Integrand size = 21, antiderivative size = 137 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (a-a \cos (c+d x))^6}{3 a^8 d}-\frac {4 (a-a \cos (c+d x))^7}{a^9 d}+\frac {19 (a-a \cos (c+d x))^8}{4 a^{10} d}-\frac {25 (a-a \cos (c+d x))^9}{9 a^{11} d}+\frac {4 (a-a \cos (c+d x))^{10}}{5 a^{12} d}-\frac {(a-a \cos (c+d x))^{11}}{11 a^{13} d} \]
4/3*(a-a*cos(d*x+c))^6/a^8/d-4*(a-a*cos(d*x+c))^7/a^9/d+19/4*(a-a*cos(d*x+ c))^8/a^10/d-25/9*(a-a*cos(d*x+c))^9/a^11/d+4/5*(a-a*cos(d*x+c))^10/a^12/d -1/11*(a-a*cos(d*x+c))^11/a^13/d
Time = 3.90 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.53 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {4 (2360+4038 \cos (c+d x)+2586 \cos (2 (c+d x))+1189 \cos (3 (c+d x))+342 \cos (4 (c+d x))+45 \cos (5 (c+d x))) \sin ^{12}\left (\frac {1}{2} (c+d x)\right )}{495 a^2 d} \]
(4*(2360 + 4038*Cos[c + d*x] + 2586*Cos[2*(c + d*x)] + 1189*Cos[3*(c + d*x )] + 342*Cos[4*(c + d*x)] + 45*Cos[5*(c + d*x)])*Sin[(c + d*x)/2]^12)/(495 *a^2*d)
Time = 0.46 (sec) , antiderivative size = 122, 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.
\(\displaystyle \int \frac {\sin ^{11}(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^{11}}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\sin ^{11}(c+d x) \cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^{11}}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^{11} \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}{\left (\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle -\frac {\int \cos ^2(c+d x) (a-a \cos (c+d x))^5 (\cos (c+d x) a+a)^3d(a \cos (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int a^2 \cos ^2(c+d x) (a-a \cos (c+d x))^5 (\cos (c+d x) a+a)^3d(a \cos (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-(a-a \cos (c+d x))^{10}+8 a (a-a \cos (c+d x))^9-25 a^2 (a-a \cos (c+d x))^8+38 a^3 (a-a \cos (c+d x))^7-28 a^4 (a-a \cos (c+d x))^6+8 a^5 (a-a \cos (c+d x))^5\right )d(a \cos (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {4}{3} a^5 (a-a \cos (c+d x))^6+4 a^4 (a-a \cos (c+d x))^7-\frac {19}{4} a^3 (a-a \cos (c+d x))^8+\frac {25}{9} a^2 (a-a \cos (c+d x))^9+\frac {1}{11} (a-a \cos (c+d x))^{11}-\frac {4}{5} a (a-a \cos (c+d x))^{10}}{a^{13} d}\) |
-(((-4*a^5*(a - a*Cos[c + d*x])^6)/3 + 4*a^4*(a - a*Cos[c + d*x])^7 - (19* a^3*(a - a*Cos[c + d*x])^8)/4 + (25*a^2*(a - a*Cos[c + d*x])^9)/9 - (4*a*( a - a*Cos[c + d*x])^10)/5 + (a - a*Cos[c + d*x])^11/11)/(a^13*d))
3.1.74.3.1 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 = 1.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(\frac {\frac {\cos \left (d x +c \right )^{11}}{11}-\frac {\cos \left (d x +c \right )^{10}}{5}-\frac {2 \cos \left (d x +c \right )^{9}}{9}+\frac {3 \cos \left (d x +c \right )^{8}}{4}-\cos \left (d x +c \right )^{6}+\frac {2 \cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\) | \(89\) |
default | \(\frac {\frac {\cos \left (d x +c \right )^{11}}{11}-\frac {\cos \left (d x +c \right )^{10}}{5}-\frac {2 \cos \left (d x +c \right )^{9}}{9}+\frac {3 \cos \left (d x +c \right )^{8}}{4}-\cos \left (d x +c \right )^{6}+\frac {2 \cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{4}}{2}-\frac {\cos \left (d x +c \right )^{3}}{3}}{d \,a^{2}}\) | \(89\) |
parallelrisch | \(\frac {-42470-3960 \cos \left (4 d x +4 c \right )-990 \cos \left (3 d x +3 c \right )+45 \cos \left (11 d x +11 c \right )-34650 \cos \left (d x +c \right )+990 \cos \left (8 d x +8 c \right )-1485 \cos \left (7 d x +7 c \right )-990 \cos \left (6 d x +6 c \right )+4257 \cos \left (5 d x +5 c \right )+13860 \cos \left (2 d x +2 c \right )+55 \cos \left (9 d x +9 c \right )-198 \cos \left (10 d x +10 c \right )}{506880 a^{2} d}\) | \(129\) |
