Integrand size = 21, antiderivative size = 145 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^2(c+d x)}{2 a^2 d}-\frac {2 \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^4(c+d x)}{2 a^2 d}+\frac {6 \csc ^5(c+d x)}{5 a^2 d}-\frac {6 \csc ^7(c+d x)}{7 a^2 d}+\frac {\csc ^8(c+d x)}{4 a^2 d}+\frac {2 \csc ^9(c+d x)}{9 a^2 d}-\frac {\csc ^{10}(c+d x)}{10 a^2 d} \]
1/2*csc(d*x+c)^2/a^2/d-2/3*csc(d*x+c)^3/a^2/d-1/2*csc(d*x+c)^4/a^2/d+6/5*c sc(d*x+c)^5/a^2/d-6/7*csc(d*x+c)^7/a^2/d+1/4*csc(d*x+c)^8/a^2/d+2/9*csc(d* x+c)^9/a^2/d-1/10*csc(d*x+c)^10/a^2/d
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^2(c+d x) \left (630-840 \csc (c+d x)-630 \csc ^2(c+d x)+1512 \csc ^3(c+d x)-1080 \csc ^5(c+d x)+315 \csc ^6(c+d x)+280 \csc ^7(c+d x)-126 \csc ^8(c+d x)\right )}{1260 a^2 d} \]
(Csc[c + d*x]^2*(630 - 840*Csc[c + d*x] - 630*Csc[c + d*x]^2 + 1512*Csc[c + d*x]^3 - 1080*Csc[c + d*x]^5 + 315*Csc[c + d*x]^6 + 280*Csc[c + d*x]^7 - 126*Csc[c + d*x]^8))/(1260*a^2*d)
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 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.
\(\displaystyle \int \frac {\cot ^{11}(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^{11} (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^{11}(c+d x) (a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^3}{a^{11}}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\csc ^{11}(c+d x)}{a^3}-\frac {2 \csc ^{10}(c+d x)}{a^3}-\frac {2 \csc ^9(c+d x)}{a^3}+\frac {6 \csc ^8(c+d x)}{a^3}-\frac {6 \csc ^6(c+d x)}{a^3}+\frac {2 \csc ^5(c+d x)}{a^3}+\frac {2 \csc ^4(c+d x)}{a^3}-\frac {\csc ^3(c+d x)}{a^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^{10}(c+d x)}{10 a^2}+\frac {2 \csc ^9(c+d x)}{9 a^2}+\frac {\csc ^8(c+d x)}{4 a^2}-\frac {6 \csc ^7(c+d x)}{7 a^2}+\frac {6 \csc ^5(c+d x)}{5 a^2}-\frac {\csc ^4(c+d x)}{2 a^2}-\frac {2 \csc ^3(c+d x)}{3 a^2}+\frac {\csc ^2(c+d x)}{2 a^2}}{d}\) |
(Csc[c + d*x]^2/(2*a^2) - (2*Csc[c + d*x]^3)/(3*a^2) - Csc[c + d*x]^4/(2*a ^2) + (6*Csc[c + d*x]^5)/(5*a^2) - (6*Csc[c + d*x]^7)/(7*a^2) + Csc[c + d* x]^8/(4*a^2) + (2*Csc[c + d*x]^9)/(9*a^2) - Csc[c + d*x]^10/(10*a^2))/d
3.1.70.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((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] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 31.77 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{4}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{10 \sin \left (d x +c \right )^{10}}+\frac {1}{4 \sin \left (d x +c \right )^{8}}-\frac {6}{7 \sin \left (d x +c \right )^{7}}+\frac {2}{9 \sin \left (d x +c \right )^{9}}+\frac {6}{5 \sin \left (d x +c \right )^{5}}}{d \,a^{2}}\) | \(89\) |
default | \(\frac {-\frac {1}{2 \sin \left (d x +c \right )^{4}}+\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{10 \sin \left (d x +c \right )^{10}}+\frac {1}{4 \sin \left (d x +c \right )^{8}}-\frac {6}{7 \sin \left (d x +c \right )^{7}}+\frac {2}{9 \sin \left (d x +c \right )^{9}}+\frac {6}{5 \sin \left (d x +c \right )^{5}}}{d \,a^{2}}\) | \(89\) |
