Integrand size = 19, antiderivative size = 85 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \tan (c+d x)}{d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b \tan ^3(c+d x)}{3 d} \] Output:
3/8*a*arctanh(sin(d*x+c))/d+b*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+1/3*b*tan(d*x+c)^3/d
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {9 a \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 a \sec (c+d x)+6 a \sec ^3(c+d x)+8 b \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
(9*a*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(9*a*Sec[c + d*x] + 6*a*Sec[c + d*x]^3 + 8*b*(3 + Tan[c + d*x]^2)))/(24*d)
Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle a \int \sec ^5(c+d x)dx+b \int \sec ^4(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {b \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
Input:
Int[(a + b*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
-((b*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d) + a*((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)
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 = 4.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(73\) |
| default | \(\frac {a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(73\) |
| parts | \(\frac {a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(75\) |
| risch | \(-\frac {i \left (9 a \,{\mathrm e}^{7 i \left (d x +c \right )}+33 a \,{\mathrm e}^{5 i \left (d x +c \right )}-48 b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{3 i \left (d x +c \right )} a -64 b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a \,{\mathrm e}^{i \left (d x +c \right )}-16 b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(135\) |
| parallelrisch | \(\frac {-18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9 a \sin \left (3 d x +3 c \right )+33 a \sin \left (d x +c \right )+8 b \sin \left (4 d x +4 c \right )+32 b \sin \left (2 d x +2 c \right )}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(150\) |
| norman | \(\frac {\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}+\frac {2 \left (3 a -2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (3 a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {\left (5 a -8 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {\left (5 a +8 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(177\) |
Input:
int((a+cos(d*x+c)*b)*sec(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
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)))-b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {9 \, a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, b \cos \left (d x + c\right )^{3} + 9 \, a \cos \left (d x + c\right )^{2} + 8 \, b \cos \left (d x + c\right ) + 6 \, a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")
Output:
1/48*(9*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*a*cos(d*x + c)^4*log(-s in(d*x + c) + 1) + 2*(16*b*cos(d*x + c)^3 + 9*a*cos(d*x + c)^2 + 8*b*cos(d *x + c) + 6*a)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)**5,x)
Output:
Integral((a + b*cos(c + d*x))*sec(c + d*x)**5, x)
Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b - 3 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*b - 3*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
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (77) = 154\).
Time = 0.55 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {9 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate((a+b*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")
Output:
1/24*(9*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*a*log(abs(tan(1/2*d*x + 1 /2*c) - 1)) + 2*(15*a*tan(1/2*d*x + 1/2*c)^7 - 24*b*tan(1/2*d*x + 1/2*c)^7 + 9*a*tan(1/2*d*x + 1/2*c)^5 + 40*b*tan(1/2*d*x + 1/2*c)^5 + 9*a*tan(1/2* d*x + 1/2*c)^3 - 40*b*tan(1/2*d*x + 1/2*c)^3 + 15*a*tan(1/2*d*x + 1/2*c) + 24*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 42.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.76 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (\frac {5\,a}{4}-2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a}{4}+\frac {10\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a}{4}-\frac {10\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a}{4}+2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \] Input:
int((a + b*cos(c + d*x))/cos(c + d*x)^5,x)
Output:
(tan(c/2 + (d*x)/2)*((5*a)/4 + 2*b) + tan(c/2 + (d*x)/2)^7*((5*a)/4 - 2*b) + tan(c/2 + (d*x)/2)^3*((3*a)/4 - (10*b)/3) + tan(c/2 + (d*x)/2)^5*((3*a) /4 + (10*b)/3))/(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*a*atanh(tan(c/2 + (d* x)/2)))/(4*d)
Time = 0.16 (sec) , antiderivative size = 257, normalized size of antiderivative = 3.02 \[ \int (a+b \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a +18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a -18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +16 \sin \left (d x +c \right )^{5} b -40 \sin \left (d x +c \right )^{3} b +24 \sin \left (d x +c \right ) b}{24 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a+b*cos(d*x+c))*sec(d*x+c)^5,x)
Output:
( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a + 18*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 9*cos(c + d*x)*log(ta n((c + d*x)/2) - 1)*a + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d *x)**4*a - 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 9 *cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a - 9*cos(c + d*x)*sin(c + d*x)**3 *a + 15*cos(c + d*x)*sin(c + d*x)*a + 16*sin(c + d*x)**5*b - 40*sin(c + d* x)**3*b + 24*sin(c + d*x)*b)/(24*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))