Integrand size = 21, antiderivative size = 157 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\log (\cos (c+d x))}{a d}-\frac {(2 a+3 b) \log (1-\sec (c+d x))}{4 (a+b)^2 d}-\frac {(2 a-3 b) \log (1+\sec (c+d x))}{4 (a-b)^2 d}-\frac {b^4 \log (a+b \sec (c+d x))}{a \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sec (c+d x))}+\frac {1}{4 (a-b) d (1+\sec (c+d x))} \] Output:
-ln(cos(d*x+c))/a/d-1/4*(2*a+3*b)*ln(1-sec(d*x+c))/(a+b)^2/d-1/4*(2*a-3*b) *ln(1+sec(d*x+c))/(a-b)^2/d-b^4*ln(a+b*sec(d*x+c))/a/(a^2-b^2)^2/d+1/4/(a+ b)/d/(1-sec(d*x+c))+1/4/(a-b)/d/(1+sec(d*x+c))
Time = 0.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {b^4 \left (-\frac {4 \log (\cos (c+d x))}{a b^4}-\frac {(2 a+3 b) \log (1-\sec (c+d x))}{b^4 (a+b)^2}-\frac {(2 a-3 b) \log (1+\sec (c+d x))}{(a-b)^2 b^4}-\frac {4 \log (a+b \sec (c+d x))}{a (a-b)^2 (a+b)^2}-\frac {1}{b^4 (a+b) (-1+\sec (c+d x))}+\frac {1}{(a-b) b^4 (1+\sec (c+d x))}\right )}{4 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + b*Sec[c + d*x]),x]
Output:
(b^4*((-4*Log[Cos[c + d*x]])/(a*b^4) - ((2*a + 3*b)*Log[1 - Sec[c + d*x]]) /(b^4*(a + b)^2) - ((2*a - 3*b)*Log[1 + Sec[c + d*x]])/((a - b)^2*b^4) - ( 4*Log[a + b*Sec[c + d*x]])/(a*(a - b)^2*(a + b)^2) - 1/(b^4*(a + b)*(-1 + Sec[c + d*x])) + 1/((a - b)*b^4*(1 + Sec[c + d*x]))))/(4*d)
Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 25, 4373, 615, 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.
| \(\displaystyle \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle \frac {b^4 \int \frac {\cos (c+d x)}{b (a+b \sec (c+d x)) \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^4 \int \left (\frac {3 b-2 a}{4 (a-b)^2 b^4 (\sec (c+d x) b+b)}+\frac {\cos (c+d x)}{a b^5}+\frac {2 a+3 b}{4 b^4 (a+b)^2 (b-b \sec (c+d x))}-\frac {1}{a (a-b)^2 (a+b)^2 (a+b \sec (c+d x))}+\frac {1}{4 b^3 (a+b) (b-b \sec (c+d x))^2}-\frac {1}{4 (a-b) b^3 (\sec (c+d x) b+b)^2}\right )d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (-\frac {\log (a+b \sec (c+d x))}{a \left (a^2-b^2\right )^2}+\frac {\log (b \sec (c+d x))}{a b^4}-\frac {(2 a+3 b) \log (b-b \sec (c+d x))}{4 b^4 (a+b)^2}-\frac {(2 a-3 b) \log (b \sec (c+d x)+b)}{4 b^4 (a-b)^2}+\frac {1}{4 b^3 (a+b) (b-b \sec (c+d x))}+\frac {1}{4 b^3 (a-b) (b \sec (c+d x)+b)}\right )}{d}\) |
Input:
Int[Cot[c + d*x]^3/(a + b*Sec[c + d*x]),x]
Output:
(b^4*(Log[b*Sec[c + d*x]]/(a*b^4) - ((2*a + 3*b)*Log[b - b*Sec[c + d*x]])/ (4*b^4*(a + b)^2) - Log[a + b*Sec[c + d*x]]/(a*(a^2 - b^2)^2) - ((2*a - 3* b)*Log[b + b*Sec[c + d*x]])/(4*(a - b)^2*b^4) + 1/(4*b^3*(a + b)*(b - b*Se c[c + d*x])) + 1/(4*(a - b)*b^3*(b + b*Sec[c + d*x]))))/d
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.40 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\) | \(126\) |
| default | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{4 \left (a +b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\) | \(126\) |
| risch | \(\frac {i a c}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b^{4} c}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i a c}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i b c}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{4} x}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {3 i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{a^{2}-2 a b +b^{2}}-\frac {3 i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i a x}{a^{2}+2 a b +b^{2}}-\frac {i x}{a}-\frac {b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (a^{2}-b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(465\) |
Input:
int(cot(d*x+c)^3/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(4*a+4*b)/(-1+cos(d*x+c))+1/4/(a+b)^2*(-2*a-3*b)*ln(-1+cos(d*x+c))- 1/(4*a-4*b)/(1+cos(d*x+c))+1/4/(a-b)^2*(-2*a+3*b)*ln(1+cos(d*x+c))-b^4/(a+ b)^2/(a-b)^2/a*ln(b+a*cos(d*x+c)))
Time = 0.18 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.68 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \, a^{4} - 2 \, a^{2} b^{2} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) - 4 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3} - {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3} - {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{4 \, {\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
