Integrand size = 21, antiderivative size = 103 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {1}{8 a d (1-\cos (c+d x))}+\frac {1}{8 a d (1+\cos (c+d x))^2}-\frac {3}{4 a d (1+\cos (c+d x))}-\frac {5 \log (1-\cos (c+d x))}{16 a d}-\frac {11 \log (1+\cos (c+d x))}{16 a d} \] Output:
-1/8/a/d/(1-cos(d*x+c))+1/8/a/d/(1+cos(d*x+c))^2-3/4/a/d/(1+cos(d*x+c))-5/ 16*ln(1-cos(d*x+c))/a/d-11/16*ln(1+cos(d*x+c))/a/d
Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\left (12+2 \cot ^2\left (\frac {1}{2} (c+d x)\right )+4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (11 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x)}{16 a d (1+\sec (c+d x))} \] Input:
Integrate[Cot[c + d*x]^3/(a + a*Sec[c + d*x]),x]
Output:
-1/16*((12 + 2*Cot[(c + d*x)/2]^2 + 4*Cos[(c + d*x)/2]^2*(11*Log[Cos[(c + d*x)/2]] + 5*Log[Sin[(c + d*x)/2]]) - Sec[(c + d*x)/2]^2)*Sec[c + d*x])/(a *d*(1 + Sec[c + d*x]))
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79, 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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {a^4 \int \frac {\cos ^4(c+d x)}{a^5 (1-\cos (c+d x))^2 (\cos (c+d x)+1)^3}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^4(c+d x)}{(1-\cos (c+d x))^2 (\cos (c+d x)+1)^3}d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\frac {11}{16 (\cos (c+d x)+1)}-\frac {3}{4 (\cos (c+d x)+1)^2}+\frac {1}{4 (\cos (c+d x)+1)^3}+\frac {5}{16 (\cos (c+d x)-1)}+\frac {1}{8 (\cos (c+d x)-1)^2}\right )d\cos (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {1}{8 (1-\cos (c+d x))}+\frac {3}{4 (\cos (c+d x)+1)}-\frac {1}{8 (\cos (c+d x)+1)^2}+\frac {5}{16} \log (1-\cos (c+d x))+\frac {11}{16} \log (\cos (c+d x)+1)}{a d}\) |
Input:
Int[Cot[c + d*x]^3/(a + a*Sec[c + d*x]),x]
Output:
-((1/(8*(1 - Cos[c + d*x])) - 1/(8*(1 + Cos[c + d*x])^2) + 3/(4*(1 + Cos[c + d*x])) + (5*Log[1 - Cos[c + d*x]])/16 + (11*Log[1 + Cos[c + d*x]])/16)/ (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 = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.65
method | result | size |
derivativedivides | \(\frac {\frac {1}{-8+8 \cos \left (d x +c \right )}-\frac {5 \ln \left (-1+\cos \left (d x +c \right )\right )}{16}+\frac {1}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )}-\frac {11 \ln \left (1+\cos \left (d x +c \right )\right )}{16}}{d a}\) | \(67\) |
default | \(\frac {\frac {1}{-8+8 \cos \left (d x +c \right )}-\frac {5 \ln \left (-1+\cos \left (d x +c \right )\right )}{16}+\frac {1}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}-\frac {3}{4 \left (1+\cos \left (d x +c \right )\right )}-\frac {11 \ln \left (1+\cos \left (d x +c \right )\right )}{16}}{d a}\) | \(67\) |
risch | \(\frac {i x}{a}+\frac {2 i c}{d a}-\frac {5 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 \,{\mathrm e}^{4 i \left (d x +c \right )}-14 \,{\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}}{4 d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d a}-\frac {11 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d a}\) | \(149\) |
Input:
int(cot(d*x+c)^3/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(1/8/(-1+cos(d*x+c))-5/16*ln(-1+cos(d*x+c))+1/8/(1+cos(d*x+c))^2-3/4 /(1+cos(d*x+c))-11/16*ln(1+cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.35 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {10 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, \cos \left (d x + c\right ) - 12}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2} - a d \cos \left (d x + c\right ) - a d\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
-1/16*(10*cos(d*x + c)^2 + 11*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(1/2*cos(d*x + c) + 1/2) + 5*(cos(d*x + c)^3 + cos(d*x + c)^2 - cos(d*x + c) - 1)*log(-1/2*cos(d*x + c) + 1/2) - 6*cos(d*x + c) - 12)/(a *d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2 - a*d*cos(d*x + c) - a*d)
\[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate(cot(d*x+c)**3/(a+a*sec(d*x+c)),x)
Output:
Integral(cot(c + d*x)**3/(sec(c + d*x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (5 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 6\right )}}{a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - a} + \frac {11 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {5 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/16*(2*(5*cos(d*x + c)^2 - 3*cos(d*x + c) - 6)/(a*cos(d*x + c)^3 + a*cos (d*x + c)^2 - a*cos(d*x + c) - a) + 11*log(cos(d*x + c) + 1)/a + 5*log(cos (d*x + c) - 1)/a)/d
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {11 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{16 \, a d} - \frac {5 \, \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{16 \, a d} - \frac {5 \, \cos \left (d x + c\right )^{2} - 3 \, \cos \left (d x + c\right ) - 6}{8 \, a d {\left (\cos \left (d x + c\right ) + 1\right )}^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
-11/16*log(abs(cos(d*x + c) + 1))/(a*d) - 5/16*log(abs(cos(d*x + c) - 1))/ (a*d) - 1/8*(5*cos(d*x + c)^2 - 3*cos(d*x + c) - 6)/(a*d*(cos(d*x + c) + 1 )^2*(cos(d*x + c) - 1))
Time = 13.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=-\frac {\frac {5\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}}{a\,d} \] Input:
int(cot(c + d*x)^3/(a + a/cos(c + d*x)),x)
Output:
-((5*log(tan(c/2 + (d*x)/2)))/8 - log(tan(c/2 + (d*x)/2)^2 + 1) + cot(c/2 + (d*x)/2)^2/16 + (5*tan(c/2 + (d*x)/2)^2)/16 - tan(c/2 + (d*x)/2)^4/32)/( a*d)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^3(c+d x)}{a+a \sec (c+d x)} \, dx=\frac {32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-20 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2}{32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a d} \] Input:
int(cot(d*x+c)^3/(a+a*sec(d*x+c)),x)
Output:
(32*log(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)**2 - 20*log(tan((c + d*x )/2))*tan((c + d*x)/2)**2 + tan((c + d*x)/2)**6 - 10*tan((c + d*x)/2)**4 - 2)/(32*tan((c + d*x)/2)**2*a*d)