Integrand size = 21, antiderivative size = 145 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {1}{32 a d (1-\cos (c+d x))^2}+\frac {1}{4 a d (1-\cos (c+d x))}+\frac {1}{24 a d (1+\cos (c+d x))^3}-\frac {9}{32 a d (1+\cos (c+d x))^2}+\frac {15}{16 a d (1+\cos (c+d x))}+\frac {11 \log (1-\cos (c+d x))}{32 a d}+\frac {21 \log (1+\cos (c+d x))}{32 a d} \] Output:
-1/32/a/d/(1-cos(d*x+c))^2+1/4/a/d/(1-cos(d*x+c))+1/24/a/d/(1+cos(d*x+c))^ 3-9/32/a/d/(1+cos(d*x+c))^2+15/16/a/d/(1+cos(d*x+c))+11/32*ln(1-cos(d*x+c) )/a/d+21/32*ln(1+cos(d*x+c))/a/d
Time = 0.35 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-48 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3 \csc ^4\left (\frac {1}{2} (c+d x)\right )-504 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-264 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-180 \sec ^2\left (\frac {1}{2} (c+d x)\right )+27 \sec ^4\left (\frac {1}{2} (c+d x)\right )-2 \sec ^6\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{192 a d (1+\sec (c+d x))} \] Input:
Integrate[Cot[c + d*x]^5/(a + a*Sec[c + d*x]),x]
Output:
-1/192*(Cos[(c + d*x)/2]^2*(-48*Csc[(c + d*x)/2]^2 + 3*Csc[(c + d*x)/2]^4 - 504*Log[Cos[(c + d*x)/2]] - 264*Log[Sin[(c + d*x)/2]] - 180*Sec[(c + d*x )/2]^2 + 27*Sec[(c + d*x)/2]^4 - 2*Sec[(c + d*x)/2]^6)*Sec[c + d*x])/(a*d* (1 + Sec[c + d*x]))
Time = 0.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77, 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.
\(\displaystyle \int \frac {\cot ^5(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {a^6 \int \frac {\cos ^6(c+d x)}{a^7 (1-\cos (c+d x))^3 (\cos (c+d x)+1)^4}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^6(c+d x)}{(1-\cos (c+d x))^3 (\cos (c+d x)+1)^4}d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {21}{32 (\cos (c+d x)+1)}+\frac {15}{16 (\cos (c+d x)+1)^2}-\frac {9}{16 (\cos (c+d x)+1)^3}+\frac {1}{8 (\cos (c+d x)+1)^4}-\frac {11}{32 (\cos (c+d x)-1)}-\frac {1}{4 (\cos (c+d x)-1)^2}-\frac {1}{16 (\cos (c+d x)-1)^3}\right )d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{4 (1-\cos (c+d x))}-\frac {15}{16 (\cos (c+d x)+1)}+\frac {1}{32 (1-\cos (c+d x))^2}+\frac {9}{32 (\cos (c+d x)+1)^2}-\frac {1}{24 (\cos (c+d x)+1)^3}-\frac {11}{32} \log (1-\cos (c+d x))-\frac {21}{32} \log (\cos (c+d x)+1)}{a d}\) |
Input:
Int[Cot[c + d*x]^5/(a + a*Sec[c + d*x]),x]
Output:
-((1/(32*(1 - Cos[c + d*x])^2) - 1/(4*(1 - Cos[c + d*x])) - 1/(24*(1 + Cos [c + d*x])^3) + 9/(32*(1 + Cos[c + d*x])^2) - 15/(16*(1 + Cos[c + d*x])) - (11*Log[1 - Cos[c + d*x]])/32 - (21*Log[1 + Cos[c + d*x]])/32)/(a*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 = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(\frac {\frac {1}{24 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {9}{32 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {15}{16 \left (1+\cos \left (d x +c \right )\right )}+\frac {21 \ln \left (1+\cos \left (d x +c \right )\right )}{32}-\frac {1}{32 \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (-1+\cos \left (d x +c \right )\right )}+\frac {11 \ln \left (-1+\cos \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
default | \(\frac {\frac {1}{24 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {9}{32 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {15}{16 \left (1+\cos \left (d x +c \right )\right )}+\frac {21 \ln \left (1+\cos \left (d x +c \right )\right )}{32}-\frac {1}{32 \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (-1+\cos \left (d x +c \right )\right )}+\frac {11 \ln \left (-1+\cos \left (d x +c \right )\right )}{32}}{d a}\) | \(91\) |
risch | \(-\frac {i x}{a}-\frac {2 i c}{d a}+\frac {33 \,{\mathrm e}^{9 i \left (d x +c \right )}-78 \,{\mathrm e}^{8 i \left (d x +c \right )}-184 \,{\mathrm e}^{7 i \left (d x +c \right )}-2 \,{\mathrm e}^{6 i \left (d x +c \right )}+270 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{4 i \left (d x +c \right )}-184 \,{\mathrm e}^{3 i \left (d x +c \right )}-78 \,{\mathrm e}^{2 i \left (d x +c \right )}+33 \,{\mathrm e}^{i \left (d x +c \right )}}{24 d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{4}}+\frac {11 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{16 d a}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{16 d a}\) | \(193\) |
Input:
