Integrand size = 39, antiderivative size = 127 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+4 B+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (3 A+2 (B+C)) \tan (c+d x)}{3 d}+\frac {a (4 A+4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (B+C) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/8*a*(4*A+4*B+3*C)*arctanh(sin(d*x+c))/d+1/3*a*(3*A+2*B+2*C)*tan(d*x+c)/d +1/8*a*(4*A+4*B+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*(B+C)*sec(d*x+c)^2*tan( d*x+c)/d+1/4*a*C*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.66 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (3 (4 A+4 B+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 (A+B+C)+3 (4 A+4 B+3 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 (B+C) \tan ^2(c+d x)\right )\right )}{24 d} \]
(a*(3*(4*A + 4*B + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(A + B + C) + 3*(4*A + 4*B + 3*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 + 8*(B + C)*Tan [c + d*x]^2)))/(24*d)
Time = 0.83 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4564, 3042, 4535, 3042, 4255, 3042, 4257, 4534, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4564 |
| \(\displaystyle \frac {1}{4} \int \sec ^2(c+d x) \left (4 a (B+C) \sec ^2(c+d x)+a (4 A+4 B+3 C) \sec (c+d x)+4 a A\right )dx+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+4 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a A\right )dx+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+4 B+3 C) \int \sec ^3(c+d x)dx+\int \sec ^2(c+d x) \left (4 a (B+C) \sec ^2(c+d x)+4 a A\right )dx\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+4 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a A\right )dx\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a A\right )dx+a (4 A+4 B+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a A\right )dx+a (4 A+4 B+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a (B+C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 a A\right )dx+a (4 A+4 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{4} \left (\frac {4}{3} a (3 A+2 (B+C)) \int \sec ^2(c+d x)dx+a (4 A+4 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 a (B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {4}{3} a (3 A+2 (B+C)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+a (4 A+4 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 a (B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (-\frac {4 a (3 A+2 (B+C)) \int 1d(-\tan (c+d x))}{3 d}+a (4 A+4 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 a (B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (a (4 A+4 B+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 a (3 A+2 (B+C)) \tan (c+d x)}{3 d}+\frac {4 a (B+C) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
(a*C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*a*(3*A + 2*(B + C))*Tan[c + d*x])/(3*d) + (4*a*(B + C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + a*(4*A + 4 *B + 3*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d) ))/4
3.5.9.3.1 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_)]*(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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-b)*C*Csc[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*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && !LtQ[n, -1]
Time = 0.64 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07
| method | result | size |
| parts | \(\frac {\left (a A +a B \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (a B +C a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}\) | \(136\) |
| derivativedivides | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \tan \left (d x +c \right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(174\) |
| default | \(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a A \tan \left (d x +c \right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(174\) |
| norman | \(\frac {-\frac {a \left (4 A +4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a \left (12 A +12 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (60 A +28 B +49 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {a \left (84 A +52 B +31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a \left (4 A +4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(181\) |
| parallelrisch | \(\frac {\left (-2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +B +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +B +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (2 A +\frac {8 C}{3}+\frac {8 B}{3}\right ) \sin \left (2 d x +2 c \right )+\left (A +B +\frac {3 C}{4}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {2 C}{3}+A +\frac {2 B}{3}\right ) \sin \left (4 d x +4 c \right )+\sin \left (d x +c \right ) \left (B +\frac {11 C}{4}+A \right )\right ) a}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(182\) |
| risch | \(-\frac {i a \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C \,{\mathrm e}^{7 i \left (d x +c \right )}-24 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-72 A \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-72 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-64 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-12 B \,{\mathrm e}^{i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}-24 A -16 B -16 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\) | \(380\) |
(A*a+B*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(B*a +C*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+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)))+a*A/d*tan(d*x+c)
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A + 2 \, B + 2 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 6 \, C a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/48*(3*(4*A + 4*B + 3*C)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 4*B + 3*C)*a*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(3*A + 2*B + 2 *C)*a*cos(d*x + c)^3 + 3*(4*A + 4*B + 3*C)*a*cos(d*x + c)^2 + 8*(B + C)*a* cos(d*x + c) + 6*C*a)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]
a*(Integral(A*sec(c + d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integ ral(B*sec(c + d*x)**3, x) + Integral(B*sec(c + d*x)**4, x) + Integral(C*se c(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))
Time = 0.21 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.72 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a \tan \left (d x + c\right )}{48 \, d} \]
integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*ta n(d*x + c))*C*a - 3*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)) - 12*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)) - 12*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*A*a*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (118) = 236\).
Time = 0.32 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.00 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A a + 4 \, B a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A a + 4 \, B a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 49 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 84 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 31 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 39 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(4*A*a + 4*B*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4* A*a + 4*B*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(12*A*a*tan(1/ 2*d*x + 1/2*c)^7 + 12*B*a*tan(1/2*d*x + 1/2*c)^7 + 9*C*a*tan(1/2*d*x + 1/2 *c)^7 - 60*A*a*tan(1/2*d*x + 1/2*c)^5 - 28*B*a*tan(1/2*d*x + 1/2*c)^5 - 49 *C*a*tan(1/2*d*x + 1/2*c)^5 + 84*A*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1 /2*d*x + 1/2*c)^3 + 31*C*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x + 1 /2*c) - 36*B*a*tan(1/2*d*x + 1/2*c) - 39*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 21.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.66 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (-A\,a-B\,a-\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (5\,A\,a+\frac {7\,B\,a}{3}+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-7\,A\,a-\frac {13\,B\,a}{3}-\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+3\,B\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+4\,B+3\,C\right )}{2\,\left (2\,A\,a+2\,B\,a+\frac {3\,C\,a}{2}\right )}\right )\,\left (4\,A+4\,B+3\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(3*A*a + 3*B*a + (13*C*a)/4) - tan(c/2 + (d*x)/2)^7*(A *a + B*a + (3*C*a)/4) - tan(c/2 + (d*x)/2)^3*(7*A*a + (13*B*a)/3 + (31*C*a )/12) + tan(c/2 + (d*x)/2)^5*(5*A*a + (7*B*a)/3 + (49*C*a)/12))/(d*(6*tan( c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c /2 + (d*x)/2)^8 + 1)) + (a*atanh((a*tan(c/2 + (d*x)/2)*(4*A + 4*B + 3*C))/ (2*(2*A*a + 2*B*a + (3*C*a)/2)))*(4*A + 4*B + 3*C))/(4*d)