Integrand size = 21, antiderivative size = 68 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cot ^6(c+d x)}{6 a d}+\frac {\csc (c+d x)}{a d}-\frac {2 \csc ^3(c+d x)}{3 a d}+\frac {\csc ^5(c+d x)}{5 a d} \]
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\csc ^6(c+d x) (-15 \cos (4 (c+d x))+78 \sin (c+d x)-5 (5+7 \sin (3 (c+d x))-3 \sin (5 (c+d x))))}{240 a d} \]
(Csc[c + d*x]^6*(-15*Cos[4*(c + d*x)] + 78*Sin[c + d*x] - 5*(5 + 7*Sin[3*( c + d*x)] - 3*Sin[5*(c + d*x)])))/(240*a*d)
Time = 0.41 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3185, 3042, 25, 3086, 210, 2009, 3087, 15}
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.
\(\displaystyle \int \frac {\cot ^7(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^7 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int \cot ^5(c+d x) \csc ^2(c+d x)dx}{a}-\frac {\int \cot ^5(c+d x) \csc (c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}-\frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^5dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right ) \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \left (\csc ^2(c+d x)-1\right )^2d\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \left (\csc ^4(c+d x)-2 \csc ^2(c+d x)+1\right )d\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)}{a d}-\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)}{a d}-\frac {\int -\cot ^5(c+d x)d(-\cot (c+d x))}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)}{a d}-\frac {\cot ^6(c+d x)}{6 a d}\) |
3.1.51.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Time = 2.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )}-\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{6 \sin \left (d x +c \right )^{6}}}{d a}\) | \(67\) |
default | \(\frac {\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {2}{3 \sin \left (d x +c \right )^{3}}+\frac {1}{2 \sin \left (d x +c \right )^{4}}+\frac {1}{\sin \left (d x +c \right )}-\frac {1}{2 \sin \left (d x +c \right )^{2}}-\frac {1}{6 \sin \left (d x +c \right )^{6}}}{d a}\) | \(67\) |
risch | \(\frac {2 i \left (-15 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}-35 \,{\mathrm e}^{9 i \left (d x +c \right )}-50 i {\mathrm e}^{6 i \left (d x +c \right )}+78 \,{\mathrm e}^{7 i \left (d x +c \right )}-78 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 i {\mathrm e}^{2 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}\) | \(126\) |
1/d/a*(1/5/sin(d*x+c)^5-2/3/sin(d*x+c)^3+1/2/sin(d*x+c)^4+1/sin(d*x+c)-1/2 /sin(d*x+c)^2-1/6/sin(d*x+c)^6)
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.41 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {15 \, \cos \left (d x + c\right )^{4} - 15 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 5}{30 \, {\left (a d \cos \left (d x + c\right )^{6} - 3 \, a d \cos \left (d x + c\right )^{4} + 3 \, a d \cos \left (d x + c\right )^{2} - a d\right )}} \]
1/30*(15*cos(d*x + c)^4 - 15*cos(d*x + c)^2 - 2*(15*cos(d*x + c)^4 - 20*co s(d*x + c)^2 + 8)*sin(d*x + c) + 5)/(a*d*cos(d*x + c)^6 - 3*a*d*cos(d*x + c)^4 + 3*a*d*cos(d*x + c)^2 - a*d)
\[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\cot ^{7}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 5}{30 \, a d \sin \left (d x + c\right )^{6}} \]
1/30*(30*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 20*sin(d*x + c)^3 + 15*sin(d *x + c)^2 + 6*sin(d*x + c) - 5)/(a*d*sin(d*x + c)^6)
Time = 0.60 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {30 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} - 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} + 6 \, \sin \left (d x + c\right ) - 5}{30 \, a d \sin \left (d x + c\right )^{6}} \]
1/30*(30*sin(d*x + c)^5 - 15*sin(d*x + c)^4 - 20*sin(d*x + c)^3 + 15*sin(d *x + c)^2 + 6*sin(d*x + c) - 5)/(a*d*sin(d*x + c)^6)
Time = 6.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^5-\frac {{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{5}-\frac {1}{6}}{a\,d\,{\sin \left (c+d\,x\right )}^6} \]