risch | \(-\frac {2 \left (315 \,{\mathrm e}^{18 i \left (d x +c \right )}-1260 \,{\mathrm e}^{16 i \left (d x +c \right )}+840 i {\mathrm e}^{3 i \left (d x +c \right )}+1260 \,{\mathrm e}^{14 i \left (d x +c \right )}-840 i {\mathrm e}^{17 i \left (d x +c \right )}-8820 \,{\mathrm e}^{12 i \left (d x +c \right )}+168 i {\mathrm e}^{5 i \left (d x +c \right )}+882 \,{\mathrm e}^{10 i \left (d x +c \right )}-2840 i {\mathrm e}^{9 i \left (d x +c \right )}-8820 \,{\mathrm e}^{8 i \left (d x +c \right )}+2840 i {\mathrm e}^{11 i \left (d x +c \right )}+1260 \,{\mathrm e}^{6 i \left (d x +c \right )}-168 i {\mathrm e}^{15 i \left (d x +c \right )}-1260 \,{\mathrm e}^{4 i \left (d x +c \right )}-4680 i {\mathrm e}^{13 i \left (d x +c \right )}+315 \,{\mathrm e}^{2 i \left (d x +c \right )}+4680 i {\mathrm e}^{7 i \left (d x +c \right )}\right )}{315 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{10}}\) | \(218\) |
1/d/a^2*(-1/2/sin(d*x+c)^4+1/2/sin(d*x+c)^2-2/3/sin(d*x+c)^3-1/10/sin(d*x+ c)^10+1/4/sin(d*x+c)^8-6/7/sin(d*x+c)^7+2/9/sin(d*x+c)^9+6/5/sin(d*x+c)^5)
Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.12 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {630 \, \cos \left (d x + c\right )^{8} - 1890 \, \cos \left (d x + c\right )^{6} + 1890 \, \cos \left (d x + c\right )^{4} - 945 \, \cos \left (d x + c\right )^{2} + 8 \, {\left (105 \, \cos \left (d x + c\right )^{6} - 126 \, \cos \left (d x + c\right )^{4} + 72 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) + 189}{1260 \, {\left (a^{2} d \cos \left (d x + c\right )^{10} - 5 \, a^{2} d \cos \left (d x + c\right )^{8} + 10 \, a^{2} d \cos \left (d x + c\right )^{6} - 10 \, a^{2} d \cos \left (d x + c\right )^{4} + 5 \, a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )}} \]
-1/1260*(630*cos(d*x + c)^8 - 1890*cos(d*x + c)^6 + 1890*cos(d*x + c)^4 - 945*cos(d*x + c)^2 + 8*(105*cos(d*x + c)^6 - 126*cos(d*x + c)^4 + 72*cos(d *x + c)^2 - 16)*sin(d*x + c) + 189)/(a^2*d*cos(d*x + c)^10 - 5*a^2*d*cos(d *x + c)^8 + 10*a^2*d*cos(d*x + c)^6 - 10*a^2*d*cos(d*x + c)^4 + 5*a^2*d*co s(d*x + c)^2 - a^2*d)
Timed out. \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {630 \, \sin \left (d x + c\right )^{8} - 840 \, \sin \left (d x + c\right )^{7} - 630 \, \sin \left (d x + c\right )^{6} + 1512 \, \sin \left (d x + c\right )^{5} - 1080 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )^{2} + 280 \, \sin \left (d x + c\right ) - 126}{1260 \, a^{2} d \sin \left (d x + c\right )^{10}} \]
1/1260*(630*sin(d*x + c)^8 - 840*sin(d*x + c)^7 - 630*sin(d*x + c)^6 + 151 2*sin(d*x + c)^5 - 1080*sin(d*x + c)^3 + 315*sin(d*x + c)^2 + 280*sin(d*x + c) - 126)/(a^2*d*sin(d*x + c)^10)
Time = 0.48 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {630 \, \sin \left (d x + c\right )^{8} - 840 \, \sin \left (d x + c\right )^{7} - 630 \, \sin \left (d x + c\right )^{6} + 1512 \, \sin \left (d x + c\right )^{5} - 1080 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )^{2} + 280 \, \sin \left (d x + c\right ) - 126}{1260 \, a^{2} d \sin \left (d x + c\right )^{10}} \]
1/1260*(630*sin(d*x + c)^8 - 840*sin(d*x + c)^7 - 630*sin(d*x + c)^6 + 151 2*sin(d*x + c)^5 - 1080*sin(d*x + c)^3 + 315*sin(d*x + c)^2 + 280*sin(d*x + c) - 126)/(a^2*d*sin(d*x + c)^10)
Time = 6.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^8}{2}-\frac {2\,{\sin \left (c+d\,x\right )}^7}{3}-\frac {{\sin \left (c+d\,x\right )}^6}{2}+\frac {6\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {6\,{\sin \left (c+d\,x\right )}^3}{7}+\frac {{\sin \left (c+d\,x\right )}^2}{4}+\frac {2\,\sin \left (c+d\,x\right )}{9}-\frac {1}{10}}{a^2\,d\,{\sin \left (c+d\,x\right )}^{10}} \]