1/4*(2*a^4 - 2*a^2*b^2 - 2*(a^3*b - a*b^3)*cos(d*x + c) - 4*(b^4*cos(d*x + c)^2 - b^4)*log(a*cos(d*x + c) + b) + (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^ 3 - (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (2*a^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3 - (2*a^4 - a^3*b - 4*a^ 2*b^2 + 3*a*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^5 - 2*a ^3*b^2 + a*b^4)*d*cos(d*x + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d)
\[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**3/(a+b*sec(d*x+c)),x)
Output:
Integral(cot(c + d*x)**3/(a + b*sec(c + d*x)), x)
Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {4 \, b^{4} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (b \cos \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/4*(4*b^4*log(a*cos(d*x + c) + b)/(a^5 - 2*a^3*b^2 + a*b^4) + (2*a - 3*b )*log(cos(d*x + c) + 1)/(a^2 - 2*a*b + b^2) + (2*a + 3*b)*log(cos(d*x + c) - 1)/(a^2 + 2*a*b + b^2) + 2*(b*cos(d*x + c) - a)/((a^2 - b^2)*cos(d*x + c)^2 - a^2 + b^2))/d
Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {b^{4} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{5} d - 2 \, a^{3} b^{2} d + a b^{4} d} - \frac {{\left (2 \, a + 3 \, b\right )} \log \left ({\left | -\cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d + 2 \, a b d + b^{2} d\right )}} - \frac {{\left (2 \, a - 3 \, b\right )} \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
-b^4*log(abs(a*cos(d*x + c) + b))/(a^5*d - 2*a^3*b^2*d + a*b^4*d) - 1/4*(2 *a + 3*b)*log(abs(-cos(d*x + c) + 1))/(a^2*d + 2*a*b*d + b^2*d) - 1/4*(2*a - 3*b)*log(abs(-cos(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) + 1/2*(a^3 - a*b^2 - (a^2*b - b^3)*cos(d*x + c))/((a + b)^2*(a - b)^2*d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1))
Time = 11.59 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2\,d\,\left (4\,a-4\,b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a+3\,b\right )}{d\,\left (2\,a^2+4\,a\,b+2\,b^2\right )}-\frac {a-b}{2\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a+b\right )\,\left (4\,a-4\,b\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a\,d\,{\left (a^2-b^2\right )}^2} \] Input:
int(cot(c + d*x)^3/(a + b/cos(c + d*x)),x)
Output:
log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - tan(c/2 + (d*x)/2)^2/(2*d*(4*a - 4*b )) - (log(tan(c/2 + (d*x)/2))*(2*a + 3*b))/(d*(4*a*b + 2*a^2 + 2*b^2)) - ( a - b)/(2*d*tan(c/2 + (d*x)/2)^2*(a + b)*(4*a - 4*b)) - (b^4*log(a + b - a *tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(a*d*(a^2 - b^2)^2)
Time = 0.16 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.09 \[ \int \frac {\cot ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \cos \left (d x +c \right ) a^{3} b -2 \cos \left (d x +c \right ) a \,b^{3}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{4}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} b^{4}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) \sin \left (d x +c \right )^{2} b^{4}-4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{3} b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a \,b^{3}+\sin \left (d x +c \right )^{2} a^{4}-\sin \left (d x +c \right )^{2} a^{2} b^{2}-2 a^{4}+2 a^{2} b^{2}}{4 \sin \left (d x +c \right )^{2} a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(cot(d*x+c)^3/(a+b*sec(d*x+c)),x)
Output:
(2*cos(c + d*x)*a**3*b - 2*cos(c + d*x)*a*b**3 + 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**4 - 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)** 2*a**2*b**2 + 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**4 - 4*log( tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*b** 4 - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4 + 2*log(tan((c + d*x)/2)) *sin(c + d*x)**2*a**3*b + 8*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**2*b** 2 - 6*log(tan((c + d*x)/2))*sin(c + d*x)**2*a*b**3 + sin(c + d*x)**2*a**4 - sin(c + d*x)**2*a**2*b**2 - 2*a**4 + 2*a**2*b**2)/(4*sin(c + d*x)**2*a*d *(a**4 - 2*a**2*b**2 + b**4))