Integrand size = 19, antiderivative size = 159 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {5 x}{a^5}+\frac {496 \sin (c+d x)}{63 a^5 d}-\frac {\sin (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sin (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sin (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sin (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}-\frac {5 \sin (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )} \] Output:
-5*x/a^5+496/63*sin(d*x+c)/a^5/d-1/9*sin(d*x+c)/d/(a+a*sec(d*x+c))^5-5/21* sin(d*x+c)/a/d/(a+a*sec(d*x+c))^4-29/63*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^ 3-67/63*sin(d*x+c)/a^3/d/(a+a*sec(d*x+c))^2-5*sin(d*x+c)/d/(a^5+a^5*sec(d* x+c))
Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(159)=318\).
Time = 4.38 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.01 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (79380 d x \cos \left (\frac {d x}{2}\right )+79380 d x \cos \left (c+\frac {d x}{2}\right )+52920 d x \cos \left (c+\frac {3 d x}{2}\right )+52920 d x \cos \left (2 c+\frac {3 d x}{2}\right )+22680 d x \cos \left (2 c+\frac {5 d x}{2}\right )+22680 d x \cos \left (3 c+\frac {5 d x}{2}\right )+5670 d x \cos \left (3 c+\frac {7 d x}{2}\right )+5670 d x \cos \left (4 c+\frac {7 d x}{2}\right )+630 d x \cos \left (4 c+\frac {9 d x}{2}\right )+630 d x \cos \left (5 c+\frac {9 d x}{2}\right )-175014 \sin \left (\frac {d x}{2}\right )+143010 \sin \left (c+\frac {d x}{2}\right )-138726 \sin \left (c+\frac {3 d x}{2}\right )+73290 \sin \left (2 c+\frac {3 d x}{2}\right )-70389 \sin \left (2 c+\frac {5 d x}{2}\right )+20475 \sin \left (3 c+\frac {5 d x}{2}\right )-21141 \sin \left (3 c+\frac {7 d x}{2}\right )+1575 \sin \left (4 c+\frac {7 d x}{2}\right )-3091 \sin \left (4 c+\frac {9 d x}{2}\right )-567 \sin \left (5 c+\frac {9 d x}{2}\right )-63 \sin \left (5 c+\frac {11 d x}{2}\right )-63 \sin \left (6 c+\frac {11 d x}{2}\right )\right )}{64512 a^5 d} \] Input:
Integrate[Cos[c + d*x]/(a + a*Sec[c + d*x])^5,x]
Output:
-1/64512*(Sec[c/2]*Sec[(c + d*x)/2]^9*(79380*d*x*Cos[(d*x)/2] + 79380*d*x* Cos[c + (d*x)/2] + 52920*d*x*Cos[c + (3*d*x)/2] + 52920*d*x*Cos[2*c + (3*d *x)/2] + 22680*d*x*Cos[2*c + (5*d*x)/2] + 22680*d*x*Cos[3*c + (5*d*x)/2] + 5670*d*x*Cos[3*c + (7*d*x)/2] + 5670*d*x*Cos[4*c + (7*d*x)/2] + 630*d*x*C os[4*c + (9*d*x)/2] + 630*d*x*Cos[5*c + (9*d*x)/2] - 175014*Sin[(d*x)/2] + 143010*Sin[c + (d*x)/2] - 138726*Sin[c + (3*d*x)/2] + 73290*Sin[2*c + (3* d*x)/2] - 70389*Sin[2*c + (5*d*x)/2] + 20475*Sin[3*c + (5*d*x)/2] - 21141* Sin[3*c + (7*d*x)/2] + 1575*Sin[4*c + (7*d*x)/2] - 3091*Sin[4*c + (9*d*x)/ 2] - 567*Sin[5*c + (9*d*x)/2] - 63*Sin[5*c + (11*d*x)/2] - 63*Sin[6*c + (1 1*d*x)/2]))/(a^5*d)
Time = 1.31 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.20, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.895, Rules used = {3042, 4304, 27, 3042, 4508, 3042, 4508, 27, 3042, 4508, 3042, 4508, 3042, 4274, 24, 3042, 3117}
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.
| \(\displaystyle \int \frac {\cos (c+d x)}{(a \sec (c+d x)+a)^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int -\frac {5 \cos (c+d x) (2 a-a \sec (c+d x))}{(\sec (c+d x) a+a)^4}dx}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \int \frac {\cos (c+d x) (2 a-a \sec (c+d x))}{(\sec (c+d x) a+a)^4}dx}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {2 a-a \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {\cos (c+d x) \left (17 a^2-12 a^2 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {17 a^2-12 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {5 \left (\frac {\frac {\int \frac {3 \cos (c+d x) \left (38 a^3-29 a^3 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \int \frac {\cos (c+d x) \left (38 a^3-29 a^3 \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \int \frac {38 a^3-29 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\int \frac {\cos (c+d x) \left (181 a^4-134 a^4 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\int \frac {181 a^4-134 a^4 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\int \cos (c+d x) \left (496 a^5-315 a^5 \sec (c+d x)\right )dx}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\int \frac {496 a^5-315 a^5 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {496 a^5 \int \cos (c+d x)dx-315 a^5 \int 1dx}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {496 a^5 \int \cos (c+d x)dx-315 a^5 x}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {496 a^5 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx-315 a^5 x}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {5 \left (\frac {\frac {3 \left (\frac {\frac {\frac {496 a^5 \sin (c+d x)}{d}-315 a^5 x}{a^2}-\frac {315 a^4 \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {67 a^3 \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a^2}-\frac {29 a^2 \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {3 a \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\right )}{9 a^2}-\frac {\sin (c+d x)}{9 d (a \sec (c+d x)+a)^5}\) |
Input:
Int[Cos[c + d*x]/(a + a*Sec[c + d*x])^5,x]
Output:
-1/9*Sin[c + d*x]/(d*(a + a*Sec[c + d*x])^5) + (5*((-3*a*Sin[c + d*x])/(7* d*(a + a*Sec[c + d*x])^4) + ((-29*a^2*Sin[c + d*x])/(5*d*(a + a*Sec[c + d* x])^3) + (3*((-67*a^3*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((-315* a^4*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (-315*a^5*x + (496*a^5*Sin[c + d*x])/d)/a^2)/(3*a^2)))/(5*a^2))/(7*a^2)))/(9*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ [m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.57
| method | result | size |
| parallelrisch | \(-\frac {80640 d x -\left (42676+63 \cos \left (5 d x +5 c \right )+1892 \cos \left (4 d x +4 c \right )+11675 \cos \left (3 d x +3 c \right )+36632 \cos \left (2 d x +2 c \right )+69350 \cos \left (d x +c \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16128 d \,a^{5}}\) | \(90\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-160 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{5}}\) | \(111\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-24 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-160 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{5}}\) | \(111\) |
| norman | \(\frac {-\frac {5 x}{a}+\frac {161 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16 a d}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 a d}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}-\frac {65 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{1008 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{144 a d}-\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{4}}\) | \(156\) |
| risch | \(-\frac {5 x}{a^{5}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{5}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{5}}+\frac {2 i \left (945 \,{\mathrm e}^{8 i \left (d x +c \right )}+6300 \,{\mathrm e}^{7 i \left (d x +c \right )}+19740 \,{\mathrm e}^{6 i \left (d x +c \right )}+36414 \,{\mathrm e}^{5 i \left (d x +c \right )}+43092 \,{\mathrm e}^{4 i \left (d x +c \right )}+33264 \,{\mathrm e}^{3 i \left (d x +c \right )}+16416 \,{\mathrm e}^{2 i \left (d x +c \right )}+4734 \,{\mathrm e}^{i \left (d x +c \right )}+631\right )}{63 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\) | \(156\) |
Input:
int(cos(d*x+c)/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
-1/16128*(80640*d*x-(42676+63*cos(5*d*x+5*c)+1892*cos(4*d*x+4*c)+11675*cos (3*d*x+3*c)+36632*cos(2*d*x+2*c)+69350*cos(d*x+c))*tan(1/2*d*x+1/2*c)*sec( 1/2*d*x+1/2*c)^8)/d/a^5
Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.25 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {315 \, d x \cos \left (d x + c\right )^{5} + 1575 \, d x \cos \left (d x + c\right )^{4} + 3150 \, d x \cos \left (d x + c\right )^{3} + 3150 \, d x \cos \left (d x + c\right )^{2} + 1575 \, d x \cos \left (d x + c\right ) + 315 \, d x - {\left (63 \, \cos \left (d x + c\right )^{5} + 946 \, \cos \left (d x + c\right )^{4} + 2840 \, \cos \left (d x + c\right )^{3} + 3633 \, \cos \left (d x + c\right )^{2} + 2165 \, \cos \left (d x + c\right ) + 496\right )} \sin \left (d x + c\right )}{63 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \] Input:
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="fricas")
Output:
-1/63*(315*d*x*cos(d*x + c)^5 + 1575*d*x*cos(d*x + c)^4 + 3150*d*x*cos(d*x + c)^3 + 3150*d*x*cos(d*x + c)^2 + 1575*d*x*cos(d*x + c) + 315*d*x - (63* cos(d*x + c)^5 + 946*cos(d*x + c)^4 + 2840*cos(d*x + c)^3 + 3633*cos(d*x + c)^2 + 2165*cos(d*x + c) + 496)*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)
\[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \] Input:
integrate(cos(d*x+c)/(a+a*sec(d*x+c))**5,x)
Output:
Integral(cos(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 = 0.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {2016 \, \sin \left (d x + c\right )}{{\left (a^{5} + \frac {a^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {8127 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {1512 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {72 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {10080 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{5}}}{1008 \, d} \] Input:
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="maxima")
Output:
1/1008*(2016*sin(d*x + c)/((a^5 + a^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) *(cos(d*x + c) + 1)) + (8127*sin(d*x + c)/(cos(d*x + c) + 1) - 1512*sin(d* x + c)^3/(cos(d*x + c) + 1)^3 + 378*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 72*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^5 - 10080*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^5)/d
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=-\frac {\frac {5040 \, {\left (d x + c\right )}}{a^{5}} - \frac {2016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{5}} - \frac {7 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 72 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1512 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8127 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{1008 \, d} \] Input:
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^5,x, algorithm="giac")
Output:
-1/1008*(5040*(d*x + c)/a^5 - 2016*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/ 2*c)^2 + 1)*a^5) - (7*a^40*tan(1/2*d*x + 1/2*c)^9 - 72*a^40*tan(1/2*d*x + 1/2*c)^7 + 378*a^40*tan(1/2*d*x + 1/2*c)^5 - 1512*a^40*tan(1/2*d*x + 1/2*c )^3 + 8127*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d
Time = 9.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-100\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+636\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2512\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+10096\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2016\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-5040\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (c+d\,x\right )}{1008\,a^5\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \] Input:
int(cos(c + d*x)/(a + a/cos(c + d*x))^5,x)
Output:
(7*sin(c/2 + (d*x)/2) - 100*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 636* cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2) - 2512*cos(c/2 + (d*x)/2)^6*sin(c/ 2 + (d*x)/2) + 10096*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) + 2016*cos(c/ 2 + (d*x)/2)^10*sin(c/2 + (d*x)/2) - 5040*cos(c/2 + (d*x)/2)^9*(c + d*x))/ (1008*a^5*d*cos(c/2 + (d*x)/2)^9)
Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^5} \, dx=\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}-65 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+306 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-1134 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+6615 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-5040 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} d x +10143 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-5040 d x}{1008 a^{5} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:
int(cos(d*x+c)/(a+a*sec(d*x+c))^5,x)
Output:
(7*tan((c + d*x)/2)**11 - 65*tan((c + d*x)/2)**9 + 306*tan((c + d*x)/2)**7 - 1134*tan((c + d*x)/2)**5 + 6615*tan((c + d*x)/2)**3 - 5040*tan((c + d*x )/2)**2*d*x + 10143*tan((c + d*x)/2) - 5040*d*x)/(1008*a**5*d*(tan((c + d* x)/2)**2 + 1))