Integrand size = 21, antiderivative size = 127 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\csc ^2(c+d x)}{2 a^2 d}+\frac {2 \csc ^3(c+d x)}{3 a^2 d}+\frac {\csc ^4(c+d x)}{4 a^2 d}-\frac {4 \csc ^5(c+d x)}{5 a^2 d}+\frac {\csc ^6(c+d x)}{6 a^2 d}+\frac {2 \csc ^7(c+d x)}{7 a^2 d}-\frac {\csc ^8(c+d x)}{8 a^2 d} \]
-1/2*csc(d*x+c)^2/a^2/d+2/3*csc(d*x+c)^3/a^2/d+1/4*csc(d*x+c)^4/a^2/d-4/5* csc(d*x+c)^5/a^2/d+1/6*csc(d*x+c)^6/a^2/d+2/7*csc(d*x+c)^7/a^2/d-1/8*csc(d *x+c)^8/a^2/d
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.61 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\csc ^2(c+d x) \left (-420+560 \csc (c+d x)+210 \csc ^2(c+d x)-672 \csc ^3(c+d x)+140 \csc ^4(c+d x)+240 \csc ^5(c+d x)-105 \csc ^6(c+d x)\right )}{840 a^2 d} \]
(Csc[c + d*x]^2*(-420 + 560*Csc[c + d*x] + 210*Csc[c + d*x]^2 - 672*Csc[c + d*x]^3 + 140*Csc[c + d*x]^4 + 240*Csc[c + d*x]^5 - 105*Csc[c + d*x]^6))/ (840*a^2*d)
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \frac {\cot ^9(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^9 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^9(c+d x) (a-a \sin (c+d x))^4 (\sin (c+d x) a+a)^2}{a^9}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\csc ^9(c+d x)}{a^3}-\frac {2 \csc ^8(c+d x)}{a^3}-\frac {\csc ^7(c+d x)}{a^3}+\frac {4 \csc ^6(c+d x)}{a^3}-\frac {\csc ^5(c+d x)}{a^3}-\frac {2 \csc ^4(c+d x)}{a^3}+\frac {\csc ^3(c+d x)}{a^3}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {\csc ^8(c+d x)}{8 a^2}+\frac {2 \csc ^7(c+d x)}{7 a^2}+\frac {\csc ^6(c+d x)}{6 a^2}-\frac {4 \csc ^5(c+d x)}{5 a^2}+\frac {\csc ^4(c+d x)}{4 a^2}+\frac {2 \csc ^3(c+d x)}{3 a^2}-\frac {\csc ^2(c+d x)}{2 a^2}}{d}\) |
(-1/2*Csc[c + d*x]^2/a^2 + (2*Csc[c + d*x]^3)/(3*a^2) + Csc[c + d*x]^4/(4* a^2) - (4*Csc[c + d*x]^5)/(5*a^2) + Csc[c + d*x]^6/(6*a^2) + (2*Csc[c + d* x]^7)/(7*a^2) - Csc[c + d*x]^8/(8*a^2))/d
3.1.69.3.1 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 = 17.33 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{8 \sin \left (d x +c \right )^{8}}-\frac {4}{5 \sin \left (d x +c \right )^{5}}+\frac {2}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{6 \sin \left (d x +c \right )^{6}}}{d \,a^{2}}\) | \(79\) |
default | \(\frac {\frac {1}{4 \sin \left (d x +c \right )^{4}}+\frac {2}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{8 \sin \left (d x +c \right )^{8}}-\frac {4}{5 \sin \left (d x +c \right )^{5}}+\frac {2}{7 \sin \left (d x +c \right )^{7}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\frac {1}{6 \sin \left (d x +c \right )^{6}}}{d \,a^{2}}\) | \(79\) |
