Integrand size = 33, antiderivative size = 145 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {3 C \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {(2 A+27 C) \tan (c+d x)}{15 a^3 d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(A-9 C) \sec ^2(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {3 C \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]
-3*C*arctanh(sin(d*x+c))/a^3/d+1/15*(2*A+27*C)*tan(d*x+c)/a^3/d-1/5*(A+C)* sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(A-9*C)*sec(d*x+c)^2*tan (d*x+c)/a/d/(a+a*sec(d*x+c))^2+3*C*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(457\) vs. \(2(145)=290\).
Time = 5.56 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.15 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (2880 C \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-5 (4 A+51 C) \sin \left (\frac {d x}{2}\right )+(22 A+567 C) \sin \left (\frac {3 d x}{2}\right )-10 A \sin \left (c-\frac {d x}{2}\right )-600 C \sin \left (c-\frac {d x}{2}\right )+10 A \sin \left (c+\frac {d x}{2}\right )+375 C \sin \left (c+\frac {d x}{2}\right )-20 A \sin \left (2 c+\frac {d x}{2}\right )-480 C \sin \left (2 c+\frac {d x}{2}\right )-60 C \sin \left (c+\frac {3 d x}{2}\right )+22 A \sin \left (2 c+\frac {3 d x}{2}\right )+402 C \sin \left (2 c+\frac {3 d x}{2}\right )-225 C \sin \left (3 c+\frac {3 d x}{2}\right )+10 A \sin \left (c+\frac {5 d x}{2}\right )+315 C \sin \left (c+\frac {5 d x}{2}\right )+30 C \sin \left (2 c+\frac {5 d x}{2}\right )+10 A \sin \left (3 c+\frac {5 d x}{2}\right )+240 C \sin \left (3 c+\frac {5 d x}{2}\right )-45 C \sin \left (4 c+\frac {5 d x}{2}\right )+2 A \sin \left (2 c+\frac {7 d x}{2}\right )+72 C \sin \left (2 c+\frac {7 d x}{2}\right )+15 C \sin \left (3 c+\frac {7 d x}{2}\right )+2 A \sin \left (4 c+\frac {7 d x}{2}\right )+57 C \sin \left (4 c+\frac {7 d x}{2}\right )\right )\right )}{60 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(2880*C*Cos[(c + d*x )/2]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]*(-5*(4*A + 51*C)*Sin[(d* x)/2] + (22*A + 567*C)*Sin[(3*d*x)/2] - 10*A*Sin[c - (d*x)/2] - 600*C*Sin[ c - (d*x)/2] + 10*A*Sin[c + (d*x)/2] + 375*C*Sin[c + (d*x)/2] - 20*A*Sin[2 *c + (d*x)/2] - 480*C*Sin[2*c + (d*x)/2] - 60*C*Sin[c + (3*d*x)/2] + 22*A* Sin[2*c + (3*d*x)/2] + 402*C*Sin[2*c + (3*d*x)/2] - 225*C*Sin[3*c + (3*d*x )/2] + 10*A*Sin[c + (5*d*x)/2] + 315*C*Sin[c + (5*d*x)/2] + 30*C*Sin[2*c + (5*d*x)/2] + 10*A*Sin[3*c + (5*d*x)/2] + 240*C*Sin[3*c + (5*d*x)/2] - 45* C*Sin[4*c + (5*d*x)/2] + 2*A*Sin[2*c + (7*d*x)/2] + 72*C*Sin[2*c + (7*d*x) /2] + 15*C*Sin[3*c + (7*d*x)/2] + 2*A*Sin[4*c + (7*d*x)/2] + 57*C*Sin[4*c + (7*d*x)/2])))/(60*a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x] )^3)
Time = 1.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4573, 25, 3042, 4507, 3042, 4496, 3042, 4274, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^3(c+d x) (a (2 A-3 C)+a (A+6 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec ^3(c+d x) (a (2 A-3 C)+a (A+6 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (2 A-3 C)+a (A+6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^2(c+d x) \left (2 (A-9 C) a^2+(2 A+27 C) \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 (A-9 C) a^2+(2 A+27 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\int \sec (c+d x) \left (45 a^3 C-a^3 (2 A+27 C) \sec (c+d x)\right )dx}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (45 a^3 C-a^3 (2 A+27 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {45 a^3 C \int \sec (c+d x)dx-a^3 (2 A+27 C) \int \sec ^2(c+d x)dx}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {45 a^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a^3 (2 A+27 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\frac {a^3 (2 A+27 C) \int 1d(-\tan (c+d x))}{d}+45 a^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {45 a^3 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (2 A+27 C) \tan (c+d x)}{d}}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {45 a^2 C \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {\frac {45 a^3 C \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (2 A+27 C) \tan (c+d x)}{d}}{a^2}}{3 a^2}+\frac {a (A-9 C) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (( a*(A - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + (( 45*a^2*C*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) - ((45*a^3*C*ArcTanh[Sin[c + d*x]])/d - (a^3*(2*A + 27*C)*Tan[c + d*x])/d)/a^2)/(3*a^2))/(5*a^2)
