (9*d*x)/2])))/(a^5*d*(1 + Cos[c + d*x])^5)
Time = 1.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.737, Rules used = {3042, 3245, 3042, 3457, 27, 3042, 3457, 3042, 3457, 3042, 3457, 27, 3042, 4257}
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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \cos (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {(9 a-4 a \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^4}dx}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {9 a-4 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (21 a^2-13 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {\left (21 a^2-13 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {21 a^2-13 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {\left (105 a^3-68 a^3 \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {105 a^3-68 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {\left (315 a^4-173 a^4 \cos (c+d x)\right ) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {315 a^4-173 a^4 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\frac {\int 315 a^5 \sec (c+d x)dx}{a^2}-\frac {488 a^4 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {315 a^3 \int \sec (c+d x)dx-\frac {488 a^4 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {315 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {488 a^4 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\frac {315 a^3 \text {arctanh}(\sin (c+d x))}{d}-\frac {488 a^4 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {173 a^3 \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
-1/9*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^5) + ((-13*a*Sin[c + d*x])/(7*d* (a + a*Cos[c + d*x])^4) + (3*((-34*a^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d *x])^3) + ((-173*a^3*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((315*a^ 3*ArcTanh[Sin[c + d*x]])/d - (488*a^4*Sin[c + d*x])/(d*(a + a*Cos[c + d*x] )))/(3*a^2))/(5*a^2)))/(7*a^2))/(9*a^2)
3.1.90.3.1 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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.66
| method | result | size |
| derivativedivides | \(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {6 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {26 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(101\) |
| default | \(\frac {-\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}-\frac {6 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {26 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(101\) |
| parallelrisch | \(\frac {-35 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-270 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1008 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2730 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-5040 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+5040 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{5040 a^{5} d}\) | \(101\) |
| norman | \(\frac {-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}-\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{5 d a}-\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d a}}{a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5} d}\) | \(139\) |
| risch | \(-\frac {2 i \left (315 \,{\mathrm e}^{8 i \left (d x +c \right )}+2835 \,{\mathrm e}^{7 i \left (d x +c \right )}+11235 \,{\mathrm e}^{6 i \left (d x +c \right )}+25515 \,{\mathrm e}^{5 i \left (d x +c \right )}+35973 \,{\mathrm e}^{4 i \left (d x +c \right )}+29757 \,{\mathrm e}^{3 i \left (d x +c \right )}+14733 \,{\mathrm e}^{2 i \left (d x +c \right )}+4077 \,{\mathrm e}^{i \left (d x +c \right )}+488\right )}{315 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{5} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{5} d}\) | \(155\) |
1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9-6/7*tan(1/2*d*x+1/2*c)^7-16/5*tan(1/ 2*d*x+1/2*c)^5-26/3*tan(1/2*d*x+1/2*c)^3-31*tan(1/2*d*x+1/2*c)-16*ln(tan(1 /2*d*x+1/2*c)-1)+16*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.27 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.61 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (488 \, \cos \left (d x + c\right )^{4} + 2125 \, \cos \left (d x + c\right )^{3} + 3549 \, \cos \left (d x + c\right )^{2} + 2740 \, \cos \left (d x + c\right ) + 863\right )} \sin \left (d x + c\right )}{630 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]
1/630*(315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos (d*x + c)^2 + 5*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 315*(cos(d*x + c )^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(488*cos(d*x + c)^4 + 2125*cos(d*x + c)^3 + 3549*cos(d*x + c)^2 + 2740*cos(d*x + c) + 863)*sin(d*x + c))/(a^5* d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a ^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) + a^5*d)
\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\cos ^{5}{\left (c + d x \right )} + 5 \cos ^{4}{\left (c + d x \right )} + 10 \cos ^{3}{\left (c + d x \right )} + 10 \cos ^{2}{\left (c + d x \right )} + 5 \cos {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]
Integral(sec(c + d*x)/(cos(c + d*x)**5 + 5*cos(c + d*x)**4 + 10*cos(c + d* x)**3 + 10*cos(c + d*x)**2 + 5*cos(c + d*x) + 1), x)/a**5
Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.04 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {\frac {\frac {9765 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2730 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{5040 \, d} \]
-1/5040*((9765*sin(d*x + c)/(cos(d*x + c) + 1) + 2730*sin(d*x + c)^3/(cos( d*x + c) + 1)^3 + 1008*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 270*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^5 - 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^5 + 5040*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/a^5)/d
Time = 0.44 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} - \frac {35 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 270 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1008 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2730 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9765 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{5040 \, d} \]
1/5040*(5040*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 5040*log(abs(tan(1/2 *d*x + 1/2*c) - 1))/a^5 - (35*a^40*tan(1/2*d*x + 1/2*c)^9 + 270*a^40*tan(1 /2*d*x + 1/2*c)^7 + 1008*a^40*tan(1/2*d*x + 1/2*c)^5 + 2730*a^40*tan(1/2*d *x + 1/2*c)^3 + 9765*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d
Time = 14.54 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,a^5}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5}+\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5}}{d} \]