Integrand size = 19, antiderivative size = 120 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {32 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))}-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3} \] Output:
arctanh(sin(d*x+c))/a^4/d-11/21*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-32/21*si n(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-2/7*sin( d*x+c)/a/d/(a+a*cos(d*x+c))^3
Time = 1.51 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.54 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {-1344 \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-686 \sin \left (\frac {d x}{2}\right )+434 \sin \left (c+\frac {d x}{2}\right )-525 \sin \left (c+\frac {3 d x}{2}\right )+147 \sin \left (2 c+\frac {3 d x}{2}\right )-203 \sin \left (2 c+\frac {5 d x}{2}\right )+21 \sin \left (3 c+\frac {5 d x}{2}\right )-32 \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{84 a^4 d (1+\cos (c+d x))^4} \] Input:
Integrate[Sec[c + d*x]/(a + a*Cos[c + d*x])^4,x]
Output:
(-1344*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]*(-686*Si n[(d*x)/2] + 434*Sin[c + (d*x)/2] - 525*Sin[c + (3*d*x)/2] + 147*Sin[2*c + (3*d*x)/2] - 203*Sin[2*c + (5*d*x)/2] + 21*Sin[3*c + (5*d*x)/2] - 32*Sin[ 3*c + (7*d*x)/2]))/(84*a^4*d*(1 + Cos[c + d*x])^4)
Time = 0.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 3245, 3042, 3457, 27, 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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \cos (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle \frac {\int \frac {(7 a-3 a \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^3}dx}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {7 a-3 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {5 \left (7 a^2-4 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\left (7 a^2-4 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {7 a^2-4 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\left (21 a^3-11 a^3 \cos (c+d x)\right ) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {21 a^3-11 a^3 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int 21 a^4 \sec (c+d x)dx}{a^2}-\frac {32 a^3 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 \int \sec (c+d x)dx-\frac {32 a^3 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {21 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {32 a^3 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {21 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {32 a^3 \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {11 \sin (c+d x)}{3 d (\cos (c+d x)+1)^2}}{a^2}-\frac {2 a \sin (c+d x)}{d (a \cos (c+d x)+a)^3}}{7 a^2}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4}\) |
Input:
Int[Sec[c + d*x]/(a + a*Cos[c + d*x])^4,x]
Output:
-1/7*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^4) + ((-2*a*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + ((-11*Sin[c + d*x])/(3*d*(1 + Cos[c + d*x])^2) + (( 21*a^2*ArcTanh[Sin[c + d*x]])/d - (32*a^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2))/a^2)/(7*a^2)
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 = 0.94 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\) | \(88\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \,a^{4}}\) | \(88\) |
| parallelrisch | \(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-168 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+168 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{168 a^{4} d}\) | \(88\) |
| norman | \(\frac {-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}}{a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\) | \(120\) |
| risch | \(-\frac {2 i \left (21 \,{\mathrm e}^{6 i \left (d x +c \right )}+147 \,{\mathrm e}^{5 i \left (d x +c \right )}+434 \,{\mathrm e}^{4 i \left (d x +c \right )}+686 \,{\mathrm e}^{3 i \left (d x +c \right )}+525 \,{\mathrm e}^{2 i \left (d x +c \right )}+203 \,{\mathrm e}^{i \left (d x +c \right )}+32\right )}{21 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\) | \(133\) |
Input:
int(sec(d*x+c)/(a+a*cos(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7-tan(1/2*d*x+1/2*c)^5-11/3*tan(1/2*d*x +1/2*c)^3-15*tan(1/2*d*x+1/2*c)-8*ln(tan(1/2*d*x+1/2*c)-1)+8*ln(tan(1/2*d* x+1/2*c)+1))
Time = 0.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.68 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} + 107 \, \cos \left (d x + c\right )^{2} + 124 \, \cos \left (d x + c\right ) + 52\right )} \sin \left (d x + c\right )}{42 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^4,x, algorithm="fricas")
Output:
1/42*(21*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 21*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(32*cos (d*x + c)^3 + 107*cos(d*x + c)^2 + 124*cos(d*x + c) + 52)*sin(d*x + c))/(a ^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4* a^4*d*cos(d*x + c) + a^4*d)
\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))**4,x)
Output:
Integral(sec(c + d*x)/(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 = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.16 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{168 \, d} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^4,x, algorithm="maxima")
Output:
-1/168*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(c os(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d
Time = 0.36 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {3 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{168 \, d} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^4,x, algorithm="giac")
Output:
1/168*(168*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 168*log(abs(tan(1/2*d* x + 1/2*c) - 1))/a^4 - (3*a^24*tan(1/2*d*x + 1/2*c)^7 + 21*a^24*tan(1/2*d* x + 1/2*c)^5 + 77*a^24*tan(1/2*d*x + 1/2*c)^3 + 315*a^24*tan(1/2*d*x + 1/2 *c))/a^28)/d
Time = 40.73 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4}+\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4}}{d} \] Input:
int(1/(cos(c + d*x)*(a + a*cos(c + d*x))^4),x)
Output:
-((11*tan(c/2 + (d*x)/2)^3)/(24*a^4) + tan(c/2 + (d*x)/2)^5/(8*a^4) + tan( c/2 + (d*x)/2)^7/(56*a^4) - (2*atanh(tan(c/2 + (d*x)/2)))/a^4 + (15*tan(c/ 2 + (d*x)/2))/(8*a^4))/d
Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {-168 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+168 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{168 a^{4} d} \] Input:
int(sec(d*x+c)/(a+a*cos(d*x+c))^4,x)
Output:
( - 168*log(tan((c + d*x)/2) - 1) + 168*log(tan((c + d*x)/2) + 1) - 3*tan( (c + d*x)/2)**7 - 21*tan((c + d*x)/2)**5 - 77*tan((c + d*x)/2)**3 - 315*ta n((c + d*x)/2))/(168*a**4*d)