Integrand size = 21, antiderivative size = 165 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {1}{64 a^2 d (1-\cos (c+d x))^2}+\frac {9}{64 a^2 d (1-\cos (c+d x))}-\frac {1}{32 a^2 d (1+\cos (c+d x))^4}+\frac {11}{48 a^2 d (1+\cos (c+d x))^3}-\frac {3}{4 a^2 d (1+\cos (c+d x))^2}+\frac {51}{32 a^2 d (1+\cos (c+d x))}+\frac {29 \log (1-\cos (c+d x))}{128 a^2 d}+\frac {99 \log (1+\cos (c+d x))}{128 a^2 d} \] Output:
-1/64/a^2/d/(1-cos(d*x+c))^2+9/64/a^2/d/(1-cos(d*x+c))-1/32/a^2/d/(1+cos(d *x+c))^4+11/48/a^2/d/(1+cos(d*x+c))^3-3/4/a^2/d/(1+cos(d*x+c))^2+51/32/a^2 /d/(1+cos(d*x+c))+29/128*ln(1-cos(d*x+c))/a^2/d+99/128*ln(1+cos(d*x+c))/a^ 2/d
Time = 0.56 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-108 \csc ^2\left (\frac {1}{2} (c+d x)\right )+6 \csc ^4\left (\frac {1}{2} (c+d x)\right )-24 \left (99 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+29 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-1224 \sec ^2\left (\frac {1}{2} (c+d x)\right )+288 \sec ^4\left (\frac {1}{2} (c+d x)\right )-44 \sec ^6\left (\frac {1}{2} (c+d x)\right )+3 \sec ^8\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2(c+d x)}{384 a^2 d (1+\sec (c+d x))^2} \] Input:
Integrate[Cot[c + d*x]^5/(a + a*Sec[c + d*x])^2,x]
Output:
-1/384*(Cos[(c + d*x)/2]^4*(-108*Csc[(c + d*x)/2]^2 + 6*Csc[(c + d*x)/2]^4 - 24*(99*Log[Cos[(c + d*x)/2]] + 29*Log[Sin[(c + d*x)/2]]) - 1224*Sec[(c + d*x)/2]^2 + 288*Sec[(c + d*x)/2]^4 - 44*Sec[(c + d*x)/2]^6 + 3*Sec[(c + d*x)/2]^8)*Sec[c + d*x]^2)/(a^2*d*(1 + Sec[c + d*x])^2)
Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4367, 27, 99, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}dx\) |
| \(\Big \downarrow \) 4367 |
| \(\displaystyle -\frac {a^6 \int \frac {\cos ^7(c+d x)}{a^8 (1-\cos (c+d x))^3 (\cos (c+d x)+1)^5}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^7(c+d x)}{(1-\cos (c+d x))^3 (\cos (c+d x)+1)^5}d\cos (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (-\frac {99}{128 (\cos (c+d x)+1)}+\frac {51}{32 (\cos (c+d x)+1)^2}-\frac {3}{2 (\cos (c+d x)+1)^3}+\frac {11}{16 (\cos (c+d x)+1)^4}-\frac {1}{8 (\cos (c+d x)+1)^5}-\frac {29}{128 (\cos (c+d x)-1)}-\frac {9}{64 (\cos (c+d x)-1)^2}-\frac {1}{32 (\cos (c+d x)-1)^3}\right )d\cos (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {9}{64 (1-\cos (c+d x))}-\frac {51}{32 (\cos (c+d x)+1)}+\frac {1}{64 (1-\cos (c+d x))^2}+\frac {3}{4 (\cos (c+d x)+1)^2}-\frac {11}{48 (\cos (c+d x)+1)^3}+\frac {1}{32 (\cos (c+d x)+1)^4}-\frac {29}{128} \log (1-\cos (c+d x))-\frac {99}{128} \log (\cos (c+d x)+1)}{a^2 d}\) |
Input:
Int[Cot[c + d*x]^5/(a + a*Sec[c + d*x])^2,x]
Output:
-((1/(64*(1 - Cos[c + d*x])^2) - 9/(64*(1 - Cos[c + d*x])) + 1/(32*(1 + Co s[c + d*x])^4) - 11/(48*(1 + Cos[c + d*x])^3) + 3/(4*(1 + Cos[c + d*x])^2) - 51/(32*(1 + Cos[c + d*x])) - (29*Log[1 - Cos[c + d*x]])/128 - (99*Log[1 + Cos[c + d*x]])/128)/(a^2*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[1/(a^(m - n - 1)*b^n*d) Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.62
