Integrand size = 21, antiderivative size = 135 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac {\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}+\frac {2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{15 d}+\frac {2 a \left (4 a^2-3 b^2\right ) \tan (c+d x)}{15 d} \]
2/15*b*(2*a^2-b^2)*sec(d*x+c)/d+1/5*sec(d*x+c)^5*(b+a*sin(d*x+c))*(a+b*sin (d*x+c))^2/d+2/15*sec(d*x+c)^3*(a+b*sin(d*x+c))*(a*b+(2*a^2-b^2)*sin(d*x+c ))/d+2/15*a*(4*a^2-3*b^2)*tan(d*x+c)/d
Time = 0.94 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.41 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\sec ^5(c+d x) \left (1152 a^2 b+64 b^3+\left (-270 a^2 b+110 b^3\right ) \cos (c+d x)-320 b^3 \cos (2 (c+d x))-135 a^2 b \cos (3 (c+d x))+55 b^3 \cos (3 (c+d x))-27 a^2 b \cos (5 (c+d x))+11 b^3 \cos (5 (c+d x))+640 a^3 \sin (c+d x)+960 a b^2 \sin (c+d x)+320 a^3 \sin (3 (c+d x))-240 a b^2 \sin (3 (c+d x))+64 a^3 \sin (5 (c+d x))-48 a b^2 \sin (5 (c+d x))\right )}{1920 d} \]
(Sec[c + d*x]^5*(1152*a^2*b + 64*b^3 + (-270*a^2*b + 110*b^3)*Cos[c + d*x] - 320*b^3*Cos[2*(c + d*x)] - 135*a^2*b*Cos[3*(c + d*x)] + 55*b^3*Cos[3*(c + d*x)] - 27*a^2*b*Cos[5*(c + d*x)] + 11*b^3*Cos[5*(c + d*x)] + 640*a^3*S in[c + d*x] + 960*a*b^2*Sin[c + d*x] + 320*a^3*Sin[3*(c + d*x)] - 240*a*b^ 2*Sin[3*(c + d*x)] + 64*a^3*Sin[5*(c + d*x)] - 48*a*b^2*Sin[5*(c + d*x)])) /(1920*d)
Time = 0.66 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3170, 27, 3042, 3340, 25, 3042, 3148, 3042, 4254, 24}
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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)^6}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle \frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}-\frac {1}{5} \int -2 \sec ^4(c+d x) (a+b \sin (c+d x)) \left (2 a^2+b \sin (c+d x) a-b^2\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{5} \int \sec ^4(c+d x) (a+b \sin (c+d x)) \left (2 a^2+b \sin (c+d x) a-b^2\right )dx+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \int \frac {(a+b \sin (c+d x)) \left (2 a^2+b \sin (c+d x) a-b^2\right )}{\cos (c+d x)^4}dx+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3340 |
| \(\displaystyle \frac {2}{5} \left (\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}-\frac {1}{3} \int -\sec ^2(c+d x) \left (a \left (4 a^2-3 b^2\right )+b \left (2 a^2-b^2\right ) \sin (c+d x)\right )dx\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{3} \int \sec ^2(c+d x) \left (a \left (4 a^2-3 b^2\right )+b \left (2 a^2-b^2\right ) \sin (c+d x)\right )dx+\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{3} \int \frac {a \left (4 a^2-3 b^2\right )+b \left (2 a^2-b^2\right ) \sin (c+d x)}{\cos (c+d x)^2}dx+\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (a \left (4 a^2-3 b^2\right ) \int \sec ^2(c+d x)dx+\frac {b \left (2 a^2-b^2\right ) \sec (c+d x)}{d}\right )+\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (a \left (4 a^2-3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {b \left (2 a^2-b^2\right ) \sec (c+d x)}{d}\right )+\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (\frac {b \left (2 a^2-b^2\right ) \sec (c+d x)}{d}-\frac {a \left (4 a^2-3 b^2\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2}{5} \left (\frac {\sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{3 d}+\frac {1}{3} \left (\frac {a \left (4 a^2-3 b^2\right ) \tan (c+d x)}{d}+\frac {b \left (2 a^2-b^2\right ) \sec (c+d x)}{d}\right )\right )+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d}\) |
(Sec[c + d*x]^5*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d) + (2*(( Sec[c + d*x]^3*(a + b*Sin[c + d*x])*(a*b + (2*a^2 - b^2)*Sin[c + d*x]))/(3 *d) + ((b*(2*a^2 - b^2)*Sec[c + d*x])/d + (a*(4*a^2 - 3*b^2)*Tan[c + d*x]) /d)/3))/5
3.5.10.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x ])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g }, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* p] || IntegerQ[m])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(g* Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Si n[e + f*x])^(m - 1)*Simp[a*c*(p + 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ [m, 0] && LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])
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]
Result contains complex when optimal does not.
