Integrand size = 21, antiderivative size = 145 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\csc ^3(c+d x)}{3 a^3 d}-\frac {3 \csc ^4(c+d x)}{4 a^3 d}+\frac {\csc ^5(c+d x)}{5 a^3 d}+\frac {5 \csc ^6(c+d x)}{6 a^3 d}-\frac {5 \csc ^7(c+d x)}{7 a^3 d}-\frac {\csc ^8(c+d x)}{8 a^3 d}+\frac {\csc ^9(c+d x)}{3 a^3 d}-\frac {\csc ^{10}(c+d x)}{10 a^3 d} \]
1/3*csc(d*x+c)^3/a^3/d-3/4*csc(d*x+c)^4/a^3/d+1/5*csc(d*x+c)^5/a^3/d+5/6*c sc(d*x+c)^6/a^3/d-5/7*csc(d*x+c)^7/a^3/d-1/8*csc(d*x+c)^8/a^3/d+1/3*csc(d* x+c)^9/a^3/d-1/10*csc(d*x+c)^10/a^3/d
Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\csc ^3(c+d x) \left (280-630 \csc (c+d x)+168 \csc ^2(c+d x)+700 \csc ^3(c+d x)-600 \csc ^4(c+d x)-105 \csc ^5(c+d x)+280 \csc ^6(c+d x)-84 \csc ^7(c+d x)\right )}{840 a^3 d} \]
(Csc[c + d*x]^3*(280 - 630*Csc[c + d*x] + 168*Csc[c + d*x]^2 + 700*Csc[c + d*x]^3 - 600*Csc[c + d*x]^4 - 105*Csc[c + d*x]^5 + 280*Csc[c + d*x]^6 - 8 4*Csc[c + d*x]^7))/(840*a^3*d)
Time = 0.29 (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.
\(\displaystyle \int \frac {\cot ^{11}(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^{11} (a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^{11}(c+d x) (a-a \sin (c+d x))^5 (\sin (c+d x) a+a)^2}{a^{11}}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\csc ^{11}(c+d x)}{a^4}-\frac {3 \csc ^{10}(c+d x)}{a^4}+\frac {\csc ^9(c+d x)}{a^4}+\frac {5 \csc ^8(c+d x)}{a^4}-\frac {5 \csc ^7(c+d x)}{a^4}-\frac {\csc ^6(c+d x)}{a^4}+\frac {3 \csc ^5(c+d x)}{a^4}-\frac {\csc ^4(c+d x)}{a^4}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^{10}(c+d x)}{10 a^3}+\frac {\csc ^9(c+d x)}{3 a^3}-\frac {\csc ^8(c+d x)}{8 a^3}-\frac {5 \csc ^7(c+d x)}{7 a^3}+\frac {5 \csc ^6(c+d x)}{6 a^3}+\frac {\csc ^5(c+d x)}{5 a^3}-\frac {3 \csc ^4(c+d x)}{4 a^3}+\frac {\csc ^3(c+d x)}{3 a^3}}{d}\) |
(Csc[c + d*x]^3/(3*a^3) - (3*Csc[c + d*x]^4)/(4*a^3) + Csc[c + d*x]^5/(5*a ^3) + (5*Csc[c + d*x]^6)/(6*a^3) - (5*Csc[c + d*x]^7)/(7*a^3) - Csc[c + d* x]^8/(8*a^3) + Csc[c + d*x]^9/(3*a^3) - Csc[c + d*x]^10/(10*a^3))/d
3.1.80.3.1 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 = 69.79 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61
method | result | size |
derivativedivides | \(\frac {-\frac {5}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{10 \sin \left (d x +c \right )^{10}}+\frac {5}{6 \sin \left (d x +c \right )^{6}}+\frac {1}{3 \sin \left (d x +c \right )^{9}}+\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {3}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{8 \sin \left (d x +c \right )^{8}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d \,a^{3}}\) | \(89\) |
default | \(\frac {-\frac {5}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{10 \sin \left (d x +c \right )^{10}}+\frac {5}{6 \sin \left (d x +c \right )^{6}}+\frac {1}{3 \sin \left (d x +c \right )^{9}}+\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {3}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{8 \sin \left (d x +c \right )^{8}}+\frac {1}{3 \sin \left (d x +c \right )^{3}}}{d \,a^{3}}\) | \(89\) |
