Integrand size = 32, antiderivative size = 103 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (3 B+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 (3 B+2 C) \tan (c+d x)}{3 d}+\frac {a^2 (3 B+2 C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d} \]
1/2*a^2*(3*B+2*C)*arctanh(sin(d*x+c))/d+2/3*a^2*(3*B+2*C)*tan(d*x+c)/d+1/6 *a^2*(3*B+2*C)*sec(d*x+c)*tan(d*x+c)/d+1/3*C*(a+a*sec(d*x+c))^2*tan(d*x+c) /d
Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left ((9 B+6 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (12 (B+C)+3 (B+2 C) \sec (c+d x)+2 C \tan ^2(c+d x)\right )\right )}{6 d} \]
(a^2*((9*B + 6*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(12*(B + C) + 3*(B + 2*C)*Sec[c + d*x] + 2*C*Tan[c + d*x]^2)))/(6*d)
Time = 0.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.89, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {3042, 4542, 27, 3042, 4275, 3042, 4254, 24, 4534, 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.
| \(\displaystyle \int (a \sec (c+d x)+a)^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4542 |
| \(\displaystyle \frac {\int a (3 B+2 C) \sec (c+d x) (\sec (c+d x) a+a)^2dx}{3 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \int \sec (c+d x) (\sec (c+d x) a+a)^2dx+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2dx+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (2 a^2 \int \sec ^2(c+d x)dx+\int \sec (c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (2 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx-\frac {2 a^2 \int 1d(-\tan (c+d x))}{d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+\frac {2 a^2 \tan (c+d x)}{d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (\frac {3}{2} a^2 \int \sec (c+d x)dx+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (\frac {3}{2} a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} (3 B+2 C) \left (\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\) |
(C*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + ((3*B + 2*C)*((3*a^2*ArcTa nh[Sin[c + d*x]])/(2*d) + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3
3.4.16.3.1 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[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_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, 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[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(b*(m + 1)) I nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 0.42 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.17
| method | result | size |
| parts | \(\frac {\left (B \,a^{2}+2 C \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (2 B \,a^{2}+C \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{2}}{d}-\frac {C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(120\) |
| derivativedivides | \(\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \tan \left (d x +c \right )+2 C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \tan \left (d x +c \right )}{d}\) | \(145\) |
| default | \(\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \tan \left (d x +c \right )+2 C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \tan \left (d x +c \right )}{d}\) | \(145\) |
| parallelrisch | \(\frac {\left (-\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (B +2 C \right ) \sin \left (2 d x +2 c \right )+\left (2 B +\frac {5 C}{3}\right ) \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right ) \left (B +\frac {3 C}{2}\right )\right ) a^{2}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(148\) |
| norman | \(\frac {\frac {8 a^{2} \left (3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {a^{2} \left (3 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{2} \left (5 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a^{2} \left (3 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(149\) |
| risch | \(-\frac {i a^{2} \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}+6 C \,{\mathrm e}^{5 i \left (d x +c \right )}-12 B \,{\mathrm e}^{4 i \left (d x +c \right )}-6 C \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B \,{\mathrm e}^{2 i \left (d x +c \right )}-24 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-6 C \,{\mathrm e}^{i \left (d x +c \right )}-12 B -10 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(214\) |
(B*a^2+2*C*a^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)) )+(2*B*a^2+C*a^2)/d*tan(d*x+c)+1/d*B*ln(sec(d*x+c)+tan(d*x+c))*a^2-C*a^2/d *(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (6 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
1/12*(3*(3*B + 2*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(3*B + 2* C)*a^2*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(6*B + 5*C)*a^2*cos(d* x + c)^2 + 3*(B + 2*C)*a^2*cos(d*x + c) + 2*C*a^2)*sin(d*x + c))/(d*cos(d* x + c)^3)
\[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int B \sec {\left (c + d x \right )}\, dx + \int 2 B \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(B*sec(c + d*x), x) + Integral(2*B*sec(c + d*x)**2, x) + Int egral(B*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**2, x) + Integral(2* C*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x))
Time = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.62 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 24 \, B a^{2} \tan \left (d x + c\right ) + 12 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*B*a^2*(2*sin(d*x + c)/ (sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 6* C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(s in(d*x + c) - 1)) + 12*B*a^2*log(sec(d*x + c) + tan(d*x + c)) + 24*B*a^2*t an(d*x + c) + 12*C*a^2*tan(d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.73 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (3 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(3*B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 16*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 18*C *a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 17.74 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.41 \[ \int (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,B}{2}+C\right )}{d}-\frac {\left (3\,B\,a^2+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,B\,a^2-\frac {16\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,B\,a^2+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]