Integrand size = 21, antiderivative size = 103 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \cot ^5(c+d x)}{5 a^3 d}+\frac {\cot ^7(c+d x)}{a^3 d}+\frac {4 \cot ^9(c+d x)}{9 a^3 d}-\frac {3 \csc ^5(c+d x)}{5 a^3 d}+\frac {\csc ^7(c+d x)}{a^3 d}-\frac {4 \csc ^9(c+d x)}{9 a^3 d} \]
3/5*cot(d*x+c)^5/a^3/d+cot(d*x+c)^7/a^3/d+4/9*cot(d*x+c)^9/a^3/d-3/5*csc(d *x+c)^5/a^3/d+csc(d*x+c)^7/a^3/d-4/9*csc(d*x+c)^9/a^3/d
Time = 1.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.70 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\csc (c) \csc ^3(2 (c+d x)) (5376 \sin (c)-1152 \sin (d x)-1764 \sin (c+d x)-1323 \sin (2 (c+d x))+98 \sin (3 (c+d x))+588 \sin (4 (c+d x))+294 \sin (5 (c+d x))+49 \sin (6 (c+d x))+3456 \sin (2 c+d x)-1152 \sin (c+2 d x)+2880 \sin (3 c+2 d x)-128 \sin (2 c+3 d x)-768 \sin (3 c+4 d x)-384 \sin (4 c+5 d x)-64 \sin (5 c+6 d x))}{5760 a^3 d (1+\sec (c+d x))^3} \]
-1/5760*(Csc[c]*Csc[2*(c + d*x)]^3*(5376*Sin[c] - 1152*Sin[d*x] - 1764*Sin [c + d*x] - 1323*Sin[2*(c + d*x)] + 98*Sin[3*(c + d*x)] + 588*Sin[4*(c + d *x)] + 294*Sin[5*(c + d*x)] + 49*Sin[6*(c + d*x)] + 3456*Sin[2*c + d*x] - 1152*Sin[c + 2*d*x] + 2880*Sin[3*c + 2*d*x] - 128*Sin[2*c + 3*d*x] - 768*S in[3*c + 4*d*x] - 384*Sin[4*c + 5*d*x] - 64*Sin[5*c + 6*d*x]))/(a^3*d*(1 + Sec[c + d*x])^3)
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.
| \(\displaystyle \int \frac {\csc ^4(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\cot ^3(c+d x) \csc (c+d x)}{(a (-\cos (c+d x))-a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cot ^3(c+d x) \csc (c+d x)}{(\cos (c+d x) a+a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\cot ^3(c+d x) \csc (c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle -\frac {\int -(a-a \cos (c+d x))^3 \cot ^3(c+d x) \csc ^7(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int (a-a \cos (c+d x))^3 \cot ^3(c+d x) \csc ^7(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right )^{10}}dx}{a^6}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \left (\sin \left (\frac {1}{2} (2 c-\pi )+d x\right ) a+a\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^{10}}dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle -\frac {\int \left (a^3 \cos ^6(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )-3 a^3 \cos ^5(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )+3 a^3 \cos ^4(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )-a^3 \cos ^3(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^3 \int \cos ^6(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )dx-3 a^3 \int \cos ^5(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )dx+3 a^3 \int \cos ^4(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )dx+a^3 \left (-\int \cos ^3(c+d x) \sec ^{10}\left (\frac {1}{2} (2 c-\pi )+d x\right )dx\right )}{a^6}\) |
3.2.5.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 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 = 0.81 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.58
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{64 d \,a^{3}}\) | \(60\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{64 d \,a^{3}}\) | \(60\) |
| parallelrisch | \(\frac {-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-135 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-15}{2880 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}\) | \(61\) |
| norman | \(\frac {-\frac {1}{192 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{576 a d}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{64 d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{320 d a}}{a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}\) | \(82\) |
| risch | \(\frac {4 i \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}+54 \,{\mathrm e}^{7 i \left (d x +c \right )}+84 \,{\mathrm e}^{6 i \left (d x +c \right )}+18 \,{\mathrm e}^{5 i \left (d x +c \right )}+18 \,{\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}+12 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{45 a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3}}\) | \(126\) |
1/64/d/a^3*(-1/9*tan(1/2*d*x+1/2*c)^9+3/5*tan(1/2*d*x+1/2*c)^5-3*tan(1/2*d *x+1/2*c)-1/3/tan(1/2*d*x+1/2*c)^3)
Time = 0.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.42 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {2 \, \cos \left (d x + c\right )^{6} + 6 \, \cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} - 7 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2}{45 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} - 3 \, a^{3} d \cos \left (d x + c\right ) - a^{3} d\right )} \sin \left (d x + c\right )} \]
1/45*(2*cos(d*x + c)^6 + 6*cos(d*x + c)^5 + 3*cos(d*x + c)^4 - 7*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)/((a^3*d*cos(d*x + c)^5 + 3* a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x + c)^2 - 3 *a^3*d*cos(d*x + c) - a^3*d)*sin(d*x + c))
\[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\csc ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {\frac {135 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {27 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{3}} + \frac {15 \, {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a^{3} \sin \left (d x + c\right )^{3}}}{2880 \, d} \]
-1/2880*((135*sin(d*x + c)/(cos(d*x + c) + 1) - 27*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^3 + 15*(cos(d*x + c) + 1)^3/(a^3*sin(d*x + c)^3))/d
Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {15}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {5 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 27 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 135 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{27}}}{2880 \, d} \]
-1/2880*(15/(a^3*tan(1/2*d*x + 1/2*c)^3) + (5*a^24*tan(1/2*d*x + 1/2*c)^9 - 27*a^24*tan(1/2*d*x + 1/2*c)^5 + 135*a^24*tan(1/2*d*x + 1/2*c))/a^27)/d
Time = 13.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {15\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+135\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-27\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{2880\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]