Integrand size = 21, antiderivative size = 168 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {5 \text {arctanh}(\sin (c+d x))}{a^5 d}+\frac {496 \tan (c+d x)}{63 a^5 d}-\frac {\tan (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac {5 \tan (c+d x)}{21 a d (a+a \cos (c+d x))^4}-\frac {29 \tan (c+d x)}{63 a^2 d (a+a \cos (c+d x))^3}-\frac {67 \tan (c+d x)}{63 a^3 d (a+a \cos (c+d x))^2}-\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \cos (c+d x)\right )} \]
-5*arctanh(sin(d*x+c))/a^5/d+496/63*tan(d*x+c)/a^5/d-1/9*tan(d*x+c)/d/(a+a *cos(d*x+c))^5-5/21*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^4-29/63*tan(d*x+c)/a^2 /d/(a+a*cos(d*x+c))^3-67/63*tan(d*x+c)/a^3/d/(a+a*cos(d*x+c))^2-5*tan(d*x+ c)/d/(a^5+a^5*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(393\) vs. \(2(168)=336\).
Time = 5.52 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.34 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {322560 \cos ^{10}\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-33978 \sin \left (\frac {d x}{2}\right )+52002 \sin \left (\frac {3 d x}{2}\right )-56952 \sin \left (c-\frac {d x}{2}\right )+43722 \sin \left (c+\frac {d x}{2}\right )-47208 \sin \left (2 c+\frac {d x}{2}\right )-18144 \sin \left (c+\frac {3 d x}{2}\right )+41796 \sin \left (2 c+\frac {3 d x}{2}\right )-28350 \sin \left (3 c+\frac {3 d x}{2}\right )+34578 \sin \left (c+\frac {5 d x}{2}\right )-5691 \sin \left (2 c+\frac {5 d x}{2}\right )+28719 \sin \left (3 c+\frac {5 d x}{2}\right )-11550 \sin \left (4 c+\frac {5 d x}{2}\right )+15517 \sin \left (2 c+\frac {7 d x}{2}\right )-504 \sin \left (3 c+\frac {7 d x}{2}\right )+13186 \sin \left (4 c+\frac {7 d x}{2}\right )-2835 \sin \left (5 c+\frac {7 d x}{2}\right )+4149 \sin \left (3 c+\frac {9 d x}{2}\right )+252 \sin \left (4 c+\frac {9 d x}{2}\right )+3582 \sin \left (5 c+\frac {9 d x}{2}\right )-315 \sin \left (6 c+\frac {9 d x}{2}\right )+496 \sin \left (4 c+\frac {11 d x}{2}\right )+63 \sin \left (5 c+\frac {11 d x}{2}\right )+433 \sin \left (6 c+\frac {11 d x}{2}\right )\right )}{2016 a^5 d (1+\cos (c+d x))^5} \]
(322560*Cos[(c + d*x)/2]^10*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Lo g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c] *Sec[c + d*x]*(-33978*Sin[(d*x)/2] + 52002*Sin[(3*d*x)/2] - 56952*Sin[c - (d*x)/2] + 43722*Sin[c + (d*x)/2] - 47208*Sin[2*c + (d*x)/2] - 18144*Sin[c + (3*d*x)/2] + 41796*Sin[2*c + (3*d*x)/2] - 28350*Sin[3*c + (3*d*x)/2] + 34578*Sin[c + (5*d*x)/2] - 5691*Sin[2*c + (5*d*x)/2] + 28719*Sin[3*c + (5* d*x)/2] - 11550*Sin[4*c + (5*d*x)/2] + 15517*Sin[2*c + (7*d*x)/2] - 504*Si n[3*c + (7*d*x)/2] + 13186*Sin[4*c + (7*d*x)/2] - 2835*Sin[5*c + (7*d*x)/2 ] + 4149*Sin[3*c + (9*d*x)/2] + 252*Sin[4*c + (9*d*x)/2] + 3582*Sin[5*c + (9*d*x)/2] - 315*Sin[6*c + (9*d*x)/2] + 496*Sin[4*c + (11*d*x)/2] + 63*Sin [5*c + (11*d*x)/2] + 433*Sin[6*c + (11*d*x)/2]))/(2016*a^5*d*(1 + Cos[c + d*x])^5)
Time = 1.44 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.19, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 3245, 27, 3042, 3457, 3042, 3457, 27, 3042, 3457, 3042, 3457, 3042, 3227, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {5 (2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \int \frac {(2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {2 a-a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {\left (17 a^2-12 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {17 a^2-12 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\int \frac {3 \left (38 a^3-29 a^3 \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \int \frac {\left (38 a^3-29 a^3 \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \int \frac {38 a^3-29 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\int \frac {\left (181 a^4-134 a^4 \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\int \frac {181 a^4-134 a^4 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\int \left (496 a^5-315 a^5 \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\int \frac {496 a^5-315 a^5 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {496 a^5 \int \sec ^2(c+d x)dx-315 a^5 \int \sec (c+d x)dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {496 a^5 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-315 a^5 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {-\frac {496 a^5 \int 1d(-\tan (c+d x))}{d}-315 a^5 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\frac {496 a^5 \tan (c+d x)}{d}-315 a^5 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\frac {496 a^5 \tan (c+d x)}{d}-\frac {315 a^5 \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {315 a^4 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \tan (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {3 a \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\right )}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \cos (c+d x)+a)^5}\) |
-1/9*Tan[c + d*x]/(d*(a + a*Cos[c + d*x])^5) + (5*((-3*a*Tan[c + d*x])/(7* d*(a + a*Cos[c + d*x])^4) + ((-29*a^2*Tan[c + d*x])/(5*d*(a + a*Cos[c + d* x])^3) + (3*((-67*a^3*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + ((-315* a^4*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-315*a^5*ArcTanh[Sin[c + d* x]])/d + (496*a^5*Tan[c + d*x])/d)/a^2)/(3*a^2)))/(5*a^2))/(7*a^2)))/(9*a^ 2)
3.1.91.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}+\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(131\) |