3.2.39.3.1 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_)], 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_)]^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]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \(\frac {360 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-360 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+17 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {3 \left (2 A +57 C \right ) \cos \left (2 d x +2 c \right )}{17}+\frac {\left (A +36 C \right ) \cos \left (3 d x +3 c \right )}{17}+\left (A +\frac {342 C}{17}\right ) \cos \left (d x +c \right )+\frac {6 A}{17}+\frac {201 C}{17}\right )}{120 d \,a^{3} \cos \left (d x +c \right )}\) | \(135\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(151\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(151\) |
| risch | \(\frac {2 i \left (45 C \,{\mathrm e}^{6 i \left (d x +c \right )}+225 C \,{\mathrm e}^{5 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}+480 C \,{\mathrm e}^{4 i \left (d x +c \right )}+10 A \,{\mathrm e}^{3 i \left (d x +c \right )}+600 C \,{\mathrm e}^{3 i \left (d x +c \right )}+22 A \,{\mathrm e}^{2 i \left (d x +c \right )}+567 C \,{\mathrm e}^{2 i \left (d x +c \right )}+10 A \,{\mathrm e}^{i \left (d x +c \right )}+315 C \,{\mathrm e}^{i \left (d x +c \right )}+2 A +72 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\) | \(208\) |
| norman | \(\frac {-\frac {\left (A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{20 a d}-\frac {\left (A +81 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 a d}-\frac {\left (7 A -153 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (25 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {5 \left (27 C +A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\left (1773 C +53 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4} a^{2}}+\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {3 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\) | \(232\) |
1/120*(360*C*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)-360*C*ln(tan(1/2*d*x+1/2* c)+1)*cos(d*x+c)+17*sec(1/2*d*x+1/2*c)^4*tan(1/2*d*x+1/2*c)*(3/17*(2*A+57* C)*cos(2*d*x+2*c)+1/17*(A+36*C)*cos(3*d*x+3*c)+(A+342/17*C)*cos(d*x+c)+6/1 7*A+201/17*C))/d/a^3/cos(d*x+c)
Time = 0.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C \cos \left (d x + c\right )^{4} + 3 \, C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (A + 36 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + 57 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A + 117 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]
-1/30*(45*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C* cos(d*x + c))*log(sin(d*x + c) + 1) - 45*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2 *(A + 36*C)*cos(d*x + c)^3 + 3*(2*A + 57*C)*cos(d*x + c)^2 + (7*A + 117*C) *cos(d*x + c) + 15*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d* x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))
\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(A*sec(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.20 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} + \frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
1/60*(3*C*(40*sin(d*x + c)/((a^3 - a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 )*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60 *log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d *x + c) + 1) - 1)/a^3) + A*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*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 )/d
Time = 0.36 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {180 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {180 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
-1/60*(180*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 180*C*log(abs(tan(1/ 2*d*x + 1/2*c) - 1))/a^3 + 120*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2* c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 10*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2 *c)^3 + 15*A*a^12*tan(1/2*d*x + 1/2*c) + 255*C*a^12*tan(1/2*d*x + 1/2*c))/ a^15)/d
Time = 14.83 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{6\,a^3}+\frac {C}{3\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{4\,a^3}+\frac {2\,C}{a^3}-\frac {2\,A-6\,C}{4\,a^3}\right )}{d}-\frac {6\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]