int(cot(d*x+c)^5/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/24/(1+cos(d*x+c))^3-9/32/(1+cos(d*x+c))^2+15/16/(1+cos(d*x+c))+21 /32*ln(1+cos(d*x+c))-1/32/(-1+cos(d*x+c))^2-1/4/(-1+cos(d*x+c))+11/32*ln(- 1+cos(d*x+c)))
Time = 0.10 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.50 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {66 \, \cos \left (d x + c\right )^{4} - 78 \, \cos \left (d x + c\right )^{3} - 158 \, \cos \left (d x + c\right )^{2} + 63 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 33 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 58 \, \cos \left (d x + c\right ) + 88}{96 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4} - 2 \, a d \cos \left (d x + c\right )^{3} - 2 \, a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right ) + a d\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
1/96*(66*cos(d*x + c)^4 - 78*cos(d*x + c)^3 - 158*cos(d*x + c)^2 + 63*(cos (d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d *x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 33*(cos(d*x + c)^5 + cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + cos(d*x + c) + 1)*log(-1/2*co s(d*x + c) + 1/2) + 58*cos(d*x + c) + 88)/(a*d*cos(d*x + c)^5 + a*d*cos(d* x + c)^4 - 2*a*d*cos(d*x + c)^3 - 2*a*d*cos(d*x + c)^2 + a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\cot ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**5/(a+a*sec(d*x+c)),x)
Output:
Integral(cot(c + d*x)**5/(sec(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.90 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\frac {2 \, {\left (33 \, \cos \left (d x + c\right )^{4} - 39 \, \cos \left (d x + c\right )^{3} - 79 \, \cos \left (d x + c\right )^{2} + 29 \, \cos \left (d x + c\right ) + 44\right )}}{a \cos \left (d x + c\right )^{5} + a \cos \left (d x + c\right )^{4} - 2 \, a \cos \left (d x + c\right )^{3} - 2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) + a} + \frac {63 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {33 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
1/96*(2*(33*cos(d*x + c)^4 - 39*cos(d*x + c)^3 - 79*cos(d*x + c)^2 + 29*co s(d*x + c) + 44)/(a*cos(d*x + c)^5 + a*cos(d*x + c)^4 - 2*a*cos(d*x + c)^3 - 2*a*cos(d*x + c)^2 + a*cos(d*x + c) + a) + 63*log(cos(d*x + c) + 1)/a + 33*log(cos(d*x + c) - 1)/a)/d
Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {21 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{32 \, a d} + \frac {11 \, \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{32 \, a d} + \frac {33 \, \cos \left (d x + c\right )^{4} - 39 \, \cos \left (d x + c\right )^{3} - 79 \, \cos \left (d x + c\right )^{2} + 29 \, \cos \left (d x + c\right ) + 44}{48 \, a d {\left (\cos \left (d x + c\right ) + 1\right )}^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(cot(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
21/32*log(abs(cos(d*x + c) + 1))/(a*d) + 11/32*log(abs(cos(d*x + c) - 1))/ (a*d) + 1/48*(33*cos(d*x + c)^4 - 39*cos(d*x + c)^3 - 79*cos(d*x + c)^2 + 29*cos(d*x + c) + 44)/(a*d*(cos(d*x + c) + 1)^3*(cos(d*x + c) - 1)^2)
Time = 12.66 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32\,a\,d}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{128\,a\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{192\,a\,d}+\frac {11\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{16\,a\,d}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {1}{4}\right )}{32\,a\,d} \] Input:
int(cot(c + d*x)^5/(a + a/cos(c + d*x)),x)
Output:
(11*tan(c/2 + (d*x)/2)^2)/(32*a*d) - (7*tan(c/2 + (d*x)/2)^4)/(128*a*d) + tan(c/2 + (d*x)/2)^6/(192*a*d) + (11*log(tan(c/2 + (d*x)/2)))/(16*a*d) - l og(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) + (cot(c/2 + (d*x)/2)^4*((7*tan(c/2 + ( d*x)/2)^2)/2 - 1/4))/(32*a*d)
Time = 0.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^5(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {-384 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+264 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3}{384 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a d} \] Input:
int(cot(d*x+c)^5/(a+a*sec(d*x+c)),x)
Output:
( - 384*log(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)**4 + 264*log(tan((c + d*x)/2))*tan((c + d*x)/2)**4 + 2*tan((c + d*x)/2)**10 - 21*tan((c + d*x) /2)**8 + 132*tan((c + d*x)/2)**6 + 42*tan((c + d*x)/2)**2 - 3)/(384*tan((c + d*x)/2)**4*a*d)