Integrand size = 19, antiderivative size = 101 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {a \cot ^5(c+d x)}{5 d}-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d}-\frac {a \csc ^5(c+d x)}{5 d} \]
a*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-2/3*a*cot(d*x+c)^3/d-1/5*a*cot(d*x+ c)^5/d-a*csc(d*x+c)/d-1/3*a*csc(d*x+c)^3/d-1/5*a*csc(d*x+c)^5/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {8 a \cot (c+d x)}{15 d}-\frac {4 a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}-\frac {a \csc ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\sin ^2(c+d x)\right )}{5 d} \]
(-8*a*Cot[c + d*x])/(15*d) - (4*a*Cot[c + d*x]*Csc[c + d*x]^2)/(15*d) - (a *Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) - (a*Csc[c + d*x]^5*Hypergeometric2F1[ -5/2, 1, -3/2, Sin[c + d*x]^2])/(5*d)
Time = 0.48 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.789, Rules used = {3042, 4360, 25, 25, 3042, 25, 3317, 25, 3042, 3101, 25, 254, 2009, 4254, 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.
\(\displaystyle \int \csc ^6(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-a \csc \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^6}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\left (\csc ^6(c+d x) \sec (c+d x) (a (-\cos (c+d x))-a)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\left ((\cos (c+d x) a+a) \csc ^6(c+d x) \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \csc ^6(c+d x) \sec (c+d x) (a \cos (c+d x)+a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {a-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sin \left (c+d x-\frac {\pi }{2}\right ) \cos \left (c+d x-\frac {\pi }{2}\right )^6}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^6 \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \csc ^6(c+d x)dx-a \int -\csc ^6(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \csc ^6(c+d x)dx+a \int \csc ^6(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc (c+d x)^6dx+a \int \csc (c+d x)^6 \sec (c+d x)dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle a \int \csc (c+d x)^6dx-\frac {a \int -\frac {\csc ^6(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a \int \frac {\csc ^6(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}+a \int \csc (c+d x)^6dx\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {a \int \left (-\csc ^4(c+d x)-\csc ^2(c+d x)+\frac {1}{1-\csc ^2(c+d x)}-1\right )d\csc (c+d x)}{d}+a \int \csc (c+d x)^6dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \csc (c+d x)^6dx-\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {a \int \left (\cot ^4(c+d x)+2 \cot ^2(c+d x)+1\right )d\cot (c+d x)}{d}-\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}-\frac {a \left (\frac {1}{5} \cot ^5(c+d x)+\frac {2}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\) |
-((a*(Cot[c + d*x] + (2*Cot[c + d*x]^3)/3 + Cot[c + d*x]^5/5))/d) - (a*(-A rcTanh[Csc[c + d*x]] + Csc[c + d*x] + Csc[c + d*x]^3/3 + Csc[c + d*x]^5/5) )/d
3.1.16.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/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 = 0.89 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(83\) |
default | \(\frac {a \left (-\frac {1}{5 \sin \left (d x +c \right )^{5}}-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\csc \left (d x +c \right )^{4}}{5}-\frac {4 \csc \left (d x +c \right )^{2}}{15}\right ) \cot \left (d x +c \right )}{d}\) | \(83\) |
parallelrisch | \(-\frac {\left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+80 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-80 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )\right ) a}{80 d}\) | \(95\) |
norman | \(\frac {-\frac {a}{80 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{48 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(124\) |
risch | \(-\frac {2 i a \left (15 \,{\mathrm e}^{7 i \left (d x +c \right )}-30 \,{\mathrm e}^{6 i \left (d x +c \right )}-35 \,{\mathrm e}^{5 i \left (d x +c \right )}+100 \,{\mathrm e}^{4 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-46 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+8\right )}{15 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\) | \(151\) |
1/d*(a*(-1/5/sin(d*x+c)^5-1/3/sin(d*x+c)^3-1/sin(d*x+c)+ln(sec(d*x+c)+tan( d*x+c)))+a*(-8/15-1/5*csc(d*x+c)^4-4/15*csc(d*x+c)^2)*cot(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (93) = 186\).
Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.88 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {16 \, a \cos \left (d x + c\right )^{4} + 14 \, a \cos \left (d x + c\right )^{3} - 54 \, a \cos \left (d x + c\right )^{2} - 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 15 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 16 \, a \cos \left (d x + c\right ) + 46 \, a}{30 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right )} \]
-1/30*(16*a*cos(d*x + c)^4 + 14*a*cos(d*x + c)^3 - 54*a*cos(d*x + c)^2 - 1 5*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*log(sin(d*x + c) + 1)*sin(d*x + c) + 15*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d* x + c) + a)*log(-sin(d*x + c) + 1)*sin(d*x + c) - 16*a*cos(d*x + c) + 46*a )/((d*cos(d*x + c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d)*sin(d*x + c) )
\[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \csc ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{6}{\left (c + d x \right )}\, dx\right ) \]
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.95 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {a {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a}{\tan \left (d x + c\right )^{5}}}{30 \, d} \]
-1/30*(a*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2 + 3)/sin(d*x + c)^5 - 15 *log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 2*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a/tan(d*x + c)^5)/d
Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {5 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 240 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 90 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, {\left (80 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{240 \, d} \]
-1/240*(5*a*tan(1/2*d*x + 1/2*c)^3 - 240*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 240*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 90*a*tan(1/2*d*x + 1/2*c) + 3*(80*a*tan(1/2*d*x + 1/2*c)^4 + 10*a*tan(1/2*d*x + 1/2*c)^2 + a)/tan(1 /2*d*x + 1/2*c)^5)/d
Time = 14.53 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \csc ^6(c+d x) (a+a \sec (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{5}\right )}{16\,d} \]