Integrand size = 21, antiderivative size = 55 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\cot ^3(c+d x)}{3 a d}+\frac {\cot ^5(c+d x)}{5 a d}-\frac {\csc ^5(c+d x)}{5 a d} \]
Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(55)=110\).
Time = 0.68 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.11 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\csc (c) \csc ^3(c+d x) \sec (c+d x) (240 \sin (c)-96 \sin (d x)-54 \sin (c+d x)-18 \sin (2 (c+d x))+18 \sin (3 (c+d x))+9 \sin (4 (c+d x))-32 \sin (c+2 d x)+32 \sin (2 c+3 d x)+16 \sin (3 c+4 d x))}{960 a d (1+\sec (c+d x))} \]
-1/960*(Csc[c]*Csc[c + d*x]^3*Sec[c + d*x]*(240*Sin[c] - 96*Sin[d*x] - 54* Sin[c + d*x] - 18*Sin[2*(c + d*x)] + 18*Sin[3*(c + d*x)] + 9*Sin[4*(c + d* x)] - 32*Sin[c + 2*d*x] + 32*Sin[2*c + 3*d*x] + 16*Sin[3*c + 4*d*x]))/(a*d *(1 + Sec[c + d*x]))
Time = 0.50 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4360, 25, 25, 3042, 25, 3318, 25, 3042, 25, 3086, 15, 3087, 244, 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.
\(\displaystyle \int \frac {\csc ^4(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\cot (c+d x) \csc ^3(c+d x)}{a (-\cos (c+d x))-a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cot (c+d x) \csc ^3(c+d x)}{\cos (c+d x) a+a}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\cot (c+d x) \csc ^3(c+d x)}{a \cos (c+d x)+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )}dx\) |
\(\Big \downarrow \) 3318 |
\(\displaystyle -\frac {\int \cot ^2(c+d x) \csc ^4(c+d x)dx}{a}-\frac {\int -\cot (c+d x) \csc ^5(c+d x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \cot (c+d x) \csc ^5(c+d x)dx}{a}-\frac {\int \cot ^2(c+d x) \csc ^4(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec \left (c+d x-\frac {\pi }{2}\right )^5 \tan \left (c+d x-\frac {\pi }{2}\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^5 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )dx}{a}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -\frac {\int \csc ^4(c+d x)d\csc (c+d x)}{a d}-\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\int \sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^2dx}{a}-\frac {\csc ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle -\frac {\int \cot ^2(c+d x) \left (\cot ^2(c+d x)+1\right )d(-\cot (c+d x))}{a d}-\frac {\csc ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (\cot ^4(c+d x)+\cot ^2(c+d x)\right )d(-\cot (c+d x))}{a d}-\frac {\csc ^5(c+d x)}{5 a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} \cot ^5(c+d x)-\frac {1}{3} \cot ^3(c+d x)}{a d}-\frac {\csc ^5(c+d x)}{5 a d}\) |
3.1.70.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, 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[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 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 = 0.53 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-30 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d a}\) | \(60\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{16 d a}\) | \(62\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}}{16 d a}\) | \(62\) |
norman | \(\frac {-\frac {1}{48 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{24 d a}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{80 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}\) | \(79\) |
risch | \(\frac {4 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{3 i \left (d x +c \right )}+2 \,{\mathrm e}^{2 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}-1\right )}{15 a d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3}}\) | \(82\) |
1/240*(-3*tan(1/2*d*x+1/2*c)^5-5*cot(1/2*d*x+1/2*c)^3-10*tan(1/2*d*x+1/2*c )^3-30*cot(1/2*d*x+1/2*c))/d/a
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.62 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 3}{15 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )} \sin \left (d x + c\right )} \]
-1/15*(2*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 3*cos(d*x + c)^2 - 3*cos(d*x + c) - 3)/((a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a *d)*sin(d*x + c))
\[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\csc ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.75 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {\frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a} + \frac {5 \, {\left (\frac {6 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{3}}{a \sin \left (d x + c\right )^{3}}}{240 \, d} \]
-1/240*((10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d* x + c) + 1)^5)/a + 5*(6*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)^3/(a*sin(d*x + c)^3))/d
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {5 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{a^{5}}}{240 \, d} \]
-1/240*(5*(6*tan(1/2*d*x + 1/2*c)^2 + 1)/(a*tan(1/2*d*x + 1/2*c)^3) + (3*a ^4*tan(1/2*d*x + 1/2*c)^5 + 10*a^4*tan(1/2*d*x + 1/2*c)^3)/a^5)/d
Time = 13.99 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09 \[ \int \frac {\csc ^4(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+5}{240\,a\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]