Integrand size = 38, antiderivative size = 145 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (b B+a C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d} \]
3/8*(B*b+C*a)*arctanh(sin(d*x+c))/d+1/5*(5*B*a+4*C*b)*tan(d*x+c)/d+3/8*(B* b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/4*(B*b+C*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5 *b*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*B*a+4*C*b)*tan(d*x+c)^3/d
Time = 0.66 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 (b B+a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 (b B+a C) \sec (c+d x)+30 (b B+a C) \sec ^3(c+d x)+8 \left (15 (a B+b C)+5 (a B+2 b C) \tan ^2(c+d x)+3 b C \tan ^4(c+d x)\right )\right )}{120 d} \]
(45*(b*B + a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*(b*B + a*C)*Sec[c + d*x] + 30*(b*B + a*C)*Sec[c + d*x]^3 + 8*(15*(a*B + b*C) + 5*(a*B + 2*b *C)*Tan[c + d*x]^2 + 3*b*C*Tan[c + d*x]^4)))/(120*d)
Time = 0.81 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4560, 3042, 4485, 3042, 4274, 3042, 4254, 2009, 4255, 3042, 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.
\(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4560 |
\(\displaystyle \int \sec ^4(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {1}{5} \int \sec ^4(c+d x) (5 a B+4 b C+5 (b B+a C) \sec (c+d x))dx+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (5 a B+4 b C+5 (b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \int \sec ^5(c+d x)dx+(5 a B+4 b C) \int \sec ^4(c+d x)dx\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left ((5 a B+4 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+5 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {(5 a B+4 b C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{5} \left (5 (a C+b B) \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {(5 a B+4 b C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
(b*C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (-(((5*a*B + 4*b*C)*(-Tan[c + d* x] - Tan[c + d*x]^3/3))/d) + 5*(b*B + a*C)*((Sec[c + d*x]^3*Tan[c + d*x])/ (4*d) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d )))/4))/5
3.8.65.3.1 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_))*(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[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.89 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81
method | result | size |
parts | \(\frac {\left (B b +C a \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(117\) |
derivativedivides | \(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
default | \(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
norman | \(\frac {-\frac {4 \left (25 a B +29 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (8 a B -5 B b -5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (8 a B +5 B b +5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (32 a B -3 B b -3 C a +16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (32 a B +3 B b +3 C a +16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(222\) |
parallelrisch | \(\frac {-45 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+420 \left (B b +C a \right ) \sin \left (2 d x +2 c \right )+80 \left (5 a B +4 C b \right ) \sin \left (3 d x +3 c \right )+90 \left (B b +C a \right ) \sin \left (4 d x +4 c \right )+16 \left (5 a B +4 C b \right ) \sin \left (5 d x +5 c \right )+320 \sin \left (d x +c \right ) \left (a B +2 C b \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(230\) |
risch | \(-\frac {i \left (45 B b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a \,{\mathrm e}^{9 i \left (d x +c \right )}+210 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a \,{\mathrm e}^{7 i \left (d x +c \right )}-240 B a \,{\mathrm e}^{6 i \left (d x +c \right )}-560 a B \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-210 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a \,{\mathrm e}^{3 i \left (d x +c \right )}-400 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B b \,{\mathrm e}^{i \left (d x +c \right )}-45 C a \,{\mathrm e}^{i \left (d x +c \right )}-80 a B -64 C b \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(279\) |
(B*b+C*a)/d*(-(-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)))-C*b/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+ c)-a*B/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{4} + 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{2} + 24 \, C b + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(45*(C*a + B*b)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*(C*a + B*b )*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*B*a + 4*C*b)*cos(d*x + c)^4 + 45*(C*a + B*b)*cos(d*x + c)^3 + 8*(5*B*a + 4*C*b)*cos(d*x + c)^2 + 24*C*b + 30*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.23 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(3*tan(d*x + c)^5 + 1 0*tan(d*x + c)^3 + 15*tan(d*x + c))*C*b - 15*C*a*(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)) - 15*B*b*(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)))/d
Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (133) = 266\).
Time = 0.31 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.28 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(45*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*(C*a + B*b)* log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*B*a*tan(1/2*d*x + 1/2*c)^9 - 7 5*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*b*tan(1/2*d*x + 1/2*c)^9 + 120*C*b*tan (1/2*d*x + 1/2*c)^9 - 320*B*a*tan(1/2*d*x + 1/2*c)^7 + 30*C*a*tan(1/2*d*x + 1/2*c)^7 + 30*B*b*tan(1/2*d*x + 1/2*c)^7 - 160*C*b*tan(1/2*d*x + 1/2*c)^ 7 + 400*B*a*tan(1/2*d*x + 1/2*c)^5 + 464*C*b*tan(1/2*d*x + 1/2*c)^5 - 320* B*a*tan(1/2*d*x + 1/2*c)^3 - 30*C*a*tan(1/2*d*x + 1/2*c)^3 - 30*B*b*tan(1/ 2*d*x + 1/2*c)^3 - 160*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*B*a*tan(1/2*d*x + 1/2*c) + 75*C*a*tan(1/2*d*x + 1/2*c) + 75*B*b*tan(1/2*d*x + 1/2*c) + 120*C *b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 21.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.61 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,B\,b}{4}+\frac {3\,C\,a}{4}\right )}{d}-\frac {\left (2\,B\,a-\frac {5\,B\,b}{4}-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,b}{2}-\frac {16\,B\,a}{3}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,B\,a}{3}-\frac {B\,b}{2}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a+\frac {5\,B\,b}{4}+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(atanh(tan(c/2 + (d*x)/2))*((3*B*b)/4 + (3*C*a)/4))/d - (tan(c/2 + (d*x)/2 )*(2*B*a + (5*B*b)/4 + (5*C*a)/4 + 2*C*b) + tan(c/2 + (d*x)/2)^5*((20*B*a) /3 + (116*C*b)/15) + tan(c/2 + (d*x)/2)^9*(2*B*a - (5*B*b)/4 - (5*C*a)/4 + 2*C*b) - tan(c/2 + (d*x)/2)^3*((16*B*a)/3 + (B*b)/2 + (C*a)/2 + (8*C*b)/3 ) - tan(c/2 + (d*x)/2)^7*((16*B*a)/3 - (B*b)/2 - (C*a)/2 + (8*C*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*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))