Integrand size = 21, antiderivative size = 73 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cot ^3(c+d x)}{3 a^2 d}-\frac {2 \cot ^5(c+d x)}{5 a^2 d}-\frac {2 \csc ^3(c+d x)}{3 a^2 d}+\frac {2 \csc ^5(c+d x)}{5 a^2 d} \]
-1/3*cot(d*x+c)^3/a^2/d-2/5*cot(d*x+c)^5/a^2/d-2/3*csc(d*x+c)^3/a^2/d+2/5* csc(d*x+c)^5/a^2/d
Time = 0.73 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.44 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\csc (c) \csc (c+d x) \sec ^2(c+d x) (-80 \sin (c)+80 \sin (d x)+55 \sin (c+d x)+44 \sin (2 (c+d x))+11 \sin (3 (c+d x))-60 \sin (2 c+d x)+16 \sin (c+2 d x)+4 \sin (2 c+3 d x))}{240 a^2 d (1+\sec (c+d x))^2} \]
(Csc[c]*Csc[c + d*x]*Sec[c + d*x]^2*(-80*Sin[c] + 80*Sin[d*x] + 55*Sin[c + d*x] + 44*Sin[2*(c + d*x)] + 11*Sin[3*(c + d*x)] - 60*Sin[2*c + d*x] + 16 *Sin[c + 2*d*x] + 4*Sin[2*c + 3*d*x]))/(240*a^2*d*(1 + Sec[c + d*x])^2)
Time = 0.46 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4359, 3042, 3190, 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.
\(\displaystyle \int \frac {\csc ^2(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4359 |
\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan \left (c+d x-\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2}dx\) |
\(\Big \downarrow \) 3190 |
\(\displaystyle \frac {\int \left (a^2 \csc ^2(c+d x) \cot ^4(c+d x)-2 a^2 \csc ^3(c+d x) \cot ^3(c+d x)+a^2 \csc ^4(c+d x) \cot ^2(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a^2 \cot ^5(c+d x)}{5 d}-\frac {a^2 \cot ^3(c+d x)}{3 d}+\frac {2 a^2 \csc ^5(c+d x)}{5 d}-\frac {2 a^2 \csc ^3(c+d x)}{3 d}}{a^4}\) |
(-1/3*(a^2*Cot[c + d*x]^3)/d - (2*a^2*Cot[c + d*x]^5)/(5*d) - (2*a^2*Csc[c + d*x]^3)/(3*d) + (2*a^2*Csc[c + d*x]^5)/(5*d))/a^4
3.1.87.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x _)])^(p_.), x_Symbol] :> Simp[a^(2*m) Int[ExpandIntegrand[(g*Tan[e + f*x] )^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x] /; F reeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _.), x_Symbol] :> Int[Cot[e + f*x]^p*(b + a*Sin[e + f*x])^m, x] /; FreeQ[{a , b, e, f, p}, x] && IntegerQ[m] && EqQ[m, p]
Time = 0.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79
method | result | size |
parallelrisch | \(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-15 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{120 a^{2} d}\) | \(58\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{2}}\) | \(60\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{8 d \,a^{2}}\) | \(60\) |
norman | \(\frac {-\frac {1}{8 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{24 d a}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{40 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a}\) | \(82\) |
risch | \(-\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+20 \,{\mathrm e}^{3 i \left (d x +c \right )}+20 \,{\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{15 a^{2} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}\) | \(82\) |
1/120*(3*tan(1/2*d*x+1/2*c)^5-5*tan(1/2*d*x+1/2*c)^3-15*cot(1/2*d*x+1/2*c) -15*tan(1/2*d*x+1/2*c))/a^2/d
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )^{2} + 8 \, \cos \left (d x + c\right ) + 4}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )} \]
-1/15*(cos(d*x + c)^3 + 2*cos(d*x + c)^2 + 8*cos(d*x + c) + 4)/((a^2*d*cos (d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c))
\[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\csc ^{2}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{2}} + \frac {15 \, {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} \sin \left (d x + c\right )}}{120 \, d} \]
-1/120*((15*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^2 + 15*(cos(d*x + c) + 1)/(a^2*sin(d*x + c)))/d
Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {15}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {3 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{10}}}{120 \, d} \]
-1/120*(15/(a^2*tan(1/2*d*x + 1/2*c)) - (3*a^8*tan(1/2*d*x + 1/2*c)^5 - 5* a^8*tan(1/2*d*x + 1/2*c)^3 - 15*a^8*tan(1/2*d*x + 1/2*c))/a^10)/d
Time = 13.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {\csc ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+14\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-3}{120\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]