Integrand size = 21, antiderivative size = 189 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {a \left (4 a^2+9 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \tan (c+d x)}{30 b d}-\frac {a \left (6 a^2-71 b^2\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d} \]
1/8*a*(4*a^2+9*b^2)*arctanh(sin(d*x+c))/d-1/30*(3*a^4-52*a^2*b^2-16*b^4)*t an(d*x+c)/b/d-1/120*a*(6*a^2-71*b^2)*sec(d*x+c)*tan(d*x+c)/d-1/60*(3*a^2-1 6*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d-1/20*a*(a+b*sec(d*x+c))^3*tan(d*x +c)/b/d+1/5*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d
Time = 0.65 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.63 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {15 a \left (4 a^2+9 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 a \left (4 a^2+9 b^2\right ) \sec (c+d x)+90 a b^2 \sec ^3(c+d x)+8 b \left (15 \left (3 a^2+b^2\right )+5 \left (3 a^2+2 b^2\right ) \tan ^2(c+d x)+3 b^2 \tan ^4(c+d x)\right )\right )}{120 d} \]
(15*a*(4*a^2 + 9*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*a*(4*a^2 + 9*b^2)*Sec[c + d*x] + 90*a*b^2*Sec[c + d*x]^3 + 8*b*(15*(3*a^2 + b^2) + 5* (3*a^2 + 2*b^2)*Tan[c + d*x]^2 + 3*b^2*Tan[c + d*x]^4)))/(120*d)
Time = 1.18 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4327, 3042, 4490, 3042, 4490, 3042, 4485, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3dx\) |
| \(\Big \downarrow \) 4327 |
| \(\displaystyle \frac {\int \sec (c+d x) (4 b-a \sec (c+d x)) (a+b \sec (c+d x))^3dx}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (4 b-a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3dx}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sec (c+d x)\right )dx-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (13 a b+\left (16 b^2-3 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4490 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sec (c+d x)\right )dx-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \sec (c+d x) \left (15 a b \left (4 a^2+9 b^2\right )-4 \left (3 a^4-52 b^2 a^2-16 b^4\right ) \sec (c+d x)\right )dx-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 a b \left (4 a^2+9 b^2\right )-4 \left (3 a^4-52 b^2 a^2-16 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2+9 b^2\right ) \int \sec (c+d x)dx-4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \int \sec ^2(c+d x)dx\right )-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2+9 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2+9 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \int 1d(-\tan (c+d x))}{d}\right )-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2+9 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \tan (c+d x)}{d}\right )-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 a b \left (4 a^2+9 b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \tan (c+d x)}{d}\right )-\frac {a b \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\) |
((a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*d) + (-1/4*(a*(a + b*Sec[c + d* x])^3*Tan[c + d*x])/d + (-1/3*((3*a^2 - 16*b^2)*(a + b*Sec[c + d*x])^2*Tan [c + d*x])/d + (-1/2*(a*b*(6*a^2 - 71*b^2)*Sec[c + d*x]*Tan[c + d*x])/d + ((15*a*b*(4*a^2 + 9*b^2)*ArcTanh[Sin[c + d*x]])/d - (4*(3*a^4 - 52*a^2*b^2 - 16*b^4)*Tan[c + d*x])/d)/2)/3)/4)/(5*b)
3.5.66.3.1 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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[Csc[e + f*x]* (a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 ))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Time = 1.89 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(148\) |
| default | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(148\) |
| parts | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {3 a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(156\) |
| parallelrisch | \(\frac {-600 a \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (a^{2}+\frac {9 b^{2}}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+600 a \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (a^{2}+\frac {9 b^{2}}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (240 a^{3}+1260 a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (1200 a^{2} b +320 b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (120 a^{3}+270 a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+\left (240 a^{2} b +64 b^{3}\right ) \sin \left (5 d x +5 c \right )+960 \sin \left (d x +c \right ) \left (a^{2}+\frac {2 b^{2}}{3}\right ) b}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(250\) |
