Integrand size = 21, antiderivative size = 135 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {4 \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {664 \tan (c+d x)}{105 a^4 d}-\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {4 \tan (c+d x)}{a^4 d (1+\cos (c+d x))}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]
-4*arctanh(sin(d*x+c))/a^4/d+664/105*tan(d*x+c)/a^4/d-88/105*tan(d*x+c)/a^ 4/d/(1+cos(d*x+c))^2-4*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*tan(d*x+c)/d/(a +a*cos(d*x+c))^4-12/35*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3
Leaf count is larger than twice the leaf count of optimal. \(341\) vs. \(2(135)=270\).
Time = 3.38 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {107520 \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-10780 \sin \left (\frac {d x}{2}\right )+18788 \sin \left (\frac {3 d x}{2}\right )-20524 \sin \left (c-\frac {d x}{2}\right )+14644 \sin \left (c+\frac {d x}{2}\right )-16660 \sin \left (2 c+\frac {d x}{2}\right )-4690 \sin \left (c+\frac {3 d x}{2}\right )+14378 \sin \left (2 c+\frac {3 d x}{2}\right )-9100 \sin \left (3 c+\frac {3 d x}{2}\right )+11668 \sin \left (c+\frac {5 d x}{2}\right )-630 \sin \left (2 c+\frac {5 d x}{2}\right )+9358 \sin \left (3 c+\frac {5 d x}{2}\right )-2940 \sin \left (4 c+\frac {5 d x}{2}\right )+4228 \sin \left (2 c+\frac {7 d x}{2}\right )+315 \sin \left (3 c+\frac {7 d x}{2}\right )+3493 \sin \left (4 c+\frac {7 d x}{2}\right )-420 \sin \left (5 c+\frac {7 d x}{2}\right )+664 \sin \left (3 c+\frac {9 d x}{2}\right )+105 \sin \left (4 c+\frac {9 d x}{2}\right )+559 \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{1680 a^4 d (1+\cos (c+d x))^4} \]
(107520*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log [Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]* Sec[c + d*x]*(-10780*Sin[(d*x)/2] + 18788*Sin[(3*d*x)/2] - 20524*Sin[c - ( d*x)/2] + 14644*Sin[c + (d*x)/2] - 16660*Sin[2*c + (d*x)/2] - 4690*Sin[c + (3*d*x)/2] + 14378*Sin[2*c + (3*d*x)/2] - 9100*Sin[3*c + (3*d*x)/2] + 116 68*Sin[c + (5*d*x)/2] - 630*Sin[2*c + (5*d*x)/2] + 9358*Sin[3*c + (5*d*x)/ 2] - 2940*Sin[4*c + (5*d*x)/2] + 4228*Sin[2*c + (7*d*x)/2] + 315*Sin[3*c + (7*d*x)/2] + 3493*Sin[4*c + (7*d*x)/2] - 420*Sin[5*c + (7*d*x)/2] + 664*S in[3*c + (9*d*x)/2] + 105*Sin[4*c + (9*d*x)/2] + 559*Sin[5*c + (9*d*x)/2]) )/(1680*a^4*d*(1 + Cos[c + d*x])^4)
Time = 1.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.18, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3245, 27, 3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {4 (2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 \int \frac {(2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \int \frac {2 a-a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {4 \left (\frac {\int \frac {\left (13 a^2-9 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\int \frac {13 a^2-9 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\int \frac {\left (61 a^3-44 a^3 \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\int \frac {61 a^3-44 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\int \left (166 a^4-105 a^4 \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\int \frac {166 a^4-105 a^4 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {166 a^4 \int \sec ^2(c+d x)dx-105 a^4 \int \sec (c+d x)dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {166 a^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-105 a^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {-\frac {166 a^4 \int 1d(-\tan (c+d x))}{d}-105 a^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\frac {166 a^4 \tan (c+d x)}{d}-105 a^4 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {4 \left (\frac {\frac {\frac {\frac {166 a^4 \tan (c+d x)}{d}-\frac {105 a^4 \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {105 a^3 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {22 \tan (c+d x)}{3 d (\cos (c+d x)+1)^2}}{5 a^2}-\frac {3 a \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\right )}{7 a^2}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