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{64 \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {9}{64 \left (-1+\cos \left (d x +c \right )\right )}+\frac {29 \ln \left (-1+\cos \left (d x +c \right )\right )}{128}-\frac {1}{32 \left (1+\cos \left (d x +c \right )\right )^{4}}+\frac {11}{48 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {51}{32 \left (1+\cos \left (d x +c \right )\right )}+\frac {99 \ln \left (1+\cos \left (d x +c \right )\right )}{128}}{d \,a^{2}}\) | \(103\) |
| default | \(\frac {-\frac {1}{64 \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {9}{64 \left (-1+\cos \left (d x +c \right )\right )}+\frac {29 \ln \left (-1+\cos \left (d x +c \right )\right )}{128}-\frac {1}{32 \left (1+\cos \left (d x +c \right )\right )^{4}}+\frac {11}{48 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {51}{32 \left (1+\cos \left (d x +c \right )\right )}+\frac {99 \ln \left (1+\cos \left (d x +c \right )\right )}{128}}{d \,a^{2}}\) | \(103\) |
| risch | \(-\frac {i x}{a^{2}}-\frac {2 i c}{d \,a^{2}}+\frac {279 \,{\mathrm e}^{11 i \left (d x +c \right )}+156 \,{\mathrm e}^{10 i \left (d x +c \right )}-1141 \,{\mathrm e}^{9 i \left (d x +c \right )}-2080 \,{\mathrm e}^{8 i \left (d x +c \right )}+670 \,{\mathrm e}^{7 i \left (d x +c \right )}+2696 \,{\mathrm e}^{6 i \left (d x +c \right )}+670 \,{\mathrm e}^{5 i \left (d x +c \right )}-2080 \,{\mathrm e}^{4 i \left (d x +c \right )}-1141 \,{\mathrm e}^{3 i \left (d x +c \right )}+156 \,{\mathrm e}^{2 i \left (d x +c \right )}+279 \,{\mathrm e}^{i \left (d x +c \right )}}{96 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4}}+\frac {99 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{64 d \,a^{2}}+\frac {29 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{64 d \,a^{2}}\) | \(215\) |
Input:
int(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(-1/64/(-1+cos(d*x+c))^2-9/64/(-1+cos(d*x+c))+29/128*ln(-1+cos(d*x +c))-1/32/(1+cos(d*x+c))^4+11/48/(1+cos(d*x+c))^3-3/4/(1+cos(d*x+c))^2+51/ 32/(1+cos(d*x+c))+99/128*ln(1+cos(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.72 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {558 \, \cos \left (d x + c\right )^{5} + 156 \, \cos \left (d x + c\right )^{4} - 1268 \, \cos \left (d x + c\right )^{3} - 676 \, \cos \left (d x + c\right )^{2} + 297 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 87 \, {\left (\cos \left (d x + c\right )^{6} + 2 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 686 \, \cos \left (d x + c\right ) + 448}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} + 2 \, a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{4} - 4 \, a^{2} d \cos \left (d x + c\right )^{3} - a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/384*(558*cos(d*x + c)^5 + 156*cos(d*x + c)^4 - 1268*cos(d*x + c)^3 - 676 *cos(d*x + c)^2 + 297*(cos(d*x + c)^6 + 2*cos(d*x + c)^5 - cos(d*x + c)^4 - 4*cos(d*x + c)^3 - cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 87*(cos(d*x + c)^6 + 2*cos(d*x + c)^5 - cos(d*x + c)^4 - 4*c os(d*x + c)^3 - cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 686*cos(d*x + c) + 448)/(a^2*d*cos(d*x + c)^6 + 2*a^2*d*cos(d*x + c)^5 - a^2*d*cos(d*x + c)^4 - 4*a^2*d*cos(d*x + c)^3 - a^2*d*cos(d*x + c )^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(cot(d*x+c)**5/(a+a*sec(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**5/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (279 \, \cos \left (d x + c\right )^{5} + 78 \, \cos \left (d x + c\right )^{4} - 634 \, \cos \left (d x + c\right )^{3} - 338 \, \cos \left (d x + c\right )^{2} + 343 \, \cos \left (d x + c\right ) + 224\right )}}{a^{2} \cos \left (d x + c\right )^{6} + 2 \, a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}} + \frac {297 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {87 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
