Integrand size = 31, antiderivative size = 140 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {b (5 A+4 C) \tan ^3(c+d x)}{15 d} \] Output:
1/8*a*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/5*b*(5*A+4*C)*tan(d*x+c)/d+1/8*a*( 4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/4*a*C*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*b*(5*A+4*C)*tan(d*x+c)^3/d
Time = 0.60 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 a (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 a (4 A+3 C) \sec (c+d x)+30 a C \sec ^3(c+d x)+8 b \left (15 (A+C)+5 (A+2 C) \tan ^2(c+d x)+3 C \tan ^4(c+d x)\right )\right )}{120 d} \] Input:
Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]
Output:
(15*a*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*a*(4*A + 3*C)*S ec[c + d*x] + 30*a*C*Sec[c + d*x]^3 + 8*b*(15*(A + C) + 5*(A + 2*C)*Tan[c + d*x]^2 + 3*C*Tan[c + d*x]^4)))/(120*d)
Time = 0.79 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4565, 3042, 4535, 3042, 4254, 2009, 4534, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4565 |
| \(\displaystyle \frac {1}{5} \int \sec ^3(c+d x) \left (5 a C \sec ^2(c+d x)+b (5 A+4 C) \sec (c+d x)+5 a A\right )dx+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a A\right )dx+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\int \sec ^3(c+d x) \left (5 a C \sec ^2(c+d x)+5 a A\right )dx+b (5 A+4 C) \int \sec ^4(c+d x)dx\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx+b (5 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx-\frac {b (5 A+4 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a A\right )dx-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (4 A+3 C) \int \sec ^3(c+d x)dx+\frac {5 a C \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (4 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {5 a C \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (4 A+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {5 a C \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (4 A+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {5 a C \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (4 A+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {5 a C \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {b (5 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
Input:
Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + C*Sec[c + d*x]^2),x]
Output:
(b*C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((5*a*C*Sec[c + d*x]^3*Tan[c + d *x])/(4*d) + (5*a*(4*A + 3*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x] *Tan[c + d*x])/(2*d)))/4 - (b*(5*A + 4*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/ 3))/d)/5
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_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*C sc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 2))), x] + Simp[1/(n + 2) Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + b*(C*(n + 1) + A*(n + 2))*C sc[e + f*x] + a*C*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && !LtQ[n, -1]
Time = 6.73 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(141\) |
| default | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(141\) |
| parts | \(-\frac {A b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(149\) |
| parallelrisch | \(\frac {-3 \left (A +\frac {3 C}{4}\right ) a \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (A +\frac {3 C}{4}\right ) a \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 a \left (A +\frac {7 C}{4}\right ) \sin \left (2 d x +2 c \right )+20 \left (A +\frac {4 C}{5}\right ) b \sin \left (3 d x +3 c \right )+6 \left (A +\frac {3 C}{4}\right ) a \sin \left (4 d x +4 c \right )+16 \left (\left (\frac {A}{4}+\frac {C}{5}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +2 C \right )\right ) b}{6 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(220\) |
| norman | \(\frac {\frac {\left (4 a A -8 A b +5 C a -8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (4 a A +8 A b +5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a A -32 A b +3 C a -16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (12 a A +32 A b +3 C a +16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {4 b \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(223\) |
| risch | \(-\frac {i \left (60 a A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a \,{\mathrm e}^{9 i \left (d x +c \right )}+120 A a \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A b \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-120 a A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A a \,{\mathrm e}^{i \left (d x +c \right )}-45 C a \,{\mathrm e}^{i \left (d x +c \right )}-80 A b -64 C b \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\) | \(279\) |
Input:
int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBO SE)
Output:
1/d*(a*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*a*(-( -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)) )-A*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-C*b*(-8/15-1/5*sec(d*x+c)^4-4/15* sec(d*x+c)^2)*tan(d*x+c))
Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 30 \, C a \cos \left (d x + c\right ) + 24 \, C b\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="f ricas")
Output:
1/240*(15*(4*A + 3*C)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3 *C)*a*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*A + 4*C)*b*cos(d*x + c)^4 + 15*(4*A + 3*C)*a*cos(d*x + c)^3 + 8*(5*A + 4*C)*b*cos(d*x + c)^2 + 30*C*a*cos(d*x + c) + 24*C*b)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))*sec(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="m axima")
Output:
1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b + 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)) - 60*A*a*(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. 334 vs. \(2 (128) = 256\).
Time = 0.26 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.39 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x, algorithm="g iac")
Output:
1/120*(15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(60*A*a*tan(1/2*d*x + 1/2*c )^9 + 75*C*a*tan(1/2*d*x + 1/2*c)^9 - 120*A*b*tan(1/2*d*x + 1/2*c)^9 - 120 *C*b*tan(1/2*d*x + 1/2*c)^9 - 120*A*a*tan(1/2*d*x + 1/2*c)^7 - 30*C*a*tan( 1/2*d*x + 1/2*c)^7 + 320*A*b*tan(1/2*d*x + 1/2*c)^7 + 160*C*b*tan(1/2*d*x + 1/2*c)^7 - 400*A*b*tan(1/2*d*x + 1/2*c)^5 - 464*C*b*tan(1/2*d*x + 1/2*c) ^5 + 120*A*a*tan(1/2*d*x + 1/2*c)^3 + 30*C*a*tan(1/2*d*x + 1/2*c)^3 + 320* A*b*tan(1/2*d*x + 1/2*c)^3 + 160*C*b*tan(1/2*d*x + 1/2*c)^3 - 60*A*a*tan(1 /2*d*x + 1/2*c) - 75*C*a*tan(1/2*d*x + 1/2*c) - 120*A*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 = 15.50 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.66 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {\left (2\,A\,b-A\,a-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,a-\frac {16\,A\,b}{3}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,A\,a-\frac {16\,A\,b}{3}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)))/cos(c + d*x)^3,x)
Output:
(a*atanh(tan(c/2 + (d*x)/2))*(4*A + 3*C))/(4*d) - (tan(c/2 + (d*x)/2)*(A*a + 2*A*b + (5*C*a)/4 + 2*C*b) + tan(c/2 + (d*x)/2)^5*((20*A*b)/3 + (116*C* b)/15) - tan(c/2 + (d*x)/2)^9*(A*a - 2*A*b + (5*C*a)/4 - 2*C*b) - tan(c/2 + (d*x)/2)^3*(2*A*a + (16*A*b)/3 + (C*a)/2 + (8*C*b)/3) + tan(c/2 + (d*x)/ 2)^7*(2*A*a - (16*A*b)/3 + (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))
Time = 0.16 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.63 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),x)
Output:
( - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 45*co s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*c + 120*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 90*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a*c - 60*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*a**2 - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 + 45*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*c - 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2 + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*c - 60*cos(c + d*x)*sin(c + d*x)**3*a**2 - 45*cos(c + d*x)*sin(c + d*x)**3*a*c + 60*cos(c + d*x)*sin(c + d*x)*a**2 + 75*cos(c + d*x)*sin(c + d*x)*a*c + 80*sin(c + d*x)**5*a*b + 64*sin(c + d*x)**5*b*c - 200*sin(c + d*x)**3*a*b - 160*sin(c + d*x)**3*b* c + 120*sin(c + d*x)*a*b + 120*sin(c + d*x)*b*c)/(120*cos(c + d*x)*d*(sin( c + d*x)**4 - 2*sin(c + d*x)**2 + 1))