Integrand size = 21, antiderivative size = 177 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^5 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \sec ^3(c+d x) \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \sec ^2(c+d x) \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {173 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}-\frac {661 \tan (c+d x)}{315 d \left (a^5+a^5 \sec (c+d x)\right )} \]
arctanh(sin(d*x+c))/a^5/d-1/9*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^5 -13/63*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4-34/105*sec(d*x+c)^2* tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3+173/315*tan(d*x+c)/a^3/d/(a+a*sec(d*x+ c))^2-661/315*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))
Time = 0.74 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (80640 \text {arctanh}(\sin (c+d x)) \cos ^9\left (\frac {1}{2} (c+d x)\right )-2 \left (5229 \sin \left (\frac {1}{2} (c+d x)\right )+9261 \sin \left (\frac {3}{2} (c+d x)\right )+5949 \sin \left (\frac {5}{2} (c+d x)\right )+1881 \sin \left (\frac {7}{2} (c+d x)\right )+244 \sin \left (\frac {9}{2} (c+d x)\right )\right )\right )}{2520 a^5 d (1+\sec (c+d x))^5} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]^5*(80640*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x )/2]^9 - 2*(5229*Sin[(c + d*x)/2] + 9261*Sin[(3*(c + d*x))/2] + 5949*Sin[( 5*(c + d*x))/2] + 1881*Sin[(7*(c + d*x))/2] + 244*Sin[(9*(c + d*x))/2])))/ (2520*a^5*d*(1 + Sec[c + d*x])^5)
Time = 1.33 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.16, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4303, 3042, 4507, 27, 3042, 4507, 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.
| \(\displaystyle \int \frac {\sec ^6(c+d x)}{(a \sec (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^6}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (4 a-9 a \sec (c+d x))}{(\sec (c+d x) a+a)^4}dx}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (4 a-9 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int \frac {3 \sec ^3(c+d x) \left (13 a^2-21 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^3}dx}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \frac {\sec ^3(c+d x) \left (13 a^2-21 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^3}dx}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (13 a^2-21 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\int \frac {\sec ^2(c+d x) \left (68 a^3-105 a^3 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (68 a^3-105 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {-\frac {\int -\frac {\sec (c+d x) \left (346 a^4-315 a^4 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\frac {\int \frac {\sec (c+d x) \left (346 a^4-315 a^4 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (346 a^4-315 a^4 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\frac {661 a^4 \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx-315 a^3 \int \sec (c+d x)dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\frac {661 a^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-315 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {\frac {661 a^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {315 a^3 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {34 a^2 \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {\frac {\frac {661 a^4 \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {315 a^3 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}\right )}{7 a^2}+\frac {13 a \tan (c+d x) \sec ^3(c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
-1/9*(Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^5) - ((13*a*Sec [c + d*x]^3*Tan[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) + (3*((34*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((-173*a^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((-315*a^3*ArcTanh[Sin[c + d*x]])/d + (661*a^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2)))/(7*a^ 2))/(9*a^2)
3.1.81.3.1 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 = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.57
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(101\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d \,a^{5}}\) | \(101\) |
| parallelrisch | \(\frac {-35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-270 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-1008 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-2730 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-5040 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+5040 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-9765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5040 d \,a^{5}}\) | \(101\) |
| risch | \(-\frac {2 i \left (315 \,{\mathrm e}^{8 i \left (d x +c \right )}+2835 \,{\mathrm e}^{7 i \left (d x +c \right )}+11235 \,{\mathrm e}^{6 i \left (d x +c \right )}+25515 \,{\mathrm e}^{5 i \left (d x +c \right )}+35973 \,{\mathrm e}^{4 i \left (d x +c \right )}+29757 \,{\mathrm e}^{3 i \left (d x +c \right )}+14733 \,{\mathrm e}^{2 i \left (d x +c \right )}+4077 \,{\mathrm e}^{i \left (d x +c \right )}+488\right )}{315 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{5}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \,a^{5}}\) | \(155\) |
1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9-6/7*tan(1/2*d*x+1/2*c)^7-16/5*tan(1/ 2*d*x+1/2*c)^5-26/3*tan(1/2*d*x+1/2*c)^3-31*tan(1/2*d*x+1/2*c)-16*ln(tan(1 /2*d*x+1/2*c)-1)+16*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.28 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (488 \, \cos \left (d x + c\right )^{4} + 2125 \, \cos \left (d x + c\right )^{3} + 3549 \, \cos \left (d x + c\right )^{2} + 2740 \, \cos \left (d x + c\right ) + 863\right )} \sin \left (d x + c\right )}{630 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]
1/630*(315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos (d*x + c)^2 + 5*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 315*(cos(d*x + c )^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(488*cos(d*x + c)^4 + 2125*cos(d*x + c)^3 + 3549*cos(d*x + c)^2 + 2740*cos(d*x + c) + 863)*sin(d*x + c))/(a^5* d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a ^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) + a^5*d)
\[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]
Integral(sec(c + d*x)**6/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10*sec(c + d*x)**2 + 5*sec(c + d*x) + 1), x)/a**5
Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\frac {\frac {9765 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2730 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{5040 \, d} \]
-1/5040*((9765*sin(d*x + c)/(cos(d*x + c) + 1) + 2730*sin(d*x + c)^3/(cos( d*x + c) + 1)^3 + 1008*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 270*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^5 - 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^5 + 5040*log(sin(d*x + c) /(cos(d*x + c) + 1) - 1)/a^5)/d
Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.71 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} - \frac {35 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 270 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1008 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2730 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9765 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{5040 \, d} \]
1/5040*(5040*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 5040*log(abs(tan(1/2 *d*x + 1/2*c) - 1))/a^5 - (35*a^40*tan(1/2*d*x + 1/2*c)^9 + 270*a^40*tan(1 /2*d*x + 1/2*c)^7 + 1008*a^40*tan(1/2*d*x + 1/2*c)^5 + 2730*a^40*tan(1/2*d *x + 1/2*c)^3 + 9765*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d
Time = 13.56 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5\,a^5}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^5}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5}+\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5}}{d} \]