Integrand size = 19, antiderivative size = 101 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \]
3/8*a*arctanh(sin(d*x+c))/d+a*tan(d*x+c)/d+3/8*a*sec(d*x+c)*tan(d*x+c)/d+1 /4*a*sec(d*x+c)^3*tan(d*x+c)/d+2/3*a*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a \left (45 \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (120+45 \sec (c+d x)+30 \sec ^3(c+d x)+80 \tan ^2(c+d x)+24 \tan ^4(c+d x)\right )\right )}{120 d} \]
(a*(45*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(120 + 45*Sec[c + d*x] + 30*Se c[c + d*x]^3 + 80*Tan[c + d*x]^2 + 24*Tan[c + d*x]^4)))/(120*d)
Time = 0.51 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 3227, 3042, 4254, 2009, 4255, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle a \int \sec ^6(c+d x)dx+a \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+a \int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \int \left (\tan ^4(c+d x)+2 \tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
-((a*(-Tan[c + d*x] - (2*Tan[c + d*x]^3)/3 - Tan[c + d*x]^5/5))/d) + a*((S ec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec [c + d*x]*Tan[c + d*x])/(2*d)))/4)
3.1.12.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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_)]*(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]
Time = 2.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(83\) |
| default | \(\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(83\) |
| parts | \(-\frac {a \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
| risch | \(-\frac {i a \left (45 \,{\mathrm e}^{9 i \left (d x +c \right )}+210 \,{\mathrm e}^{7 i \left (d x +c \right )}-640 \,{\mathrm e}^{4 i \left (d x +c \right )}-210 \,{\mathrm e}^{3 i \left (d x +c \right )}-320 \,{\mathrm e}^{2 i \left (d x +c \right )}-45 \,{\mathrm e}^{i \left (d x +c \right )}-64\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(128\) |
| norman | \(\frac {-\frac {13 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {137 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 d}-\frac {167 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 d}+\frac {17 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {3 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(170\) |
| parallelrisch | \(\frac {8 \left (\left (-\frac {45 \cos \left (d x +c \right )}{32}-\frac {45 \cos \left (3 d x +3 c \right )}{64}-\frac {9 \cos \left (5 d x +5 c \right )}{64}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\frac {45 \cos \left (d x +c \right )}{32}+\frac {45 \cos \left (3 d x +3 c \right )}{64}+\frac {9 \cos \left (5 d x +5 c \right )}{64}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\sin \left (3 d x +3 c \right )+\frac {9 \sin \left (4 d x +4 c \right )}{32}+\frac {\sin \left (5 d x +5 c \right )}{5}+2 \sin \left (d x +c \right )+\frac {21 \sin \left (2 d x +2 c \right )}{16}\right ) a}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(177\) |
1/d*(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)+t an(d*x+c)))-a*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {45 \, a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (64 \, a \cos \left (d x + c\right )^{4} + 45 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 30 \, a \cos \left (d x + c\right ) + 24 \, a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
1/240*(45*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*cos(d*x + c)^5*log (-sin(d*x + c) + 1) + 2*(64*a*cos(d*x + c)^4 + 45*a*cos(d*x + c)^3 + 32*a* cos(d*x + c)^2 + 30*a*cos(d*x + c) + 24*a)*sin(d*x + c))/(d*cos(d*x + c)^5 )
\[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=a \left (\int \cos {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a - 15 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a - 15* 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)))/d
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.23 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {45 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 130 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 190 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(45*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(45*a*tan(1/2*d*x + 1/2*c)^9 - 130*a*tan(1/2*d*x + 1/2* c)^7 + 464*a*tan(1/2*d*x + 1/2*c)^5 - 190*a*tan(1/2*d*x + 1/2*c)^3 + 195*a *tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 19.01 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int (a+a \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {13\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(3*a*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((13*a*tan(c/2 + (d*x)/2))/4 - (19 *a*tan(c/2 + (d*x)/2)^3)/6 + (116*a*tan(c/2 + (d*x)/2)^5)/15 - (13*a*tan(c /2 + (d*x)/2)^7)/6 + (3*a*tan(c/2 + (d*x)/2)^9)/4)/(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))