Integrand size = 21, antiderivative size = 162 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13 \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {152 \tan (c+d x)}{15 a^3 d}+\frac {13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {\sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
13/2*arctanh(sin(d*x+c))/a^3/d-152/15*tan(d*x+c)/a^3/d+13/2*sec(d*x+c)*tan (d*x+c)/a^3/d-1/5*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-11/15*sec(d *x+c)^3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-76/15*sec(d*x+c)^2*tan(d*x+c)/d/ (a^3+a^3*sec(d*x+c))
Time = 1.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {1560 \text {arctanh}(\sin (c+d x)) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x)-\frac {1}{8} \sec ^5(c+d x) (1142 \sin (c+d x)+1614 \sin (2 (c+d x))+1414 \sin (3 (c+d x))+717 \sin (4 (c+d x))+152 \sin (5 (c+d x)))}{30 a^3 d (1+\sec (c+d x))^3} \]
(1560*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/2]^6*Sec[c + d*x]^3 - (Sec[c + d *x]^5*(1142*Sin[c + d*x] + 1614*Sin[2*(c + d*x)] + 1414*Sin[3*(c + d*x)] + 717*Sin[4*(c + d*x)] + 152*Sin[5*(c + d*x)]))/8)/(30*a^3*d*(1 + Sec[c + d *x])^3)
Time = 1.11 (sec) , antiderivative size = 176, 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, 4303, 3042, 4507, 3042, 4507, 3042, 4274, 3042, 4254, 24, 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.
\(\displaystyle \int \frac {\sec ^6(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^6}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4303 |
\(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (4 a-7 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (4 a-7 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle -\frac {\frac {\int \frac {\sec ^3(c+d x) \left (33 a^2-43 a^2 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (33 a^2-43 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle -\frac {\frac {\frac {\int \sec ^2(c+d x) \left (152 a^3-195 a^3 \sec (c+d x)\right )dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (152 a^3-195 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle -\frac {\frac {\frac {152 a^3 \int \sec ^2(c+d x)dx-195 a^3 \int \sec ^3(c+d x)dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {152 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-195 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {\frac {\frac {-\frac {152 a^3 \int 1d(-\tan (c+d x))}{d}-195 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle -\frac {\frac {\frac {\frac {152 a^3 \tan (c+d x)}{d}-195 a^3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle -\frac {\frac {\frac {\frac {152 a^3 \tan (c+d x)}{d}-195 a^3 \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\frac {\frac {152 a^3 \tan (c+d x)}{d}-195 a^3 \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\frac {\frac {76 a^2 \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {152 a^3 \tan (c+d x)}{d}-195 a^3 \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}+\frac {11 a \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*(Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) - ((11*a*Sec [c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((76*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((152*a^3*Tan[c + d*x])/d - 195*a^3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d) ))/a^2)/(3*a^2))/(5*a^2)
3.1.61.3.1 Defintions of rubi rules used
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
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_)]*(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 = 0.43 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(135\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {14}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+26 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(135\) |
parallelrisch | \(\frac {\left (-1560 \cos \left (2 d x +2 c \right )-1560\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1560 \cos \left (2 d x +2 c \right )+1560\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2331 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\cos \left (d x +c \right )+\frac {174 \cos \left (2 d x +2 c \right )}{259}+\frac {239 \cos \left (3 d x +3 c \right )}{777}+\frac {152 \cos \left (4 d x +4 c \right )}{2331}+\frac {1354}{2331}\right )}{240 a^{3} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(138\) |
risch | \(-\frac {i \left (195 \,{\mathrm e}^{8 i \left (d x +c \right )}+975 \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 \,{\mathrm e}^{i \left (d x +c \right )}+304\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}\) | \(169\) |
norman | \(\frac {\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {721 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {6613 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}-\frac {1165 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}+\frac {475 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {59 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} a^{2}}-\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {13 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(212\) |
1/4/d/a^3*(-1/5*tan(1/2*d*x+1/2*c)^5-8/3*tan(1/2*d*x+1/2*c)^3-31*tan(1/2*d *x+1/2*c)+2/(tan(1/2*d*x+1/2*c)-1)^2+14/(tan(1/2*d*x+1/2*c)-1)-26*ln(tan(1 /2*d*x+1/2*c)-1)-2/(tan(1/2*d*x+1/2*c)+1)^2+14/(tan(1/2*d*x+1/2*c)+1)+26*l n(tan(1/2*d*x+1/2*c)+1))
Time = 0.29 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (304 \, \cos \left (d x + c\right )^{4} + 717 \, \cos \left (d x + c\right )^{3} + 479 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 15\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
1/60*(195*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 195*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3 *cos(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(304*cos(d*x + c)^4 + 717*cos(d*x + c)^3 + 479*cos(d*x + c)^2 + 45*cos(d*x + c) - 15)*s in(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos( d*x + c)^3 + a^3*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]
-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40* sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^ 5)/a^3 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d* x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d
Time = 0.32 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {390 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
1/60*(390*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 390*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*tan(1/2*d*x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c ))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 12.99 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^6(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {13\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20\,a^3\,d}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^3\,d}-\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {31\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^3\,d} \]