Integrand size = 21, antiderivative size = 136 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {11 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))^2}-\frac {43 \tan (c+d x)}{21 a^4 d (1+\sec (c+d x))}-\frac {\sec ^3(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 \sec ^2(c+d x) \tan (c+d x)}{7 a d (a+a \sec (c+d x))^3} \]
arctanh(sin(d*x+c))/a^4/d+11/21*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-43/21*ta n(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c ))^4-2/7*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3
Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (672 \text {arctanh}(\sin (c+d x)) \cos ^7\left (\frac {1}{2} (c+d x)\right )-126 \sin \left (\frac {1}{2} (c+d x)\right )-189 \sin \left (\frac {3}{2} (c+d x)\right )-91 \sin \left (\frac {5}{2} (c+d x)\right )-16 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{42 a^4 d (1+\sec (c+d x))^4} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]^4*(672*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/ 2]^7 - 126*Sin[(c + d*x)/2] - 189*Sin[(3*(c + d*x))/2] - 91*Sin[(5*(c + d* x))/2] - 16*Sin[(7*(c + d*x))/2]))/(42*a^4*d*(1 + Sec[c + d*x])^4)
Time = 1.00 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4303, 3042, 4507, 27, 3042, 4496, 25, 3042, 4486, 3042, 4257, 4281}
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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{(a \sec (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) (3 a-7 a \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a-7 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int \frac {5 \sec ^2(c+d x) \left (4 a^2-7 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec ^2(c+d x) \left (4 a^2-7 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^2-7 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle -\frac {\frac {-\frac {\int -\frac {\sec (c+d x) \left (22 a^3-21 a^3 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {\sec (c+d x) \left (22 a^3-21 a^3 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (22 a^3-21 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {\frac {43 a^3 \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx-21 a^2 \int \sec (c+d x)dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\frac {43 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-21 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {\frac {43 a^3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {21 a^2 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle -\frac {\frac {\frac {\frac {43 a^3 \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {21 a^2 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {11 \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{a^2}+\frac {2 a \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {\tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}\) |
-1/7*(Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) - ((2*a*Sec[ c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((-11*Tan[c + d*x])/ (3*d*(1 + Sec[c + d*x])^2) + ((-21*a^2*ArcTanh[Sin[c + d*x]])/d + (43*a^3* Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/a^2)/(7*a^2)
3.1.72.3.1 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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.65
| 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 | \(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-168 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+168 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{168 a^{4} d}\) | \(88\) |
| risch | \(-\frac {2 i \left (21 \,{\mathrm e}^{6 i \left (d x +c \right )}+147 \,{\mathrm e}^{5 i \left (d x +c \right )}+434 \,{\mathrm e}^{4 i \left (d x +c \right )}+686 \,{\mathrm e}^{3 i \left (d x +c \right )}+525 \,{\mathrm e}^{2 i \left (d x +c \right )}+203 \,{\mathrm e}^{i \left (d x +c \right )}+32\right )}{21 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}\) | \(133\) |
| norman | \(\frac {-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {169 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}-\frac {229 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{24 a d}+\frac {293 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}-\frac {121 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{168 a d}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{168 a d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{56 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{56 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}\) | \(211\) |
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 = 0.27 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} + 107 \, \cos \left (d x + c\right )^{2} + 124 \, \cos \left (d x + c\right ) + 52\right )} \sin \left (d x + c\right )}{42 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]
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)
\[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]
Integral(sec(c + d*x)**5/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)/a**4
Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{168 \, d} \]
-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 = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {3 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{168 \, d} \]
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 = 13.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4}+\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4}}{d} \]