Integrand size = 21, antiderivative size = 197 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {(a+2 b) \log (1-\sec (c+d x))}{2 (a+b)^3 d}-\frac {(a-2 b) \log (1+\sec (c+d x))}{2 (a-b)^3 d}-\frac {b^4 \left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 \left (a^2-b^2\right )^3 d}+\frac {1}{4 (a+b)^2 d (1-\sec (c+d x))}+\frac {1}{4 (a-b)^2 d (1+\sec (c+d x))}+\frac {b^4}{a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:
-ln(cos(d*x+c))/a^2/d-1/2*(a+2*b)*ln(1-sec(d*x+c))/(a+b)^3/d-1/2*(a-2*b)*l n(1+sec(d*x+c))/(a-b)^3/d-b^4*(5*a^2-b^2)*ln(a+b*sec(d*x+c))/a^2/(a^2-b^2) ^3/d+1/4/(a+b)^2/d/(1-sec(d*x+c))+1/4/(a-b)^2/d/(1+sec(d*x+c))+b^4/a/(a^2- b^2)^2/d/(a+b*sec(d*x+c))
Time = 1.33 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.95 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {b^4 \left (-\frac {4 \log (\cos (c+d x))}{a^2 b^4}-\frac {2 (a+2 b) \log (1-\sec (c+d x))}{b^4 (a+b)^3}-\frac {2 (a-2 b) \log (1+\sec (c+d x))}{(a-b)^3 b^4}-\frac {4 \left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 (a-b)^3 (a+b)^3}-\frac {1}{b^4 (a+b)^2 (-1+\sec (c+d x))}+\frac {1}{(a-b)^2 b^4 (1+\sec (c+d x))}+\frac {4}{a (a-b)^2 (a+b)^2 (a+b \sec (c+d x))}\right )}{4 d} \] Input:
Integrate[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
Output:
(b^4*((-4*Log[Cos[c + d*x]])/(a^2*b^4) - (2*(a + 2*b)*Log[1 - Sec[c + d*x] ])/(b^4*(a + b)^3) - (2*(a - 2*b)*Log[1 + Sec[c + d*x]])/((a - b)^3*b^4) - (4*(5*a^2 - b^2)*Log[a + b*Sec[c + d*x]])/(a^2*(a - b)^3*(a + b)^3) - 1/( b^4*(a + b)^2*(-1 + Sec[c + d*x])) + 1/((a - b)^2*b^4*(1 + Sec[c + d*x])) + 4/(a*(a - b)^2*(a + b)^2*(a + b*Sec[c + d*x]))))/(4*d)
Time = 0.45 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle \frac {b^4 \int \frac {\cos (c+d x)}{b (a+b \sec (c+d x))^2 \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {b^4 \int \left (\frac {2 b-a}{2 (a-b)^3 b^4 (\sec (c+d x) b+b)}+\frac {\cos (c+d x)}{a^2 b^5}+\frac {a+2 b}{2 b^4 (a+b)^3 (b-b \sec (c+d x))}+\frac {b^2-5 a^2}{a^2 (a-b)^3 (a+b)^3 (a+b \sec (c+d x))}+\frac {1}{4 b^3 (a+b)^2 (b-b \sec (c+d x))^2}-\frac {1}{a (a-b)^2 (a+b)^2 (a+b \sec (c+d x))^2}-\frac {1}{4 (a-b)^2 b^3 (\sec (c+d x) b+b)^2}\right )d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^4 \left (\frac {\log (b \sec (c+d x))}{a^2 b^4}+\frac {1}{a \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {\left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 \left (a^2-b^2\right )^3}-\frac {(a+2 b) \log (b-b \sec (c+d x))}{2 b^4 (a+b)^3}-\frac {(a-2 b) \log (b \sec (c+d x)+b)}{2 b^4 (a-b)^3}+\frac {1}{4 b^3 (a+b)^2 (b-b \sec (c+d x))}+\frac {1}{4 b^3 (a-b)^2 (b \sec (c+d x)+b)}\right )}{d}\) |
Input:
Int[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]
Output:
(b^4*(Log[b*Sec[c + d*x]]/(a^2*b^4) - ((a + 2*b)*Log[b - b*Sec[c + d*x]])/ (2*b^4*(a + b)^3) - ((5*a^2 - b^2)*Log[a + b*Sec[c + d*x]])/(a^2*(a^2 - b^ 2)^3) - ((a - 2*b)*Log[b + b*Sec[c + d*x]])/(2*(a - b)^3*b^4) + 1/(4*b^3*( a + b)^2*(b - b*Sec[c + d*x])) + 1/(a*(a^2 - b^2)^2*(a + b*Sec[c + d*x])) + 1/(4*(a - b)^2*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 = 0.91 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a +2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {b^{5}}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )}-\frac {b^{4} \left (5 a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2}}}{d}\) | \(164\) |
