Integrand size = 29, antiderivative size = 97 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x)}{d}+\frac {(4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {B \tan ^3(c+d x)}{3 d} \]
1/8*(4*A+3*C)*arctanh(sin(d*x+c))/d+B*tan(d*x+c)/d+1/8*(4*A+3*C)*sec(d*x+c )*tan(d*x+c)/d+1/4*C*sec(d*x+c)^3*tan(d*x+c)/d+1/3*B*tan(d*x+c)^3/d
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 (4 A+3 C) \sec (c+d x)+6 C \sec ^3(c+d x)+8 B \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \]
(3*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(4*A + 3*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 + 8*B*(3 + Tan[c + d*x]^2)))/(24*d)
Time = 0.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {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.
| \(\displaystyle \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \int \sec ^3(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {B \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \int \sec ^3(c+d x)dx-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} (4 A+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {B \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {C \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
(C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*A + 3*C)*(ArcTanh[Sin[c + d*x] ]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4 - (B*(-Tan[c + d*x] - Tan[ c + d*x]^3/3))/d
3.1.55.3.1 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]
Time = 0.39 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(106\) |
| default | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(106\) |
| parts | \(\frac {A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(111\) |
| parallelrisch | \(\frac {-48 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+48 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (24 A +18 C \right ) \sin \left (3 d x +3 c \right )+64 B \sin \left (2 d x +2 c \right )+16 B \sin \left (4 d x +4 c \right )+24 \left (A +\frac {11 C}{4}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(167\) |
| norman | \(\frac {\frac {\left (4 A +5 C -8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (4 A +5 C +8 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A -9 C -40 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (12 A -9 C +40 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(169\) |
| risch | \(-\frac {i \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C \,{\mathrm e}^{7 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}-16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(221\) |
1/d*(A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*(-2/3-1 /3*sec(d*x+c)^2)*tan(d*x+c)+C*(-(-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))))
Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.21 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, C\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(3*(4*A + 3*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*C)*c os(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*B*cos(d*x + c)^3 + 3*(4*A + 3 *C)*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*C)*sin(d*x + c))/(d*cos(d*x + c) ^4)
\[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.43 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*C*(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*(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. 230 vs. \(2 (89) = 178\).
Time = 0.31 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.37 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(3*(4*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A + 3*C)*log (abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(12*A*tan(1/2*d*x + 1/2*c)^7 - 24*B*ta n(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 12*A*tan(1/2*d*x + 1/ 2*c)^5 + 40*B*tan(1/2*d*x + 1/2*c)^5 + 9*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*t an(1/2*d*x + 1/2*c)^3 - 40*B*tan(1/2*d*x + 1/2*c)^3 + 9*C*tan(1/2*d*x + 1/ 2*c)^3 + 12*A*tan(1/2*d*x + 1/2*c) + 24*B*tan(1/2*d*x + 1/2*c) + 15*C*tan( 1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 17.94 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.65 \[ \int \sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{4}\right )}{d}+\frac {\left (A-2\,B+\frac {5\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,B}{3}-A+\frac {3\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C}{4}-\frac {10\,B}{3}-A\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A+2\,B+\frac {5\,C}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(atanh(tan(c/2 + (d*x)/2))*(A + (3*C)/4))/d + (tan(c/2 + (d*x)/2)*(A + 2*B + (5*C)/4) + tan(c/2 + (d*x)/2)^7*(A - 2*B + (5*C)/4) - tan(c/2 + (d*x)/2 )^3*(A + (10*B)/3 - (3*C)/4) + tan(c/2 + (d*x)/2)^5*((10*B)/3 - A + (3*C)/ 4))/(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))