| norman | \(\frac {\frac {\left (4 a^{3}-24 a^{2} b +15 a \,b^{2}-8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (4 a^{3}+24 a^{2} b +15 a \,b^{2}+8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a^{3}-96 a^{2} b +9 a \,b^{2}-16 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (12 a^{3}+96 a^{2} b +9 a \,b^{2}+16 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {4 b \left (75 a^{2}+29 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {a \left (4 a^{2}+9 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 a^{2}+9 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(259\) |
| risch | \(-\frac {i \left (60 a^{3} {\mathrm e}^{9 i \left (d x +c \right )}+135 a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+120 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+630 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-720 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-1680 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-640 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-120 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-630 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-1200 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-320 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-60 a^{3} {\mathrm e}^{i \left (d x +c \right )}-135 b^{2} a \,{\mathrm e}^{i \left (d x +c \right )}-240 a^{2} b -64 b^{3}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\) | \(308\) |
1/d*(a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-3*a^2*b *(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+3*a*b^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d *x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-b^3*(-8/15-1/5*sec(d*x+c) ^4-4/15*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.28 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.90 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (15 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 90 \, a b^{2} \cos \left (d x + c\right ) + 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, b^{3} + 8 \, {\left (15 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(15*(4*a^3 + 9*a*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*a ^3 + 9*a*b^2)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(15*a^2*b + 4* b^3)*cos(d*x + c)^4 + 90*a*b^2*cos(d*x + c) + 15*(4*a^3 + 9*a*b^2)*cos(d*x + c)^3 + 24*b^3 + 8*(15*a^2*b + 4*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*c os(d*x + c)^5)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.96 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} b^{3} - 45 \, a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*b^3 - 45*a*b^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (177) = 354\).
Time = 0.35 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.94 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1200 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 360 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 225 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(4*a^3 + 9*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3 + 9*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(60*a^3*tan(1/2*d*x + 1 /2*c)^9 - 360*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 225*a*b^2*tan(1/2*d*x + 1/2*c )^9 - 120*b^3*tan(1/2*d*x + 1/2*c)^9 - 120*a^3*tan(1/2*d*x + 1/2*c)^7 + 96 0*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 90*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 160*b^3 *tan(1/2*d*x + 1/2*c)^7 - 1200*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 464*b^3*tan( 1/2*d*x + 1/2*c)^5 + 120*a^3*tan(1/2*d*x + 1/2*c)^3 + 960*a^2*b*tan(1/2*d* x + 1/2*c)^3 + 90*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 160*b^3*tan(1/2*d*x + 1/2 *c)^3 - 60*a^3*tan(1/2*d*x + 1/2*c) - 360*a^2*b*tan(1/2*d*x + 1/2*c) - 225 *a*b^2*tan(1/2*d*x + 1/2*c) - 120*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 17.17 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.37 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^3+\frac {9\,a\,b^2}{4}\right )}{d}-\frac {\left (-a^3+6\,a^2\,b-\frac {15\,a\,b^2}{4}+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,a^3-16\,a^2\,b+\frac {3\,a\,b^2}{2}-\frac {8\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (20\,a^2\,b+\frac {116\,b^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,a^3-16\,a^2\,b-\frac {3\,a\,b^2}{2}-\frac {8\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^3+6\,a^2\,b+\frac {15\,a\,b^2}{4}+2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(atanh(tan(c/2 + (d*x)/2))*((9*a*b^2)/4 + a^3))/d - (tan(c/2 + (d*x)/2)^7* ((3*a*b^2)/2 - 16*a^2*b + 2*a^3 - (8*b^3)/3) - tan(c/2 + (d*x)/2)^3*((3*a* b^2)/2 + 16*a^2*b + 2*a^3 + (8*b^3)/3) + tan(c/2 + (d*x)/2)*((15*a*b^2)/4 + 6*a^2*b + a^3 + 2*b^3) + tan(c/2 + (d*x)/2)^5*(20*a^2*b + (116*b^3)/15) - tan(c/2 + (d*x)/2)^9*((15*a*b^2)/4 - 6*a^2*b + a^3 - 2*b^3))/(d*(5*tan(c /2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*ta n(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))