risch | \(-\frac {4 i \left (-315 i {\mathrm e}^{16 i \left (d x +c \right )}+70 \,{\mathrm e}^{17 i \left (d x +c \right )}+490 i {\mathrm e}^{14 i \left (d x +c \right )}-658 \,{\mathrm e}^{15 i \left (d x +c \right )}+35 i {\mathrm e}^{12 i \left (d x +c \right )}-90 \,{\mathrm e}^{13 i \left (d x +c \right )}+2268 i {\mathrm e}^{10 i \left (d x +c \right )}-1410 \,{\mathrm e}^{11 i \left (d x +c \right )}+35 i {\mathrm e}^{8 i \left (d x +c \right )}+1410 \,{\mathrm e}^{9 i \left (d x +c \right )}+490 i {\mathrm e}^{6 i \left (d x +c \right )}+90 \,{\mathrm e}^{7 i \left (d x +c \right )}-315 i {\mathrm e}^{4 i \left (d x +c \right )}+658 \,{\mathrm e}^{5 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{105 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{10}}\) | \(196\) |
1/d/a^3*(-5/7/sin(d*x+c)^7-1/10/sin(d*x+c)^10+5/6/sin(d*x+c)^6+1/3/sin(d*x +c)^9+1/5/sin(d*x+c)^5-3/4/sin(d*x+c)^4-1/8/sin(d*x+c)^8+1/3/sin(d*x+c)^3)
Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {630 \, \cos \left (d x + c\right )^{6} - 1190 \, \cos \left (d x + c\right )^{4} + 595 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (35 \, \cos \left (d x + c\right )^{6} - 126 \, \cos \left (d x + c\right )^{4} + 72 \, \cos \left (d x + c\right )^{2} - 16\right )} \sin \left (d x + c\right ) - 119}{840 \, {\left (a^{3} d \cos \left (d x + c\right )^{10} - 5 \, a^{3} d \cos \left (d x + c\right )^{8} + 10 \, a^{3} d \cos \left (d x + c\right )^{6} - 10 \, a^{3} d \cos \left (d x + c\right )^{4} + 5 \, a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]
-1/840*(630*cos(d*x + c)^6 - 1190*cos(d*x + c)^4 + 595*cos(d*x + c)^2 - 8* (35*cos(d*x + c)^6 - 126*cos(d*x + c)^4 + 72*cos(d*x + c)^2 - 16)*sin(d*x + c) - 119)/(a^3*d*cos(d*x + c)^10 - 5*a^3*d*cos(d*x + c)^8 + 10*a^3*d*cos (d*x + c)^6 - 10*a^3*d*cos(d*x + c)^4 + 5*a^3*d*cos(d*x + c)^2 - a^3*d)
Timed out. \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {280 \, \sin \left (d x + c\right )^{7} - 630 \, \sin \left (d x + c\right )^{6} + 168 \, \sin \left (d x + c\right )^{5} + 700 \, \sin \left (d x + c\right )^{4} - 600 \, \sin \left (d x + c\right )^{3} - 105 \, \sin \left (d x + c\right )^{2} + 280 \, \sin \left (d x + c\right ) - 84}{840 \, a^{3} d \sin \left (d x + c\right )^{10}} \]
1/840*(280*sin(d*x + c)^7 - 630*sin(d*x + c)^6 + 168*sin(d*x + c)^5 + 700* sin(d*x + c)^4 - 600*sin(d*x + c)^3 - 105*sin(d*x + c)^2 + 280*sin(d*x + c ) - 84)/(a^3*d*sin(d*x + c)^10)
Time = 0.53 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {280 \, \sin \left (d x + c\right )^{7} - 630 \, \sin \left (d x + c\right )^{6} + 168 \, \sin \left (d x + c\right )^{5} + 700 \, \sin \left (d x + c\right )^{4} - 600 \, \sin \left (d x + c\right )^{3} - 105 \, \sin \left (d x + c\right )^{2} + 280 \, \sin \left (d x + c\right ) - 84}{840 \, a^{3} d \sin \left (d x + c\right )^{10}} \]
1/840*(280*sin(d*x + c)^7 - 630*sin(d*x + c)^6 + 168*sin(d*x + c)^5 + 700* sin(d*x + c)^4 - 600*sin(d*x + c)^3 - 105*sin(d*x + c)^2 + 280*sin(d*x + c ) - 84)/(a^3*d*sin(d*x + c)^10)
Time = 6.80 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.59 \[ \int \frac {\cot ^{11}(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {280\,{\sin \left (c+d\,x\right )}^7-630\,{\sin \left (c+d\,x\right )}^6+168\,{\sin \left (c+d\,x\right )}^5+700\,{\sin \left (c+d\,x\right )}^4-600\,{\sin \left (c+d\,x\right )}^3-105\,{\sin \left (c+d\,x\right )}^2+280\,\sin \left (c+d\,x\right )-84}{840\,a^3\,d\,{\sin \left (c+d\,x\right )}^{10}} \]