-1/7*Tan[c + d*x]/(d*(a + a*Cos[c + d*x])^4) + (4*((-3*a*Tan[c + d*x])/(5* d*(a + a*Cos[c + d*x])^3) + ((-22*Tan[c + d*x])/(3*d*(1 + Cos[c + d*x])^2) + ((-105*a^3*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-105*a^4*ArcTanh[ Sin[c + d*x]])/d + (166*a^4*Tan[c + d*x])/d)/a^2)/(3*a^2))/(5*a^2)))/(7*a^ 2)
3.1.80.3.1 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 = 1.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\) | \(118\) |
| default | \(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\) | \(118\) |
| parallelrisch | \(\frac {3360 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-3360 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2861 \left (\cos \left (d x +c \right )+\frac {1650 \cos \left (2 d x +2 c \right )}{2861}+\frac {559 \cos \left (3 d x +3 c \right )}{2861}+\frac {83 \cos \left (4 d x +4 c \right )}{2861}+\frac {1672}{2861}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a^{4} d \cos \left (d x +c \right )}\) | \(121\) |
| norman | \(\frac {-\frac {65 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {31 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {47 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d a}+\frac {11 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}\) | \(155\) |
| risch | \(\frac {8 i \left (105 \,{\mathrm e}^{8 i \left (d x +c \right )}+735 \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 \,{\mathrm e}^{6 i \left (d x +c \right )}+4165 \,{\mathrm e}^{5 i \left (d x +c \right )}+5131 \,{\mathrm e}^{4 i \left (d x +c \right )}+4697 \,{\mathrm e}^{3 i \left (d x +c \right )}+2917 \,{\mathrm e}^{2 i \left (d x +c \right )}+1057 \,{\mathrm e}^{i \left (d x +c \right )}+166\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\) | \(169\) |
1/8/d/a^4*(1/7*tan(1/2*d*x+1/2*c)^7+7/5*tan(1/2*d*x+1/2*c)^5+23/3*tan(1/2* d*x+1/2*c)^3+49*tan(1/2*d*x+1/2*c)-8/(tan(1/2*d*x+1/2*c)+1)-32*ln(tan(1/2* d*x+1/2*c)+1)-8/(tan(1/2*d*x+1/2*c)-1)+32*ln(tan(1/2*d*x+1/2*c)-1))
Time = 0.26 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {210 \, {\left (\cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 210 \, {\left (\cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (664 \, \cos \left (d x + c\right )^{4} + 2236 \, \cos \left (d x + c\right )^{3} + 2636 \, \cos \left (d x + c\right )^{2} + 1184 \, \cos \left (d x + c\right ) + 105\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]
-1/105*(210*(cos(d*x + c)^5 + 4*cos(d*x + c)^4 + 6*cos(d*x + c)^3 + 4*cos( d*x + c)^2 + cos(d*x + c))*log(sin(d*x + c) + 1) - 210*(cos(d*x + c)^5 + 4 *cos(d*x + c)^4 + 6*cos(d*x + c)^3 + 4*cos(d*x + c)^2 + cos(d*x + c))*log( -sin(d*x + c) + 1) - (664*cos(d*x + c)^4 + 2236*cos(d*x + c)^3 + 2636*cos( d*x + c)^2 + 1184*cos(d*x + c) + 105)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))
\[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(sec(c + d*x)**2/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1), x)/a**4
Time = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{840 \, d} \]
1/840*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)* (cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15 *sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d
Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {3360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]
-1/840*(3360*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3360*log(abs(tan(1/2 *d*x + 1/2*c) - 1))/a^4 + 1680*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c) ^2 - 1)*a^4) - (15*a^24*tan(1/2*d*x + 1/2*c)^7 + 147*a^24*tan(1/2*d*x + 1/ 2*c)^5 + 805*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*a^24*tan(1/2*d*x + 1/2*c)) /a^28)/d
Time = 14.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^4\,d}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4\,d}-\frac {8\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {49\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4\,d} \]