risch | \(\frac {2 \,{\mathrm e}^{14 i \left (d x +c \right )}-8 \,{\mathrm e}^{12 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{13 i \left (d x +c \right )}}{3}+\frac {10 \,{\mathrm e}^{10 i \left (d x +c \right )}}{3}+\frac {16 i {\mathrm e}^{11 i \left (d x +c \right )}}{15}-\frac {80 \,{\mathrm e}^{8 i \left (d x +c \right )}}{3}-\frac {1376 i {\mathrm e}^{9 i \left (d x +c \right )}}{105}+\frac {10 \,{\mathrm e}^{6 i \left (d x +c \right )}}{3}+\frac {1376 i {\mathrm e}^{7 i \left (d x +c \right )}}{105}-8 \,{\mathrm e}^{4 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{5 i \left (d x +c \right )}}{15}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {16 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{8}}\) | \(172\) |
1/d/a^2*(1/4/sin(d*x+c)^4+2/3/sin(d*x+c)^3-1/8/sin(d*x+c)^8-4/5/sin(d*x+c) ^5+2/7/sin(d*x+c)^7-1/2/sin(d*x+c)^2+1/6/sin(d*x+c)^6)
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {420 \, \cos \left (d x + c\right )^{6} - 1050 \, \cos \left (d x + c\right )^{4} + 700 \, \cos \left (d x + c\right )^{2} + 16 \, {\left (35 \, \cos \left (d x + c\right )^{4} - 28 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 175}{840 \, {\left (a^{2} d \cos \left (d x + c\right )^{8} - 4 \, a^{2} d \cos \left (d x + c\right )^{6} + 6 \, a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d\right )}} \]
1/840*(420*cos(d*x + c)^6 - 1050*cos(d*x + c)^4 + 700*cos(d*x + c)^2 + 16* (35*cos(d*x + c)^4 - 28*cos(d*x + c)^2 + 8)*sin(d*x + c) - 175)/(a^2*d*cos (d*x + c)^8 - 4*a^2*d*cos(d*x + c)^6 + 6*a^2*d*cos(d*x + c)^4 - 4*a^2*d*co s(d*x + c)^2 + a^2*d)
\[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cot ^{9}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {420 \, \sin \left (d x + c\right )^{6} - 560 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} + 672 \, \sin \left (d x + c\right )^{3} - 140 \, \sin \left (d x + c\right )^{2} - 240 \, \sin \left (d x + c\right ) + 105}{840 \, a^{2} d \sin \left (d x + c\right )^{8}} \]
-1/840*(420*sin(d*x + c)^6 - 560*sin(d*x + c)^5 - 210*sin(d*x + c)^4 + 672 *sin(d*x + c)^3 - 140*sin(d*x + c)^2 - 240*sin(d*x + c) + 105)/(a^2*d*sin( d*x + c)^8)
Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {420 \, \sin \left (d x + c\right )^{6} - 560 \, \sin \left (d x + c\right )^{5} - 210 \, \sin \left (d x + c\right )^{4} + 672 \, \sin \left (d x + c\right )^{3} - 140 \, \sin \left (d x + c\right )^{2} - 240 \, \sin \left (d x + c\right ) + 105}{840 \, a^{2} d \sin \left (d x + c\right )^{8}} \]
-1/840*(420*sin(d*x + c)^6 - 560*sin(d*x + c)^5 - 210*sin(d*x + c)^4 + 672 *sin(d*x + c)^3 - 140*sin(d*x + c)^2 - 240*sin(d*x + c) + 105)/(a^2*d*sin( d*x + c)^8)
Time = 6.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.60 \[ \int \frac {\cot ^9(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-420\,{\sin \left (c+d\,x\right )}^6+560\,{\sin \left (c+d\,x\right )}^5+210\,{\sin \left (c+d\,x\right )}^4-672\,{\sin \left (c+d\,x\right )}^3+140\,{\sin \left (c+d\,x\right )}^2+240\,\sin \left (c+d\,x\right )-105}{840\,a^2\,d\,{\sin \left (c+d\,x\right )}^8} \]