Integrand size = 19, antiderivative size = 101 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {1}{6 a^3 d (1+\cos (c+d x))^3}-\frac {7}{8 a^3 d (1+\cos (c+d x))^2}+\frac {17}{8 a^3 d (1+\cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{16 a^3 d}+\frac {15 \log (1+\cos (c+d x))}{16 a^3 d} \] Output:
1/6/a^3/d/(1+cos(d*x+c))^3-7/8/a^3/d/(1+cos(d*x+c))^2+17/8/a^3/d/(1+cos(d* x+c))+1/16*ln(1-cos(d*x+c))/a^3/d+15/16*ln(1+cos(d*x+c))/a^3/d
Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.96 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\left (2-21 \cos ^2\left (\frac {1}{2} (c+d x)\right )+102 \cos ^4\left (\frac {1}{2} (c+d x)\right )+12 \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (15 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right ) \sec ^3(c+d x)}{12 a^3 d (1+\sec (c+d x))^3} \] Input:
Integrate[Cot[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
((2 - 21*Cos[(c + d*x)/2]^2 + 102*Cos[(c + d*x)/2]^4 + 12*Cos[(c + d*x)/2] ^6*(15*Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]]))*Sec[c + d*x]^3)/(12 *a^3*d*(1 + Sec[c + d*x])^3)
Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, 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.
\(\displaystyle \int \frac {\cot (c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^3}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -\frac {a^2 \int \frac {\cos ^4(c+d x)}{a^5 (1-\cos (c+d x)) (\cos (c+d x)+1)^4}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cos ^4(c+d x)}{(1-\cos (c+d x)) (\cos (c+d x)+1)^4}d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {15}{16 (\cos (c+d x)+1)}+\frac {17}{8 (\cos (c+d x)+1)^2}-\frac {7}{4 (\cos (c+d x)+1)^3}+\frac {1}{2 (\cos (c+d x)+1)^4}-\frac {1}{16 (\cos (c+d x)-1)}\right )d\cos (c+d x)}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {17}{8 (\cos (c+d x)+1)}+\frac {7}{8 (\cos (c+d x)+1)^2}-\frac {1}{6 (\cos (c+d x)+1)^3}-\frac {1}{16} \log (1-\cos (c+d x))-\frac {15}{16} \log (\cos (c+d x)+1)}{a^3 d}\) |
Input:
Int[Cot[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
-((-1/6*1/(1 + Cos[c + d*x])^3 + 7/(8*(1 + Cos[c + d*x])^2) - 17/(8*(1 + C os[c + d*x])) - Log[1 - Cos[c + d*x]]/16 - (15*Log[1 + Cos[c + d*x]])/16)/ (a^3*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 = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{16}+\frac {1}{6 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {7}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {17}{8 \left (1+\cos \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16}}{d \,a^{3}}\) | \(67\) |
default | \(\frac {\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{16}+\frac {1}{6 \left (1+\cos \left (d x +c \right )\right )^{3}}-\frac {7}{8 \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {17}{8 \left (1+\cos \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\cos \left (d x +c \right )\right )}{16}}{d \,a^{3}}\) | \(67\) |
risch | \(-\frac {i x}{a^{3}}-\frac {2 i c}{d \,a^{3}}+\frac {51 \,{\mathrm e}^{5 i \left (d x +c \right )}+162 \,{\mathrm e}^{4 i \left (d x +c \right )}+238 \,{\mathrm e}^{3 i \left (d x +c \right )}+162 \,{\mathrm e}^{2 i \left (d x +c \right )}+51 \,{\mathrm e}^{i \left (d x +c \right )}}{12 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{6}}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{8 d \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{8 d \,a^{3}}\) | \(136\) |
Input:
int(cot(d*x+c)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d/a^3*(1/16*ln(-1+cos(d*x+c))+1/6/(1+cos(d*x+c))^3-7/8/(1+cos(d*x+c))^2+ 17/8/(1+cos(d*x+c))+15/16*ln(1+cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.50 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {102 \, \cos \left (d x + c\right )^{2} + 45 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 162 \, \cos \left (d x + c\right ) + 68}{48 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/48*(102*cos(d*x + c)^2 + 45*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d *x + c) + 1)*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 162*cos(d*x + c ) + 68)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\cot {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))**3,x)
Output:
Integral(cot(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x ) + 1), x)/a**3
Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (51 \, \cos \left (d x + c\right )^{2} + 81 \, \cos \left (d x + c\right ) + 34\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac {45 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
1/48*(2*(51*cos(d*x + c)^2 + 81*cos(d*x + c) + 34)/(a^3*cos(d*x + c)^3 + 3 *a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3) + 45*log(cos(d*x + c) + 1) /a^3 + 3*log(cos(d*x + c) - 1)/a^3)/d
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{16 \, a^{3} d} + \frac {\log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{16 \, a^{3} d} + \frac {51 \, \cos \left (d x + c\right )^{2} + 81 \, \cos \left (d x + c\right ) + 34}{24 \, a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}^{3}} \] Input:
integrate(cot(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
15/16*log(abs(cos(d*x + c) + 1))/(a^3*d) + 1/16*log(abs(cos(d*x + c) - 1)) /(a^3*d) + 1/24*(51*cos(d*x + c)^2 + 81*cos(d*x + c) + 34)/(a^3*d*(cos(d*x + c) + 1)^3)
Time = 13.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8}-\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{32}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}}{a^3\,d} \] Input:
int(cot(c + d*x)/(a + a/cos(c + d*x))^3,x)
Output:
(log(tan(c/2 + (d*x)/2))/8 - log(tan(c/2 + (d*x)/2)^2 + 1) + (11*tan(c/2 + (d*x)/2)^2)/16 - (5*tan(c/2 + (d*x)/2)^4)/32 + tan(c/2 + (d*x)/2)^6/48)/( a^3*d)
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+66 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{96 a^{3} d} \] Input:
int(cot(d*x+c)/(a+a*sec(d*x+c))^3,x)
Output:
( - 96*log(tan((c + d*x)/2)**2 + 1) + 12*log(tan((c + d*x)/2)) + 2*tan((c + d*x)/2)**6 - 15*tan((c + d*x)/2)**4 + 66*tan((c + d*x)/2)**2)/(96*a**3*d )