Integrand size = 19, antiderivative size = 101 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^5(c+d x)}{5 d} \] Output:
3/8*b*arctanh(sin(d*x+c))/d+a*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+2/3*a*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \left (\tan (c+d x)+\frac {2}{3} \tan ^3(c+d x)+\frac {1}{5} \tan ^5(c+d x)\right )}{d} \] Input:
Integrate[(a + b*Cos[c + d*x])*Sec[c + d*x]^6,x]
Output:
(3*b*ArcTanh[Sin[c + d*x]])/(8*d) + (3*b*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*(Tan[c + d*x] + (2*Tan[c + d* x]^3)/3 + Tan[c + d*x]^5/5))/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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle a \int \sec ^6(c+d x)dx+b \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^6dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \int \left (\tan ^4(c+d x)+2 \tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{5} \tan ^5(c+d x)-\frac {2}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
Input:
Int[(a + b*Cos[c + d*x])*Sec[c + d*x]^6,x]
Output:
-((a*(-Tan[c + d*x] - (2*Tan[c + d*x]^3)/3 - Tan[c + d*x]^5/5))/d) + b*((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)
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 = 4.82 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(83\) |
| default | \(\frac {-a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(83\) |
| parts | \(-\frac {a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(85\) |
| risch | \(-\frac {i \left (45 b \,{\mathrm e}^{9 i \left (d x +c \right )}+210 b \,{\mathrm e}^{7 i \left (d x +c \right )}-640 \,{\mathrm e}^{4 i \left (d x +c \right )} a -210 b \,{\mathrm e}^{3 i \left (d x +c \right )}-320 a \,{\mathrm e}^{2 i \left (d x +c \right )}-45 \,{\mathrm e}^{i \left (d x +c \right )} b -64 a \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(135\) |
| parallelrisch | \(\frac {-225 b \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+225 b \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+210 b \sin \left (2 d x +2 c \right )+160 a \sin \left (3 d x +3 c \right )+45 b \sin \left (4 d x +4 c \right )+32 a \left (\sin \left (5 d x +5 c \right )+10 \sin \left (d x +c \right )\right )}{60 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(183\) |
| norman | \(\frac {\frac {\left (8 a -9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}-\frac {\left (8 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {\left (8 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (8 a +9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}-\frac {\left (152 a -15 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d}-\frac {\left (152 a +15 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(206\) |
Input:
int((a+cos(d*x+c)*b)*sec(d*x+c)^6,x,method=_RETURNVERBOSE)
Output:
1/d*(-a*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+b*(-(-1/4*se c(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 = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {45 \, b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (64 \, a \cos \left (d x + c\right )^{4} + 45 \, b \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 30 \, b \cos \left (d x + c\right ) + 24 \, a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="fricas")
Output:
1/240*(45*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*b*cos(d*x + c)^5*log (-sin(d*x + c) + 1) + 2*(64*a*cos(d*x + c)^4 + 45*b*cos(d*x + c)^3 + 32*a* cos(d*x + c)^2 + 30*b*cos(d*x + c) + 24*a)*sin(d*x + c))/(d*cos(d*x + c)^5 )
\[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)**6,x)
Output:
Integral((a + b*cos(c + d*x))*sec(c + d*x)**6, x)
Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a - 15 \, b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="maxima")
Output:
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a - 15* 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
Time = 0.54 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.76 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {45 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="giac")
Output:
1/120*(45*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*a*tan(1/2*d*x + 1/2*c)^9 - 75*b*tan(1/2*d*x + 1/2* c)^9 - 160*a*tan(1/2*d*x + 1/2*c)^7 + 30*b*tan(1/2*d*x + 1/2*c)^7 + 464*a* tan(1/2*d*x + 1/2*c)^5 - 160*a*tan(1/2*d*x + 1/2*c)^3 - 30*b*tan(1/2*d*x + 1/2*c)^3 + 120*a*tan(1/2*d*x + 1/2*c) + 75*b*tan(1/2*d*x + 1/2*c))/(tan(1 /2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 42.17 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.78 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\left (2\,a-\frac {5\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {b}{2}-\frac {8\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}+\left (-\frac {8\,a}{3}-\frac {b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a+\frac {5\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + b*cos(c + d*x))/cos(c + d*x)^6,x)
Output:
(3*b*atanh(tan(c/2 + (d*x)/2)))/(4*d) - (tan(c/2 + (d*x)/2)*(2*a + (5*b)/4 ) - tan(c/2 + (d*x)/2)^3*((8*a)/3 + b/2) + tan(c/2 + (d*x)/2)^9*(2*a - (5* b)/4) - tan(c/2 + (d*x)/2)^7*((8*a)/3 - b/2) + (116*a*tan(c/2 + (d*x)/2)^5 )/15)/(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))
Time = 0.16 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.54 \[ \int (a+b \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +64 \sin \left (d x +c \right )^{5} a -160 \sin \left (d x +c \right )^{3} a +120 \sin \left (d x +c \right ) a}{120 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a+b*cos(d*x+c))*sec(d*x+c)^6,x)
Output:
( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 90*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 45*cos(c + d*x)*log( tan((c + d*x)/2) - 1)*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 45*cos(c + d*x)*sin(c + d* x)**3*b + 75*cos(c + d*x)*sin(c + d*x)*b + 64*sin(c + d*x)**5*a - 160*sin( c + d*x)**3*a + 120*sin(c + d*x)*a)/(120*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))