Integrand size = 21, antiderivative size = 114 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (a-a \cos (c+d x))^5}{5 a^7 d}-\frac {2 (a-a \cos (c+d x))^6}{a^8 d}+\frac {13 (a-a \cos (c+d x))^7}{7 a^9 d}-\frac {3 (a-a \cos (c+d x))^8}{4 a^{10} d}+\frac {(a-a \cos (c+d x))^9}{9 a^{11} d} \]
4/5*(a-a*cos(d*x+c))^5/a^7/d-2*(a-a*cos(d*x+c))^6/a^8/d+13/7*(a-a*cos(d*x+ c))^7/a^9/d-3/4*(a-a*cos(d*x+c))^8/a^10/d+1/9*(a-a*cos(d*x+c))^9/a^11/d
Time = 2.93 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.54 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2 (992+1615 \cos (c+d x)+970 \cos (2 (c+d x))+385 \cos (3 (c+d x))+70 \cos (4 (c+d x))) \sin ^{10}\left (\frac {1}{2} (c+d x)\right )}{315 a^2 d} \]
(2*(992 + 1615*Cos[c + d*x] + 970*Cos[2*(c + d*x)] + 385*Cos[3*(c + d*x)] + 70*Cos[4*(c + d*x)])*Sin[(c + d*x)/2]^10)/(315*a^2*d)
Time = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 3042, 25, 3315, 27, 99, 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.
\(\displaystyle \int \frac {\sin ^9(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^9}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin ^9(c+d x) \cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^9}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^9 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}{\left (\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle -\frac {\int \cos ^2(c+d x) (a-a \cos (c+d x))^4 (\cos (c+d x) a+a)^2d(a \cos (c+d x))}{a^9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int a^2 \cos ^2(c+d x) (a-a \cos (c+d x))^4 (\cos (c+d x) a+a)^2d(a \cos (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left ((a-a \cos (c+d x))^8-6 a (a-a \cos (c+d x))^7+13 a^2 (a-a \cos (c+d x))^6-12 a^3 (a-a \cos (c+d x))^5+4 a^4 (a-a \cos (c+d x))^4\right )d(a \cos (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {4}{5} a^4 (a-a \cos (c+d x))^5+2 a^3 (a-a \cos (c+d x))^6-\frac {13}{7} a^2 (a-a \cos (c+d x))^7-\frac {1}{9} (a-a \cos (c+d x))^9+\frac {3}{4} a (a-a \cos (c+d x))^8}{a^{11} d}\) |
-(((-4*a^4*(a - a*Cos[c + d*x])^5)/5 + 2*a^3*(a - a*Cos[c + d*x])^6 - (13* a^2*(a - a*Cos[c + d*x])^7)/7 + (3*a*(a - a*Cos[c + d*x])^8)/4 - (a - a*Co s[c + d*x])^9/9)/(a^11*d))
3.1.75.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.96 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {-\frac {\cos \left (d x +c \right )^{9}}{9}+\frac {\cos \left (d x +c \right )^{8}}{4}+\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {2 \cos \left (d x +c \right )^{6}}{3}+\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\) | \(79\) |
default | \(\frac {-\frac {\cos \left (d x +c \right )^{9}}{9}+\frac {\cos \left (d x +c \right )^{8}}{4}+\frac {\cos \left (d x +c \right )^{7}}{7}-\frac {2 \cos \left (d x +c \right )^{6}}{3}+\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\) | \(79\) |
parallelrisch | \(\frac {-\frac {64}{315}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3}-\frac {64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{35}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{9} a^{2}}\) | \(91\) |
norman | \(\frac {-\frac {64}{315 a d}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 a d}-\frac {64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{35 d a}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{35 d a}-\frac {256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d a}-\frac {128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{9} a}\) | \(124\) |
risch | \(-\frac {13 \cos \left (d x +c \right )}{128 a^{2} d}-\frac {\cos \left (9 d x +9 c \right )}{2304 d \,a^{2}}+\frac {\cos \left (8 d x +8 c \right )}{512 d \,a^{2}}-\frac {3 \cos \left (7 d x +7 c \right )}{1792 d \,a^{2}}-\frac {\cos \left (6 d x +6 c \right )}{192 d \,a^{2}}+\frac {\cos \left (5 d x +5 c \right )}{80 d \,a^{2}}-\frac {\cos \left (4 d x +4 c \right )}{128 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right )}{96 d \,a^{2}}+\frac {3 \cos \left (2 d x +2 c \right )}{64 d \,a^{2}}\) | \(152\) |
1/d/a^2*(-1/9*cos(d*x+c)^9+1/4*cos(d*x+c)^8+1/7*cos(d*x+c)^7-2/3*cos(d*x+c )^6+1/5*cos(d*x+c)^5+1/2*cos(d*x+c)^4-1/3*cos(d*x+c)^3)
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {140 \, \cos \left (d x + c\right )^{9} - 315 \, \cos \left (d x + c\right )^{8} - 180 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 630 \, \cos \left (d x + c\right )^{4} + 420 \, \cos \left (d x + c\right )^{3}}{1260 \, a^{2} d} \]
-1/1260*(140*cos(d*x + c)^9 - 315*cos(d*x + c)^8 - 180*cos(d*x + c)^7 + 84 0*cos(d*x + c)^6 - 252*cos(d*x + c)^5 - 630*cos(d*x + c)^4 + 420*cos(d*x + c)^3)/(a^2*d)
Timed out. \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {140 \, \cos \left (d x + c\right )^{9} - 315 \, \cos \left (d x + c\right )^{8} - 180 \, \cos \left (d x + c\right )^{7} + 840 \, \cos \left (d x + c\right )^{6} - 252 \, \cos \left (d x + c\right )^{5} - 630 \, \cos \left (d x + c\right )^{4} + 420 \, \cos \left (d x + c\right )^{3}}{1260 \, a^{2} d} \]
-1/1260*(140*cos(d*x + c)^9 - 315*cos(d*x + c)^8 - 180*cos(d*x + c)^7 + 84 0*cos(d*x + c)^6 - 252*cos(d*x + c)^5 - 630*cos(d*x + c)^4 + 420*cos(d*x + c)^3)/(a^2*d)
Time = 0.37 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {64 \, {\left (\frac {9 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {36 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {84 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {126 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {210 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 1\right )}}{315 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{9}} \]
-64/315*(9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 36*(cos(d*x + c) - 1)^2 /(cos(d*x + c) + 1)^2 + 84*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 126 *(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 210*(cos(d*x + c) - 1)^6/(cos (d*x + c) + 1)^6 - 1)/(a^2*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^9 )
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84 \[ \int \frac {\sin ^9(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {{\cos \left (c+d\,x\right )}^4}{2\,a^2}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a^2}-\frac {2\,{\cos \left (c+d\,x\right )}^6}{3\,a^2}+\frac {{\cos \left (c+d\,x\right )}^7}{7\,a^2}+\frac {{\cos \left (c+d\,x\right )}^8}{4\,a^2}-\frac {{\cos \left (c+d\,x\right )}^9}{9\,a^2}}{d} \]