Integrand size = 28, antiderivative size = 85 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {C \tan (c+d x)}{d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {C \tan ^3(c+d x)}{3 d} \]
3/8*B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d+3/8*B*sec(d*x+c)*tan(d*x+c)/d+1 /4*B*sec(d*x+c)^3*tan(d*x+c)/d+1/3*C*tan(d*x+c)^3/d
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {9 B \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 B \sec (c+d x)+6 B \sec ^3(c+d x)+8 C \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \]
(9*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(9*B*Sec[c + d*x] + 6*B*Sec[c + d*x]^3 + 8*C*(3 + Tan[c + d*x]^2)))/(24*d)
Time = 0.56 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3489, 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.
| \(\displaystyle \int \sec ^6(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3489 |
| \(\displaystyle \int \sec ^5(c+d x) (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle B \int \sec ^5(c+d x)dx+C \int \sec ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+C \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {C \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle B \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle B \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle B \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle B \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {C \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
-((C*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d) + B*((Sec[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)
3.3.25.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b Int[(b*Sin[e + f* x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, 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 = 4.54 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(73\) |
| default | \(\frac {-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(73\) |
| parts | \(\frac {B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(75\) |
| risch | \(-\frac {i \left (9 B \,{\mathrm e}^{7 i \left (d x +c \right )}+33 B \,{\mathrm e}^{5 i \left (d x +c \right )}-48 C \,{\mathrm e}^{4 i \left (d x +c \right )}-33 B \,{\mathrm e}^{3 i \left (d x +c \right )}-64 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 B \,{\mathrm e}^{i \left (d x +c \right )}-16 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(135\) |
| parallelrisch | \(\frac {-18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9 B \sin \left (3 d x +3 c \right )+33 B \sin \left (d x +c \right )+8 \sin \left (4 d x +4 c \right ) C +32 \sin \left (2 d x +2 c \right ) C}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(150\) |
| norman | \(\frac {-\frac {8 C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (B -8 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (B +8 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {2 \left (3 B -2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (3 B +2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (5 B -8 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(219\) |
1/d*(-C*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*(-(-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 = 99, normalized size of antiderivative = 1.16 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {9 \, B \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, B \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, C \cos \left (d x + c\right )^{3} + 9 \, B \cos \left (d x + c\right )^{2} + 8 \, C \cos \left (d x + c\right ) + 6 \, B\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(9*B*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*B*cos(d*x + c)^4*log(-s in(d*x + c) + 1) + 2*(16*C*cos(d*x + c)^3 + 9*B*cos(d*x + c)^2 + 8*C*cos(d *x + c) + 6*B)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C - 3 \, B {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C - 3*B*(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
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (77) = 154\).
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {9 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(9*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*B*log(abs(tan(1/2*d*x + 1 /2*c) - 1)) + 2*(15*B*tan(1/2*d*x + 1/2*c)^7 - 24*C*tan(1/2*d*x + 1/2*c)^7 + 9*B*tan(1/2*d*x + 1/2*c)^5 + 40*C*tan(1/2*d*x + 1/2*c)^5 + 9*B*tan(1/2* d*x + 1/2*c)^3 - 40*C*tan(1/2*d*x + 1/2*c)^3 + 15*B*tan(1/2*d*x + 1/2*c) + 24*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 3.92 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.76 \[ \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (\frac {5\,B}{4}-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,B}{4}+\frac {10\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,B}{4}-\frac {10\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)^7*((5*B)/4 - 2*C) + tan(c/2 + (d*x)/2)^3*((3*B)/4 - (1 0*C)/3) + tan(c/2 + (d*x)/2)^5*((3*B)/4 + (10*C)/3) + tan(c/2 + (d*x)/2)*( (5*B)/4 + 2*C))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*ta n(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*B*atanh(tan(c/2 + (d* x)/2)))/(4*d)