Time = 1.85 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24
| method | result | size |
| risch | \(-\frac {4 \left (45 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+10 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-40 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-15 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-20 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+15 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+10 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-4 i a^{3}+3 i a \,b^{2}\right )}{15 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{5}}\) | \(168\) |
| derivativedivides | \(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{5 \cos \left (d x +c \right )^{5}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )}{d}\) | \(173\) |
| default | \(\frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{5 \cos \left (d x +c \right )^{5}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )}{d}\) | \(173\) |
| parallelrisch | \(\frac {-2 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}-6 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b +\frac {8 \left (a^{3}-3 a \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+\frac {4 \left (-29 a^{3}-12 a \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15}+\frac {4 \left (-9 a^{2} b -b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {8 \left (a^{3}-3 a \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}}{3}-2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {6 a^{2} b}{5}+\frac {4 b^{3}}{15}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(219\) |
| norman | \(\frac {-\frac {18 a^{2} b -4 b^{3}}{15 d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{2} b \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (9 a^{2} b +2 b^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (27 a^{2} b +4 b^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (45 a^{2} b +20 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (117 a^{2} b +34 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (279 a^{2} b +88 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a \left (5 a^{2}+12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (5 a^{2}+12 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (43 a^{2}+204 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a \left (43 a^{2}+204 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a \left (109 a^{2}+312 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 a \left (109 a^{2}+312 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 b \left (63 a^{2}+26 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\) | \(442\) |
-4/15*(45*I*a*b^2*exp(6*I*(d*x+c))+10*b^3*exp(7*I*(d*x+c))-40*I*a^3*exp(4* I*(d*x+c))-15*I*a*b^2*exp(4*I*(d*x+c))-72*a^2*b*exp(5*I*(d*x+c))-4*b^3*exp (5*I*(d*x+c))-20*I*a^3*exp(2*I*(d*x+c))+15*I*a*b^2*exp(2*I*(d*x+c))+10*b^3 *exp(3*I*(d*x+c))-4*I*a^3+3*I*a*b^2)/d/(1+exp(2*I*(d*x+c)))^5
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {5 \, b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 3 \, b^{3} - {\left (2 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{3} + 9 \, a b^{2} + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]
-1/15*(5*b^3*cos(d*x + c)^2 - 9*a^2*b - 3*b^3 - (2*(4*a^3 - 3*a*b^2)*cos(d *x + c)^4 + 3*a^3 + 9*a*b^2 + (4*a^3 - 3*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.78 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}} + \frac {9 \, a^{2} b}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]
1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 3*(3* tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a*b^2 - (5*cos(d*x + c)^2 - 3)*b^3/cos( d*x + c)^5 + 9*a^2*b/cos(d*x + c)^5)/d
Time = 0.35 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.80 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 58 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b - 2 \, b^{3}\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} d} \]
-2/15*(15*a^3*tan(1/2*d*x + 1/2*c)^9 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 2 0*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 30*b^3*ta n(1/2*d*x + 1/2*c)^6 + 58*a^3*tan(1/2*d*x + 1/2*c)^5 + 24*a*b^2*tan(1/2*d* x + 1/2*c)^5 + 90*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 10*b^3*tan(1/2*d*x + 1/2* c)^4 - 20*a^3*tan(1/2*d*x + 1/2*c)^3 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 1 0*b^3*tan(1/2*d*x + 1/2*c)^2 + 15*a^3*tan(1/2*d*x + 1/2*c) + 9*a^2*b - 2*b ^3)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*d)
Time = 4.98 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88 \[ \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\cos \left (c+d\,x\right )}^4\,\left (\frac {8\,a^3\,\sin \left (c+d\,x\right )}{15}-\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{5}\right )-{\cos \left (c+d\,x\right )}^2\,\left (-\frac {4\,\sin \left (c+d\,x\right )\,a^3}{15}+\frac {\sin \left (c+d\,x\right )\,a\,b^2}{5}+\frac {b^3}{3}\right )+\frac {3\,a^2\,b}{5}+\frac {a^3\,\sin \left (c+d\,x\right )}{5}+\frac {b^3}{5}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{5}}{d\,{\cos \left (c+d\,x\right )}^5} \]