Integrand size = 21, antiderivative size = 122 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {9 a^2 \tan (c+d x)}{5 d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {3 a^2 \tan ^3(c+d x)}{5 d} \]
3/4*a^2*arctanh(sin(d*x+c))/d+9/5*a^2*tan(d*x+c)/d+3/4*a^2*sec(d*x+c)*tan( d*x+c)/d+1/2*a^2*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a^2*sec(d*x+c)^4*tan(d*x+c) /d+3/5*a^2*tan(d*x+c)^3/d
Time = 0.63 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.56 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {a^2 \left (15 \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \sec (c+d x)+10 \sec ^3(c+d x)+4 \sec ^4(c+d x)+12 \left (3+\tan ^2(c+d x)\right )\right )\right )}{20 d} \]
(a^2*(15*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*Sec[c + d*x] + 10*Sec[c + d*x]^3 + 4*Sec[c + d*x]^4 + 12*(3 + Tan[c + d*x]^2))))/(20*d)
Time = 0.70 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4275, 3042, 4255, 3042, 4255, 3042, 4257, 4534, 3042, 4254, 2009}
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.
| \(\displaystyle \int \sec ^4(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle 2 a^2 \int \sec ^5(c+d x)dx+\int \sec ^4(c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+2 a^2 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {9}{5} a^2 \int \sec ^4(c+d x)dx+2 a^2 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9}{5} a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+2 a^2 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {9 a^2 \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{5 d}+2 a^2 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 a^2 \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {9 a^2 \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}\) |
(a^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) - (9*a^2*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/(5*d) + 2*a^2*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTan h[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
3.1.10.3.1 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_))^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]
Time = 0.98 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+2 a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(111\) |
| default | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+2 a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(111\) |
| parts | \(-\frac {a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(116\) |
| risch | \(-\frac {i a^{2} \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}+70 \,{\mathrm e}^{7 i \left (d x +c \right )}-40 \,{\mathrm e}^{6 i \left (d x +c \right )}-200 \,{\mathrm e}^{4 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}-120 \,{\mathrm e}^{2 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-24\right )}{10 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\) | \(145\) |
| norman | \(\frac {-\frac {13 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {9 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {72 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {7 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {3 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {3 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(152\) |
| parallelrisch | \(\frac {6 \left (\left (-\frac {5 \cos \left (d x +c \right )}{4}-\frac {5 \cos \left (3 d x +3 c \right )}{8}-\frac {\cos \left (5 d x +5 c \right )}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\frac {5 \cos \left (d x +c \right )}{4}+\frac {5 \cos \left (3 d x +3 c \right )}{8}+\frac {\cos \left (5 d x +5 c \right )}{8}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (3 d x +3 c \right )+\frac {\sin \left (4 d x +4 c \right )}{4}+\frac {\sin \left (5 d x +5 c \right )}{5}+\frac {4 \sin \left (d x +c \right )}{3}+\frac {7 \sin \left (2 d x +2 c \right )}{6}\right ) a^{2}}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(179\) |
1/d*(-a^2*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+2*a^2*(-(- 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^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, a^{2} \cos \left (d x + c\right )^{4} + 15 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{40 \, d \cos \left (d x + c\right )^{5}} \]
1/40*(15*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*a^2*cos(d*x + c)^5* log(-sin(d*x + c) + 1) + 2*(24*a^2*cos(d*x + c)^4 + 15*a^2*cos(d*x + c)^3 + 12*a^2*cos(d*x + c)^2 + 10*a^2*cos(d*x + c) + 4*a^2)*sin(d*x + c))/(d*co s(d*x + c)^5)
\[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(sec(c + d*x)**4, x) + Integral(2*sec(c + d*x)**5, x) + Inte gral(sec(c + d*x)**6, x))
Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.09 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 15 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]
1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + 40 *(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 - 15*a^2*(2*(3*sin(d*x + c)^3 - 5*s in(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
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.13 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 65 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{20 \, d} \]
1/20*(15*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^2*log(abs(tan(1/2*d *x + 1/2*c) - 1)) - 2*(15*a^2*tan(1/2*d*x + 1/2*c)^9 - 70*a^2*tan(1/2*d*x + 1/2*c)^7 + 144*a^2*tan(1/2*d*x + 1/2*c)^5 - 90*a^2*tan(1/2*d*x + 1/2*c)^ 3 + 65*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 17.89 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.39 \[ \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx=\frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}-9\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(3*a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d) - ((72*a^2*tan(c/2 + (d*x)/2)^5)/5 - 9*a^2*tan(c/2 + (d*x)/2)^3 - 7*a^2*tan(c/2 + (d*x)/2)^7 + (3*a^2*tan(c/ 2 + (d*x)/2)^9)/2 + (13*a^2*tan(c/2 + (d*x)/2))/2)/(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))