risch | \(-\frac {35 \cos \left (d x +c \right )}{512 a^{2} d}+\frac {\cos \left (11 d x +11 c \right )}{11264 d \,a^{2}}-\frac {\cos \left (10 d x +10 c \right )}{2560 d \,a^{2}}+\frac {\cos \left (9 d x +9 c \right )}{9216 d \,a^{2}}+\frac {\cos \left (8 d x +8 c \right )}{512 d \,a^{2}}-\frac {3 \cos \left (7 d x +7 c \right )}{1024 d \,a^{2}}-\frac {\cos \left (6 d x +6 c \right )}{512 d \,a^{2}}+\frac {43 \cos \left (5 d x +5 c \right )}{5120 d \,a^{2}}-\frac {\cos \left (4 d x +4 c \right )}{128 d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right )}{512 d \,a^{2}}+\frac {7 \cos \left (2 d x +2 c \right )}{256 d \,a^{2}}\) | \(186\) |
1/d/a^2*(1/11*cos(d*x+c)^11-1/5*cos(d*x+c)^10-2/9*cos(d*x+c)^9+3/4*cos(d*x +c)^8-cos(d*x+c)^6+2/5*cos(d*x+c)^5+1/2*cos(d*x+c)^4-1/3*cos(d*x+c)^3)
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {180 \, \cos \left (d x + c\right )^{11} - 396 \, \cos \left (d x + c\right )^{10} - 440 \, \cos \left (d x + c\right )^{9} + 1485 \, \cos \left (d x + c\right )^{8} - 1980 \, \cos \left (d x + c\right )^{6} + 792 \, \cos \left (d x + c\right )^{5} + 990 \, \cos \left (d x + c\right )^{4} - 660 \, \cos \left (d x + c\right )^{3}}{1980 \, a^{2} d} \]
1/1980*(180*cos(d*x + c)^11 - 396*cos(d*x + c)^10 - 440*cos(d*x + c)^9 + 1 485*cos(d*x + c)^8 - 1980*cos(d*x + c)^6 + 792*cos(d*x + c)^5 + 990*cos(d* x + c)^4 - 660*cos(d*x + c)^3)/(a^2*d)
Timed out. \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {180 \, \cos \left (d x + c\right )^{11} - 396 \, \cos \left (d x + c\right )^{10} - 440 \, \cos \left (d x + c\right )^{9} + 1485 \, \cos \left (d x + c\right )^{8} - 1980 \, \cos \left (d x + c\right )^{6} + 792 \, \cos \left (d x + c\right )^{5} + 990 \, \cos \left (d x + c\right )^{4} - 660 \, \cos \left (d x + c\right )^{3}}{1980 \, a^{2} d} \]
1/1980*(180*cos(d*x + c)^11 - 396*cos(d*x + c)^10 - 440*cos(d*x + c)^9 + 1 485*cos(d*x + c)^8 - 1980*cos(d*x + c)^6 + 792*cos(d*x + c)^5 + 990*cos(d* x + c)^4 - 660*cos(d*x + c)^3)/(a^2*d)
Time = 0.38 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.35 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {64 \, {\left (\frac {11 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {55 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {165 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {330 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {462 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {198 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {990 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 1\right )}}{495 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{11}} \]
-64/495*(11*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 55*(cos(d*x + c) - 1)^ 2/(cos(d*x + c) + 1)^2 + 165*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3 30*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 462*(cos(d*x + c) - 1)^5/(c os(d*x + c) + 1)^5 + 198*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6 + 990*( cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 - 1)/(a^2*d*((cos(d*x + c) - 1)/( cos(d*x + c) + 1) - 1)^11)
Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {\sin ^{11}(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}-\frac {{\cos \left (c+d\,x\right )}^4}{2\,a^2}-\frac {2\,{\cos \left (c+d\,x\right )}^5}{5\,a^2}+\frac {{\cos \left (c+d\,x\right )}^6}{a^2}-\frac {3\,{\cos \left (c+d\,x\right )}^8}{4\,a^2}+\frac {2\,{\cos \left (c+d\,x\right )}^9}{9\,a^2}+\frac {{\cos \left (c+d\,x\right )}^{10}}{5\,a^2}-\frac {{\cos \left (c+d\,x\right )}^{11}}{11\,a^2}}{d} \]