| default | \(\frac {\frac {\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{9}+\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {16}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(131\) |
| parallelrisch | \(\frac {40320 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-40320 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+31846 \left (\cos \left (d x +c \right )+\frac {10010 \cos \left (2 d x +2 c \right )}{15923}+\frac {4253 \cos \left (3 d x +3 c \right )}{15923}+\frac {2165 \cos \left (4 d x +4 c \right )}{31846}+\frac {124 \cos \left (5 d x +5 c \right )}{15923}+\frac {18359}{31846}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8064 a^{5} d \cos \left (d x +c \right )}\) | \(132\) |
| norman | \(\frac {-\frac {161 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {105 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {17 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d a}+\frac {65 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{1008 d a}+\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{5} d}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{5} d}\) | \(174\) |
| risch | \(\frac {2 i \left (315 \,{\mathrm e}^{10 i \left (d x +c \right )}+2835 \,{\mathrm e}^{9 i \left (d x +c \right )}+11550 \,{\mathrm e}^{8 i \left (d x +c \right )}+28350 \,{\mathrm e}^{7 i \left (d x +c \right )}+47208 \,{\mathrm e}^{6 i \left (d x +c \right )}+56952 \,{\mathrm e}^{5 i \left (d x +c \right )}+52002 \,{\mathrm e}^{4 i \left (d x +c \right )}+34578 \,{\mathrm e}^{3 i \left (d x +c \right )}+15517 \,{\mathrm e}^{2 i \left (d x +c \right )}+4149 \,{\mathrm e}^{i \left (d x +c \right )}+496\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{5} d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{5} d}\) | \(191\) |
1/16/d/a^5*(1/9*tan(1/2*d*x+1/2*c)^9+8/7*tan(1/2*d*x+1/2*c)^7+6*tan(1/2*d* x+1/2*c)^5+24*tan(1/2*d*x+1/2*c)^3+129*tan(1/2*d*x+1/2*c)-16/(tan(1/2*d*x+ 1/2*c)-1)+80*ln(tan(1/2*d*x+1/2*c)-1)-16/(tan(1/2*d*x+1/2*c)+1)-80*ln(tan( 1/2*d*x+1/2*c)+1))
Time = 0.29 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.65 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (496 \, \cos \left (d x + c\right )^{5} + 2165 \, \cos \left (d x + c\right )^{4} + 3633 \, \cos \left (d x + c\right )^{3} + 2840 \, \cos \left (d x + c\right )^{2} + 946 \, \cos \left (d x + c\right ) + 63\right )} \sin \left (d x + c\right )}{126 \, {\left (a^{5} d \cos \left (d x + c\right )^{6} + 5 \, a^{5} d \cos \left (d x + c\right )^{5} + 10 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 5 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d \cos \left (d x + c\right )\right )}} \]
-1/126*(315*(cos(d*x + c)^6 + 5*cos(d*x + c)^5 + 10*cos(d*x + c)^4 + 10*co s(d*x + c)^3 + 5*cos(d*x + c)^2 + cos(d*x + c))*log(sin(d*x + c) + 1) - 31 5*(cos(d*x + c)^6 + 5*cos(d*x + c)^5 + 10*cos(d*x + c)^4 + 10*cos(d*x + c) ^3 + 5*cos(d*x + c)^2 + cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(496*cos( d*x + c)^5 + 2165*cos(d*x + c)^4 + 3633*cos(d*x + c)^3 + 2840*cos(d*x + c) ^2 + 946*cos(d*x + c) + 63)*sin(d*x + c))/(a^5*d*cos(d*x + c)^6 + 5*a^5*d* cos(d*x + c)^5 + 10*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 5*a^5 *d*cos(d*x + c)^2 + a^5*d*cos(d*x + c))
\[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos ^{5}{\left (c + d x \right )} + 5 \cos ^{4}{\left (c + d x \right )} + 10 \cos ^{3}{\left (c + d x \right )} + 10 \cos ^{2}{\left (c + d x \right )} + 5 \cos {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]
Integral(sec(c + d*x)**2/(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 = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {\frac {2016 \, \sin \left (d x + c\right )}{{\left (a^{5} - \frac {a^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {8127 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1512 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {72 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{1008 \, d} \]
1/1008*(2016*sin(d*x + c)/((a^5 - a^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) *(cos(d*x + c) + 1)) + (8127*sin(d*x + c)/(cos(d*x + c) + 1) + 1512*sin(d* x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 72*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*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(s in(d*x + c)/(cos(d*x + c) + 1) - 1)/a^5)/d
Time = 0.44 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=-\frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{5}} - \frac {7 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 72 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1512 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8127 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{1008 \, d} \]
-1/1008*(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 + 2016*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c )^2 - 1)*a^5) - (7*a^40*tan(1/2*d*x + 1/2*c)^9 + 72*a^40*tan(1/2*d*x + 1/2 *c)^7 + 378*a^40*tan(1/2*d*x + 1/2*c)^5 + 1512*a^40*tan(1/2*d*x + 1/2*c)^3 + 8127*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d
Time = 14.78 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^5} \, dx=\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^5\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{14\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5\,d}-\frac {10\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^5\right )}+\frac {129\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5\,d} \]