Output:
1/384*(2*(279*cos(d*x + c)^5 + 78*cos(d*x + c)^4 - 634*cos(d*x + c)^3 - 33 8*cos(d*x + c)^2 + 343*cos(d*x + c) + 224)/(a^2*cos(d*x + c)^6 + 2*a^2*cos (d*x + c)^5 - a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^3 - a^2*cos(d*x + c) ^2 + 2*a^2*cos(d*x + c) + a^2) + 297*log(cos(d*x + c) + 1)/a^2 + 87*log(co s(d*x + c) - 1)/a^2)/d
Time = 0.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.70 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {99 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{128 \, a^{2} d} + \frac {29 \, \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{128 \, a^{2} d} + \frac {279 \, \cos \left (d x + c\right )^{5} + 78 \, \cos \left (d x + c\right )^{4} - 634 \, \cos \left (d x + c\right )^{3} - 338 \, \cos \left (d x + c\right )^{2} + 343 \, \cos \left (d x + c\right ) + 224}{192 \, a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )}^{4} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x, algorithm="giac")
Output:
99/128*log(abs(cos(d*x + c) + 1))/(a^2*d) + 29/128*log(abs(cos(d*x + c) - 1))/(a^2*d) + 1/192*(279*cos(d*x + c)^5 + 78*cos(d*x + c)^4 - 634*cos(d*x + c)^3 - 338*cos(d*x + c)^2 + 343*cos(d*x + c) + 224)/(a^2*d*(cos(d*x + c) + 1)^4*(cos(d*x + c) - 1)^2)
Time = 11.49 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,a^2\,d}-\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{256\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{512\,a^2\,d}+\frac {29\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{64\,a^2\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {1}{4}\right )}{64\,a^2\,d} \] Input:
int(cot(c + d*x)^5/(a + a/cos(c + d*x))^2,x)
Output:
tan(c/2 + (d*x)/2)^2/(2*a^2*d) - (29*tan(c/2 + (d*x)/2)^4)/(256*a^2*d) + t an(c/2 + (d*x)/2)^6/(48*a^2*d) - tan(c/2 + (d*x)/2)^8/(512*a^2*d) + (29*lo g(tan(c/2 + (d*x)/2)))/(64*a^2*d) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) + (cot(c/2 + (d*x)/2)^4*(4*tan(c/2 + (d*x)/2)^2 - 1/4))/(64*a^2*d)
Time = 1.47 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.82 \[ \int \frac {\cot ^5(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {-1536 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+696 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-174 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+768 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+96 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6}{1536 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a^{2} d} \] Input:
int(cot(d*x+c)^5/(a+a*sec(d*x+c))^2,x)
Output:
( - 1536*log(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)**4 + 696*log(tan((c + d*x)/2))*tan((c + d*x)/2)**4 - 3*tan((c + d*x)/2)**12 + 32*tan((c + d*x )/2)**10 - 174*tan((c + d*x)/2)**8 + 768*tan((c + d*x)/2)**6 + 96*tan((c + d*x)/2)**2 - 6)/(1536*tan((c + d*x)/2)**4*a**2*d)