Integrand size = 21, antiderivative size = 112 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {24 \tan (c+d x)}{5 a^3 d}-\frac {\tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {3 \tan (c+d x)}{5 a d (a+a \cos (c+d x))^2}-\frac {3 \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \]
-3*arctanh(sin(d*x+c))/a^3/d+24/5*tan(d*x+c)/a^3/d-1/5*tan(d*x+c)/d/(a+a*c os(d*x+c))^3-3/5*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^2-3*tan(d*x+c)/d/(a^3+a^3 *cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(286\) vs. \(2(112)=224\).
Time = 1.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.55 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+76 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+20 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+8 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{5 a^3 d (1+\cos (c+d x))^3} \]
(2*Cos[(c + d*x)/2]*(Sec[c/2]*Sin[(d*x)/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2] *Sin[(d*x)/2] + 76*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 3*Log[Cos[(c + d*x )/2] + Sin[(c + d*x)/2]] + Sin[d*x]/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin [c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) + Cos[(c + d*x)/2]*Tan[c/2] + 8*Cos[(c + d*x)/2]^3*Tan[c/2]))/( 5*a^3*d*(1 + Cos[c + d*x])^3)
Time = 0.89 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3245, 27, 3042, 3457, 27, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {3 (2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \int \frac {(2 a-a \cos (c+d x)) \sec ^2(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {2 a-a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {3 \left (3 a^2-2 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {\left (3 a^2-2 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{\cos (c+d x) a+a}dx}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {3 a^2-2 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {3 \left (\frac {\frac {\int \left (8 a^3-5 a^3 \cos (c+d x)\right ) \sec ^2(c+d x)dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {\frac {\int \frac {8 a^3-5 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {3 \left (\frac {\frac {8 a^3 \int \sec ^2(c+d x)dx-5 a^3 \int \sec (c+d x)dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {\frac {8 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-5 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {3 \left (\frac {\frac {-\frac {8 a^3 \int 1d(-\tan (c+d x))}{d}-5 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {3 \left (\frac {\frac {\frac {8 a^3 \tan (c+d x)}{d}-5 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {3 \left (\frac {\frac {\frac {8 a^3 \tan (c+d x)}{d}-\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{d}}{a^2}-\frac {5 a^2 \tan (c+d x)}{d (a \cos (c+d x)+a)}}{a^2}-\frac {a \tan (c+d x)}{d (a \cos (c+d x)+a)^2}\right )}{5 a^2}-\frac {\tan (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
-1/5*Tan[c + d*x]/(d*(a + a*Cos[c + d*x])^3) + (3*(-((a*Tan[c + d*x])/(d*( a + a*Cos[c + d*x])^2)) + ((-5*a^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-5*a^3*ArcTanh[Sin[c + d*x]])/d + (8*a^3*Tan[c + d*x])/d)/a^2)/a^2))/( 5*a^2)
3.1.70.3.1 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 = 1.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d \,a^{3}}\) | \(105\) |
| default | \(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d \,a^{3}}\) | \(105\) |
| parallelrisch | \(\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\frac {57 \left (\cos \left (d x +c \right )+\frac {\cos \left (2 d x +2 c \right )}{2}+\frac {2 \cos \left (3 d x +3 c \right )}{19}+\frac {67}{114}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20}}{a^{3} d \cos \left (d x +c \right )}\) | \(110\) |
| norman | \(\frac {-\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d a}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\) | \(136\) |
| risch | \(\frac {2 i \left (15 \,{\mathrm e}^{6 i \left (d x +c \right )}+75 \,{\mathrm e}^{5 i \left (d x +c \right )}+160 \,{\mathrm e}^{4 i \left (d x +c \right )}+200 \,{\mathrm e}^{3 i \left (d x +c \right )}+189 \,{\mathrm e}^{2 i \left (d x +c \right )}+105 \,{\mathrm e}^{i \left (d x +c \right )}+24\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\) | \(147\) |
1/4/d/a^3*(1/5*tan(1/2*d*x+1/2*c)^5+2*tan(1/2*d*x+1/2*c)^3+17*tan(1/2*d*x+ 1/2*c)-4/(tan(1/2*d*x+1/2*c)+1)-12*ln(tan(1/2*d*x+1/2*c)+1)-4/(tan(1/2*d*x +1/2*c)-1)+12*ln(tan(1/2*d*x+1/2*c)-1))
Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.70 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (24 \, \cos \left (d x + c\right )^{3} + 57 \, \cos \left (d x + c\right )^{2} + 39 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{10 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
-1/10*(15*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*co s(d*x + c)^2 + cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(24*cos(d*x + c)^3 + 57*cos(d*x + c)^2 + 39*cos(d*x + c) + 5)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
\[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{2}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.34 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{20 \, d} \]
1/20*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(co s(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3 /(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log( sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d
Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{20 \, d} \]
-1/20*(60*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 40*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1) *a^3) - (a^12*tan(1/2*d*x + 1/2*c)^5 + 10*a^12*tan(1/2*d*x + 1/2*c)^3 + 85 *a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 14.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {6\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {17\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \]