Integrand size = 19, antiderivative size = 196 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{256 a^8 d}-\frac {1}{16 d (a+a \sin (c+d x))^8}-\frac {1}{28 a d (a+a \sin (c+d x))^7}-\frac {1}{48 a^2 d (a+a \sin (c+d x))^6}-\frac {1}{80 a^3 d (a+a \sin (c+d x))^5}-\frac {1}{128 d \left (a^2+a^2 \sin (c+d x)\right )^4}-\frac {a}{192 d \left (a^3+a^3 \sin (c+d x)\right )^3}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )^2}-\frac {1}{256 d \left (a^8+a^8 \sin (c+d x)\right )} \] Output:
1/256*arctanh(sin(d*x+c))/a^8/d-1/16/d/(a+a*sin(d*x+c))^8-1/28/a/d/(a+a*si n(d*x+c))^7-1/48/a^2/d/(a+a*sin(d*x+c))^6-1/80/a^3/d/(a+a*sin(d*x+c))^5-1/ 128/d/(a^2+a^2*sin(d*x+c))^4-1/192*a/d/(a^3+a^3*sin(d*x+c))^3-1/256/d/(a^4 +a^4*sin(d*x+c))^2-1/256/d/(a^8+a^8*sin(d*x+c))
Time = 0.83 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.62 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=-\frac {4096-105 \text {arctanh}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^{16}+5993 \sin (c+d x)+8008 \sin ^2(c+d x)+8351 \sin ^3(c+d x)+6160 \sin ^4(c+d x)+2975 \sin ^5(c+d x)+840 \sin ^6(c+d x)+105 \sin ^7(c+d x)}{26880 a^8 d (1+\sin (c+d x))^8} \] Input:
Integrate[Sec[c + d*x]/(a + a*Sin[c + d*x])^8,x]
Output:
-1/26880*(4096 - 105*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] + Sin[(c + d* x)/2])^16 + 5993*Sin[c + d*x] + 8008*Sin[c + d*x]^2 + 8351*Sin[c + d*x]^3 + 6160*Sin[c + d*x]^4 + 2975*Sin[c + d*x]^5 + 840*Sin[c + d*x]^6 + 105*Sin [c + d*x]^7)/(a^8*d*(1 + Sin[c + d*x])^8)
Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3146, 54, 2009}
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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+a)^8} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x) (a \sin (c+d x)+a)^8}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^9}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a \int \left (\frac {1}{2 (\sin (c+d x) a+a)^9 a}+\frac {1}{4 (\sin (c+d x) a+a)^8 a^2}+\frac {1}{8 (\sin (c+d x) a+a)^7 a^3}+\frac {1}{16 (\sin (c+d x) a+a)^6 a^4}+\frac {1}{32 (\sin (c+d x) a+a)^5 a^5}+\frac {1}{64 (\sin (c+d x) a+a)^4 a^6}+\frac {1}{128 (\sin (c+d x) a+a)^3 a^7}+\frac {1}{256 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^8}+\frac {1}{256 (\sin (c+d x) a+a)^2 a^8}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{256 a^9}-\frac {1}{256 a^8 (a \sin (c+d x)+a)}-\frac {1}{256 a^7 (a \sin (c+d x)+a)^2}-\frac {1}{192 a^6 (a \sin (c+d x)+a)^3}-\frac {1}{128 a^5 (a \sin (c+d x)+a)^4}-\frac {1}{80 a^4 (a \sin (c+d x)+a)^5}-\frac {1}{48 a^3 (a \sin (c+d x)+a)^6}-\frac {1}{28 a^2 (a \sin (c+d x)+a)^7}-\frac {1}{16 a (a \sin (c+d x)+a)^8}\right )}{d}\) |
Input:
Int[Sec[c + d*x]/(a + a*Sin[c + d*x])^8,x]
Output:
(a*(ArcTanh[Sin[c + d*x]]/(256*a^9) - 1/(16*a*(a + a*Sin[c + d*x])^8) - 1/ (28*a^2*(a + a*Sin[c + d*x])^7) - 1/(48*a^3*(a + a*Sin[c + d*x])^6) - 1/(8 0*a^4*(a + a*Sin[c + d*x])^5) - 1/(128*a^5*(a + a*Sin[c + d*x])^4) - 1/(19 2*a^6*(a + a*Sin[c + d*x])^3) - 1/(256*a^7*(a + a*Sin[c + d*x])^2) - 1/(25 6*a^8*(a + a*Sin[c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 2.45 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{8}}-\frac {1}{28 \left (1+\sin \left (d x +c \right )\right )^{7}}-\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{128 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{192 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d \,a^{8}}\) | \(127\) |
| default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{512}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{8}}-\frac {1}{28 \left (1+\sin \left (d x +c \right )\right )^{7}}-\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {1}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{128 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{192 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{512}}{d \,a^{8}}\) | \(127\) |
| risch | \(-\frac {i \left (1680 i {\mathrm e}^{14 i \left (d x +c \right )}+105 \,{\mathrm e}^{15 i \left (d x +c \right )}-59360 i {\mathrm e}^{12 i \left (d x +c \right )}-12635 \,{\mathrm e}^{13 i \left (d x +c \right )}+478576 i {\mathrm e}^{10 i \left (d x +c \right )}+195321 \,{\mathrm e}^{11 i \left (d x +c \right )}-1366080 i {\mathrm e}^{8 i \left (d x +c \right )}-907075 \,{\mathrm e}^{9 i \left (d x +c \right )}+478576 i {\mathrm e}^{6 i \left (d x +c \right )}+907075 \,{\mathrm e}^{7 i \left (d x +c \right )}-59360 i {\mathrm e}^{4 i \left (d x +c \right )}-195321 \,{\mathrm e}^{5 i \left (d x +c \right )}+1680 i {\mathrm e}^{2 i \left (d x +c \right )}+12635 \,{\mathrm e}^{3 i \left (d x +c \right )}-105 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{13440 d \,a^{8} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{16}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{256 d \,a^{8}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{256 d \,a^{8}}\) | \(240\) |
| norman | \(\frac {\frac {127 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a d}+\frac {127 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8 d a}+\frac {255 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 a d}+\frac {255 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{128 d a}+\frac {1049 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d a}+\frac {1049 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{4 d a}+\frac {10205 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128 d a}+\frac {10205 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{128 d a}+\frac {26609 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{14 d a}+\frac {141263 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{120 d a}+\frac {141263 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{120 d a}+\frac {409771 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{640 d a}+\frac {409771 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{640 d a}+\frac {4547161 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2688 d a}+\frac {4547161 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2688 d a}}{a^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{16}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{256 d \,a^{8}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{256 d \,a^{8}}\) | \(343\) |
| parallelrisch | \(\frac {\left (-3669120 \sin \left (3 d x +3 c \right )+470400 \sin \left (5 d x +5 c \right )-13440 \sin \left (7 d x +7 c \right )-6726720 \cos \left (2 d x +2 c \right )+1528800 \cos \left (4 d x +4 c \right )-100800 \cos \left (6 d x +6 c \right )+840 \cos \left (8 d x +8 c \right )+9609600 \sin \left (d x +c \right )+5405400\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (3669120 \sin \left (3 d x +3 c \right )-470400 \sin \left (5 d x +5 c \right )+13440 \sin \left (7 d x +7 c \right )+6726720 \cos \left (2 d x +2 c \right )-1528800 \cos \left (4 d x +4 c \right )+100800 \cos \left (6 d x +6 c \right )-840 \cos \left (8 d x +8 c \right )-9609600 \sin \left (d x +c \right )-5405400\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+23052288 \sin \left (3 d x +3 c \right )-3153920 \sin \left (5 d x +5 c \right )+94208 \sin \left (7 d x +7 c \right )+40334728 \cos \left (2 d x +2 c \right )-9957500 \cos \left (4 d x +4 c \right )+692280 \cos \left (6 d x +6 c \right )-5993 \cos \left (8 d x +8 c \right )-54046720 \sin \left (d x +c \right )-27636315}{215040 d \,a^{8} \left (-6435-1820 \cos \left (4 d x +4 c \right )-11440 \sin \left (d x +c \right )+4368 \sin \left (3 d x +3 c \right )+8008 \cos \left (2 d x +2 c \right )+16 \sin \left (7 d x +7 c \right )+120 \cos \left (6 d x +6 c \right )-\cos \left (8 d x +8 c \right )-560 \sin \left (5 d x +5 c \right )\right )}\) | \(385\) |
Input:
int(sec(d*x+c)/(a+a*sin(d*x+c))^8,x,method=_RETURNVERBOSE)
Output:
1/d/a^8*(-1/512*ln(sin(d*x+c)-1)-1/16/(1+sin(d*x+c))^8-1/28/(1+sin(d*x+c)) ^7-1/48/(1+sin(d*x+c))^6-1/80/(1+sin(d*x+c))^5-1/128/(1+sin(d*x+c))^4-1/19 2/(1+sin(d*x+c))^3-1/256/(1+sin(d*x+c))^2-1/256/(1+sin(d*x+c))+1/512*ln(1+ sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (178) = 356\).
Time = 0.15 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.91 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {1680 \, \cos \left (d x + c\right )^{6} - 17360 \, \cos \left (d x + c\right )^{4} + 45696 \, \cos \left (d x + c\right )^{2} + 105 \, {\left (\cos \left (d x + c\right )^{8} - 32 \, \cos \left (d x + c\right )^{6} + 160 \, \cos \left (d x + c\right )^{4} - 256 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (\cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} + 24 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 128\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left (\cos \left (d x + c\right )^{8} - 32 \, \cos \left (d x + c\right )^{6} + 160 \, \cos \left (d x + c\right )^{4} - 256 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (\cos \left (d x + c\right )^{6} - 10 \, \cos \left (d x + c\right )^{4} + 24 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 128\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (105 \, \cos \left (d x + c\right )^{6} - 3290 \, \cos \left (d x + c\right )^{4} + 14616 \, \cos \left (d x + c\right )^{2} - 17424\right )} \sin \left (d x + c\right ) - 38208}{53760 \, {\left (a^{8} d \cos \left (d x + c\right )^{8} - 32 \, a^{8} d \cos \left (d x + c\right )^{6} + 160 \, a^{8} d \cos \left (d x + c\right )^{4} - 256 \, a^{8} d \cos \left (d x + c\right )^{2} + 128 \, a^{8} d - 8 \, {\left (a^{8} d \cos \left (d x + c\right )^{6} - 10 \, a^{8} d \cos \left (d x + c\right )^{4} + 24 \, a^{8} d \cos \left (d x + c\right )^{2} - 16 \, a^{8} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)/(a+a*sin(d*x+c))^8,x, algorithm="fricas")
Output:
1/53760*(1680*cos(d*x + c)^6 - 17360*cos(d*x + c)^4 + 45696*cos(d*x + c)^2 + 105*(cos(d*x + c)^8 - 32*cos(d*x + c)^6 + 160*cos(d*x + c)^4 - 256*cos( d*x + c)^2 - 8*(cos(d*x + c)^6 - 10*cos(d*x + c)^4 + 24*cos(d*x + c)^2 - 1 6)*sin(d*x + c) + 128)*log(sin(d*x + c) + 1) - 105*(cos(d*x + c)^8 - 32*co s(d*x + c)^6 + 160*cos(d*x + c)^4 - 256*cos(d*x + c)^2 - 8*(cos(d*x + c)^6 - 10*cos(d*x + c)^4 + 24*cos(d*x + c)^2 - 16)*sin(d*x + c) + 128)*log(-si n(d*x + c) + 1) + 2*(105*cos(d*x + c)^6 - 3290*cos(d*x + c)^4 + 14616*cos( d*x + c)^2 - 17424)*sin(d*x + c) - 38208)/(a^8*d*cos(d*x + c)^8 - 32*a^8*d *cos(d*x + c)^6 + 160*a^8*d*cos(d*x + c)^4 - 256*a^8*d*cos(d*x + c)^2 + 12 8*a^8*d - 8*(a^8*d*cos(d*x + c)^6 - 10*a^8*d*cos(d*x + c)^4 + 24*a^8*d*cos (d*x + c)^2 - 16*a^8*d)*sin(d*x + c))
Timed out. \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)/(a+a*sin(d*x+c))**8,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.09 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=-\frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{7} + 840 \, \sin \left (d x + c\right )^{6} + 2975 \, \sin \left (d x + c\right )^{5} + 6160 \, \sin \left (d x + c\right )^{4} + 8351 \, \sin \left (d x + c\right )^{3} + 8008 \, \sin \left (d x + c\right )^{2} + 5993 \, \sin \left (d x + c\right ) + 4096\right )}}{a^{8} \sin \left (d x + c\right )^{8} + 8 \, a^{8} \sin \left (d x + c\right )^{7} + 28 \, a^{8} \sin \left (d x + c\right )^{6} + 56 \, a^{8} \sin \left (d x + c\right )^{5} + 70 \, a^{8} \sin \left (d x + c\right )^{4} + 56 \, a^{8} \sin \left (d x + c\right )^{3} + 28 \, a^{8} \sin \left (d x + c\right )^{2} + 8 \, a^{8} \sin \left (d x + c\right ) + a^{8}} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{8}} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{8}}}{53760 \, d} \] Input:
integrate(sec(d*x+c)/(a+a*sin(d*x+c))^8,x, algorithm="maxima")
Output:
-1/53760*(2*(105*sin(d*x + c)^7 + 840*sin(d*x + c)^6 + 2975*sin(d*x + c)^5 + 6160*sin(d*x + c)^4 + 8351*sin(d*x + c)^3 + 8008*sin(d*x + c)^2 + 5993* sin(d*x + c) + 4096)/(a^8*sin(d*x + c)^8 + 8*a^8*sin(d*x + c)^7 + 28*a^8*s in(d*x + c)^6 + 56*a^8*sin(d*x + c)^5 + 70*a^8*sin(d*x + c)^4 + 56*a^8*sin (d*x + c)^3 + 28*a^8*sin(d*x + c)^2 + 8*a^8*sin(d*x + c) + a^8) - 105*log( sin(d*x + c) + 1)/a^8 + 105*log(sin(d*x + c) - 1)/a^8)/d
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.64 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{512 \, a^{8} d} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{512 \, a^{8} d} - \frac {105 \, \sin \left (d x + c\right )^{7} + 840 \, \sin \left (d x + c\right )^{6} + 2975 \, \sin \left (d x + c\right )^{5} + 6160 \, \sin \left (d x + c\right )^{4} + 8351 \, \sin \left (d x + c\right )^{3} + 8008 \, \sin \left (d x + c\right )^{2} + 5993 \, \sin \left (d x + c\right ) + 4096}{26880 \, a^{8} d {\left (\sin \left (d x + c\right ) + 1\right )}^{8}} \] Input:
integrate(sec(d*x+c)/(a+a*sin(d*x+c))^8,x, algorithm="giac")
Output:
1/512*log(abs(sin(d*x + c) + 1))/(a^8*d) - 1/512*log(abs(sin(d*x + c) - 1) )/(a^8*d) - 1/26880*(105*sin(d*x + c)^7 + 840*sin(d*x + c)^6 + 2975*sin(d* x + c)^5 + 6160*sin(d*x + c)^4 + 8351*sin(d*x + c)^3 + 8008*sin(d*x + c)^2 + 5993*sin(d*x + c) + 4096)/(a^8*d*(sin(d*x + c) + 1)^8)
Time = 0.33 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.01 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{256\,a^8\,d}-\frac {\frac {{\sin \left (c+d\,x\right )}^7}{256}+\frac {{\sin \left (c+d\,x\right )}^6}{32}+\frac {85\,{\sin \left (c+d\,x\right )}^5}{768}+\frac {11\,{\sin \left (c+d\,x\right )}^4}{48}+\frac {1193\,{\sin \left (c+d\,x\right )}^3}{3840}+\frac {143\,{\sin \left (c+d\,x\right )}^2}{480}+\frac {5993\,\sin \left (c+d\,x\right )}{26880}+\frac {16}{105}}{d\,\left (a^8\,{\sin \left (c+d\,x\right )}^8+8\,a^8\,{\sin \left (c+d\,x\right )}^7+28\,a^8\,{\sin \left (c+d\,x\right )}^6+56\,a^8\,{\sin \left (c+d\,x\right )}^5+70\,a^8\,{\sin \left (c+d\,x\right )}^4+56\,a^8\,{\sin \left (c+d\,x\right )}^3+28\,a^8\,{\sin \left (c+d\,x\right )}^2+8\,a^8\,\sin \left (c+d\,x\right )+a^8\right )} \] Input:
int(1/(cos(c + d*x)*(a + a*sin(c + d*x))^8),x)
Output:
atanh(sin(c + d*x))/(256*a^8*d) - ((5993*sin(c + d*x))/26880 + (143*sin(c + d*x)^2)/480 + (1193*sin(c + d*x)^3)/3840 + (11*sin(c + d*x)^4)/48 + (85* sin(c + d*x)^5)/768 + sin(c + d*x)^6/32 + sin(c + d*x)^7/256 + 16/105)/(d* (8*a^8*sin(c + d*x) + a^8 + 28*a^8*sin(c + d*x)^2 + 56*a^8*sin(c + d*x)^3 + 70*a^8*sin(c + d*x)^4 + 56*a^8*sin(c + d*x)^5 + 28*a^8*sin(c + d*x)^6 + 8*a^8*sin(c + d*x)^7 + a^8*sin(c + d*x)^8))
Time = 0.28 (sec) , antiderivative size = 536, normalized size of antiderivative = 2.73 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^8} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)/(a+a*sin(d*x+c))^8,x)
Output:
( - 840*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 - 6720*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**7 - 23520*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 47040*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 58800*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**4 - 47040*log(tan((c + d*x)/2) - 1)*sin(c + d*x)* *3 - 23520*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 6720*log(tan((c + d *x)/2) - 1)*sin(c + d*x) - 840*log(tan((c + d*x)/2) - 1) + 840*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**8 + 6720*log(tan((c + d*x)/2) + 1)*sin(c + d* x)**7 + 23520*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6 + 47040*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 + 58800*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 47040*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 + 23520*log(tan( (c + d*x)/2) + 1)*sin(c + d*x)**2 + 6720*log(tan((c + d*x)/2) + 1)*sin(c + d*x) + 840*log(tan((c + d*x)/2) + 1) + 5993*sin(c + d*x)**8 + 47104*sin(c + d*x)**7 + 161084*sin(c + d*x)**6 + 311808*sin(c + d*x)**5 + 370230*sin( c + d*x)**4 + 268800*sin(c + d*x)**3 + 103740*sin(c + d*x)**2 - 26775)/(21 5040*a**8*d*(sin(c + d*x)**8 + 8*sin(c + d*x)**7 + 28*sin(c + d*x)**6 + 56 *sin(c + d*x)**5 + 70*sin(c + d*x)**4 + 56*sin(c + d*x)**3 + 28*sin(c + d* x)**2 + 8*sin(c + d*x) + 1))