Integrand size = 21, antiderivative size = 133 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {5 \text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {12 \tan (c+d x)}{a^2 d}-\frac {5 \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac {10 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 \tan ^3(c+d x)}{a^2 d} \]
-5*arctanh(sin(d*x+c))/a^2/d+12*tan(d*x+c)/a^2/d-5*sec(d*x+c)*tan(d*x+c)/a ^2/d-10/3*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*sec(d*x+c)^2*ta n(d*x+c)/d/(a+a*cos(d*x+c))^2+4*tan(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(133)=266\).
Time = 2.91 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.58 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {960 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (-3 \sin \left (\frac {d x}{2}\right )+155 \sin \left (\frac {3 d x}{2}\right )-153 \sin \left (c-\frac {d x}{2}\right )+21 \sin \left (c+\frac {d x}{2}\right )-135 \sin \left (2 c+\frac {d x}{2}\right )+25 \sin \left (c+\frac {3 d x}{2}\right )+45 \sin \left (2 c+\frac {3 d x}{2}\right )-85 \sin \left (3 c+\frac {3 d x}{2}\right )+99 \sin \left (c+\frac {5 d x}{2}\right )+21 \sin \left (2 c+\frac {5 d x}{2}\right )+33 \sin \left (3 c+\frac {5 d x}{2}\right )-45 \sin \left (4 c+\frac {5 d x}{2}\right )+57 \sin \left (2 c+\frac {7 d x}{2}\right )+18 \sin \left (3 c+\frac {7 d x}{2}\right )+24 \sin \left (4 c+\frac {7 d x}{2}\right )-15 \sin \left (5 c+\frac {7 d x}{2}\right )+24 \sin \left (3 c+\frac {9 d x}{2}\right )+11 \sin \left (4 c+\frac {9 d x}{2}\right )+13 \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]
(960*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Co s[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec [c + d*x]^3*(-3*Sin[(d*x)/2] + 155*Sin[(3*d*x)/2] - 153*Sin[c - (d*x)/2] + 21*Sin[c + (d*x)/2] - 135*Sin[2*c + (d*x)/2] + 25*Sin[c + (3*d*x)/2] + 45 *Sin[2*c + (3*d*x)/2] - 85*Sin[3*c + (3*d*x)/2] + 99*Sin[c + (5*d*x)/2] + 21*Sin[2*c + (5*d*x)/2] + 33*Sin[3*c + (5*d*x)/2] - 45*Sin[4*c + (5*d*x)/2 ] + 57*Sin[2*c + (7*d*x)/2] + 18*Sin[3*c + (7*d*x)/2] + 24*Sin[4*c + (7*d* x)/2] - 15*Sin[5*c + (7*d*x)/2] + 24*Sin[3*c + (9*d*x)/2] + 11*Sin[4*c + ( 9*d*x)/2] + 13*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)
Time = 0.87 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09, 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, 3227, 3042, 4254, 2009, 4255, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {2 (3 a-2 a \cos (c+d x)) \sec ^4(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \frac {(3 a-2 a \cos (c+d x)) \sec ^4(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {3 a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {2 \left (\frac {\int 3 \left (6 a^2-5 a^2 \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \left (6 a^2-5 a^2 \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \frac {6 a^2-5 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (6 a^2 \int \sec ^4(c+d x)dx-5 a^2 \int \sec ^3(c+d x)dx\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (6 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx-5 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-\frac {6 a^2 \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}-5 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {6 a^2 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {6 a^2 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {6 a^2 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {2 \left (\frac {3 \left (-5 a^2 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {6 a^2 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {5 \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}\right )}{3 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*(Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (2*((-5*Se c[c + d*x]^2*Tan[c + d*x])/(d*(1 + Cos[c + d*x])) + (3*(-5*a^2*(ArcTanh[Si n[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (6*a^2*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/a^2))/(3*a^2)
3.1.62.3.1 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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\) | \(152\) |
| default | \(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {10}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+10 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \,a^{2}}\) | \(152\) |
| norman | \(\frac {-\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {80 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {23 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}\) | \(155\) |
| parallelrisch | \(\frac {\left (90 \cos \left (d x +c \right )+30 \cos \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-90 \cos \left (d x +c \right )-30 \cos \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+95 \left (\cos \left (d x +c \right )+\frac {12 \cos \left (2 d x +2 c \right )}{19}+\frac {33 \cos \left (3 d x +3 c \right )}{95}+\frac {12 \cos \left (4 d x +4 c \right )}{95}+\frac {52}{95}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a^{2} d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(159\) |
| risch | \(\frac {2 i \left (15 \,{\mathrm e}^{8 i \left (d x +c \right )}+45 \,{\mathrm e}^{7 i \left (d x +c \right )}+85 \,{\mathrm e}^{6 i \left (d x +c \right )}+135 \,{\mathrm e}^{5 i \left (d x +c \right )}+153 \,{\mathrm e}^{4 i \left (d x +c \right )}+155 \,{\mathrm e}^{3 i \left (d x +c \right )}+99 \,{\mathrm e}^{2 i \left (d x +c \right )}+57 \,{\mathrm e}^{i \left (d x +c \right )}+24\right )}{3 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\) | \(169\) |
1/2/d/a^2*(1/3*tan(1/2*d*x+1/2*c)^3+9*tan(1/2*d*x+1/2*c)-2/3/(tan(1/2*d*x+ 1/2*c)+1)^3+3/(tan(1/2*d*x+1/2*c)+1)^2-10/(tan(1/2*d*x+1/2*c)+1)-10*ln(tan (1/2*d*x+1/2*c)+1)-2/3/(tan(1/2*d*x+1/2*c)-1)^3-3/(tan(1/2*d*x+1/2*c)-1)^2 -10/(tan(1/2*d*x+1/2*c)-1)+10*ln(tan(1/2*d*x+1/2*c)-1))
Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {15 \, {\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (24 \, \cos \left (d x + c\right )^{4} + 33 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]
-1/6*(15*(cos(d*x + c)^5 + 2*cos(d*x + c)^4 + cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^5 + 2*cos(d*x + c)^4 + cos(d*x + c)^3)*log(-s in(d*x + c) + 1) - 2*(24*cos(d*x + c)^4 + 33*cos(d*x + c)^3 + 6*cos(d*x + c)^2 - cos(d*x + c) + 1)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos (d*x + c)^4 + a^2*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.24 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \]
1/6*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c ) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a^ 2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d *x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d
Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {30 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {30 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
-1/6*(30*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 30*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 4*(15*tan(1/2*d*x + 1/2*c)^5 - 20*tan(1/2*d*x + 1/2*c) ^3 + 9*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (a^4*t an(1/2*d*x + 1/2*c)^3 + 27*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 14.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d}-\frac {10\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d} \]