Integrand size = 19, antiderivative size = 143 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}+\frac {4 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}+\frac {4 \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}+\frac {8 \tan (c+d x)}{315 a d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {8 \tan (c+d x)}{315 d \left (a^5+a^5 \sec (c+d x)\right )} \] Output:
1/9*tan(d*x+c)/d/(a+a*sec(d*x+c))^5+4/63*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^4 +4/105*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3+8/315*tan(d*x+c)/a/d/(a^2+a^2*s ec(d*x+c))^2+8/315*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))
Time = 3.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (5418 \sin \left (\frac {d x}{2}\right )-5040 \sin \left (c+\frac {d x}{2}\right )+3612 \sin \left (c+\frac {3 d x}{2}\right )-3360 \sin \left (2 c+\frac {3 d x}{2}\right )+1728 \sin \left (2 c+\frac {5 d x}{2}\right )-1260 \sin \left (3 c+\frac {5 d x}{2}\right )+432 \sin \left (3 c+\frac {7 d x}{2}\right )-315 \sin \left (4 c+\frac {7 d x}{2}\right )+83 \sin \left (4 c+\frac {9 d x}{2}\right )\right )}{80640 a^5 d} \] Input:
Integrate[Sec[c + d*x]/(a + a*Sec[c + d*x])^5,x]
Output:
(Sec[c/2]*Sec[(c + d*x)/2]^9*(5418*Sin[(d*x)/2] - 5040*Sin[c + (d*x)/2] + 3612*Sin[c + (3*d*x)/2] - 3360*Sin[2*c + (3*d*x)/2] + 1728*Sin[2*c + (5*d* x)/2] - 1260*Sin[3*c + (5*d*x)/2] + 432*Sin[3*c + (7*d*x)/2] - 315*Sin[4*c + (7*d*x)/2] + 83*Sin[4*c + (9*d*x)/2]))/(80640*a^5*d)
Time = 0.72 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4283, 3042, 4283, 3042, 4283, 3042, 4283, 3042, 4281}
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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \sec (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {4 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^4}dx}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {4 \left (\frac {3 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {4 \left (\frac {3 \left (\frac {2 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {3 \left (\frac {2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {4 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {4 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a}+\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}+\frac {4 \left (\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}+\frac {3 \left (\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {2 \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}\right )}{7 a}\right )}{9 a}\) |
Input:
Int[Sec[c + d*x]/(a + a*Sec[c + d*x])^5,x]
Output:
Tan[c + d*x]/(9*d*(a + a*Sec[c + d*x])^5) + (4*(Tan[c + d*x]/(7*d*(a + a*S ec[c + d*x])^4) + (3*(Tan[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) + (2*(Tan[ c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[c + d*x]))))/(5*a)))/(7*a)))/(9*a)
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(m + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) ] && IntegerQ[2*m]
Time = 0.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.50
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) | \(71\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}\) | \(71\) |
| parallelrisch | \(\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-180 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+378 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-420 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+315 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5040 d \,a^{5}}\) | \(73\) |
| norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{40 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{28 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{144 a d}}{a^{4}}\) | \(99\) |
| risch | \(\frac {2 i \left (315 \,{\mathrm e}^{8 i \left (d x +c \right )}+1260 \,{\mathrm e}^{7 i \left (d x +c \right )}+3360 \,{\mathrm e}^{6 i \left (d x +c \right )}+5040 \,{\mathrm e}^{5 i \left (d x +c \right )}+5418 \,{\mathrm e}^{4 i \left (d x +c \right )}+3612 \,{\mathrm e}^{3 i \left (d x +c \right )}+1728 \,{\mathrm e}^{2 i \left (d x +c \right )}+432 \,{\mathrm e}^{i \left (d x +c \right )}+83\right )}{315 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\) | \(113\) |
Input:
int(sec(d*x+c)/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
1/16/d/a^5*(1/9*tan(1/2*d*x+1/2*c)^9-4/7*tan(1/2*d*x+1/2*c)^7+6/5*tan(1/2* d*x+1/2*c)^5-4/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))
Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {{\left (83 \, \cos \left (d x + c\right )^{4} + 100 \, \cos \left (d x + c\right )^{3} + 84 \, \cos \left (d x + c\right )^{2} + 40 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{315 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="fricas")
Output:
1/315*(83*cos(d*x + c)^4 + 100*cos(d*x + c)^3 + 84*cos(d*x + c)^2 + 40*cos (d*x + c) + 8)*sin(d*x + c)/(a^5*d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c ) + a^5*d)
\[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))**5,x)
Output:
Integral(sec(c + d*x)/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d* x)**3 + 10*sec(c + d*x)**2 + 5*sec(c + d*x) + 1), x)/a**5
Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {420 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {180 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{5040 \, a^{5} d} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="maxima")
Output:
1/5040*(315*sin(d*x + c)/(cos(d*x + c) + 1) - 420*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 180*sin(d*x + c)^7 /(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)
Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.50 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 420 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{5040 \, a^{5} d} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="giac")
Output:
1/5040*(35*tan(1/2*d*x + 1/2*c)^9 - 180*tan(1/2*d*x + 1/2*c)^7 + 378*tan(1 /2*d*x + 1/2*c)^5 - 420*tan(1/2*d*x + 1/2*c)^3 + 315*tan(1/2*d*x + 1/2*c)) /(a^5*d)
Time = 9.43 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (315\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+378\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-180\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\right )}{5040\,a^5\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \] Input:
int(1/(cos(c + d*x)*(a + a/cos(c + d*x))^5),x)
Output:
(sin(c/2 + (d*x)/2)*(315*cos(c/2 + (d*x)/2)^8 + 35*sin(c/2 + (d*x)/2)^8 - 180*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^6 + 378*cos(c/2 + (d*x)/2)^4*s in(c/2 + (d*x)/2)^4 - 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2))/(504 0*a^5*d*cos(c/2 + (d*x)/2)^9)
Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.50 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-180 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+378 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-420 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+315\right )}{5040 a^{5} d} \] Input:
int(sec(d*x+c)/(a+a*sec(d*x+c))^5,x)
Output:
(tan((c + d*x)/2)*(35*tan((c + d*x)/2)**8 - 180*tan((c + d*x)/2)**6 + 378* tan((c + d*x)/2)**4 - 420*tan((c + d*x)/2)**2 + 315))/(5040*a**5*d)