| default | \(\frac {\frac {1}{4 \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (d x +c \right )\right )}{2 \left (a +b \right )^{3}}-\frac {1}{4 \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}+\frac {\left (-a +2 b \right ) \ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{3}}-\frac {b^{5}}{a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )}-\frac {b^{4} \left (5 a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3} a^{2}}}{d}\) | \(164\) |
| risch | \(\frac {10 i b^{4} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {2 i b x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {10 i b^{4} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {2 i b c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {2 i b^{6} x}{a^{2} \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {2 i b c}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}+\frac {i a c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {2 i b x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}-\frac {2 i b^{6} c}{a^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}-\frac {i x}{a^{2}}+\frac {-2 a^{4} b \,{\mathrm e}^{5 i \left (d x +c \right )}-2 b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+2 a^{5} {\mathrm e}^{4 i \left (d x +c \right )}-2 a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{2} b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+4 b^{5} {\mathrm e}^{3 i \left (d x +c \right )}+2 a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}-2 b^{5} {\mathrm e}^{i \left (d x +c \right )}}{\left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right ) \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {5 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 )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {b^{6} \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^{2} d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(886\) |
Input:
int(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/4/(a+b)^2/(-1+cos(d*x+c))+1/2*(-a-2*b)/(a+b)^3*ln(-1+cos(d*x+c))-1/ 4/(a-b)^2/(1+cos(d*x+c))+1/2*(-a+2*b)/(a-b)^3*ln(1+cos(d*x+c))-b^5/a^2/(a+ b)^2/(a-b)^2/(b+a*cos(d*x+c))-b^4*(5*a^2-b^2)/(a+b)^3/(a-b)^3/a^2*ln(b+a*c os(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (187) = 374\).
Time = 0.27 (sec) , antiderivative size = 693, normalized size of antiderivative = 3.52 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {a^{6} b + a^{2} b^{5} - 2 \, b^{7} - 2 \, {\left (a^{6} b - a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right ) + 2 \, {\left (5 \, a^{2} b^{5} - b^{7} - {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (5 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (a^{6} b + a^{5} b^{2} - 3 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5} - {\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b + a^{5} b^{2} - 3 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{6} b - a^{5} b^{2} - 3 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5} - {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b - a^{5} b^{2} - 3 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right ) - {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} d\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
1/2*(a^6*b + a^2*b^5 - 2*b^7 - 2*(a^6*b - a^4*b^3 + a^2*b^5 - b^7)*cos(d*x + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c) + 2*(5*a^2*b^5 - b^7 - (5*a^3*b^4 - a*b^6)*cos(d*x + c)^3 - (5*a^2*b^5 - b^7)*cos(d*x + c)^2 + (5 *a^3*b^4 - a*b^6)*cos(d*x + c))*log(a*cos(d*x + c) + b) + (a^6*b + a^5*b^2 - 3*a^4*b^3 - 5*a^3*b^4 - 2*a^2*b^5 - (a^7 + a^6*b - 3*a^5*b^2 - 5*a^4*b^ 3 - 2*a^3*b^4)*cos(d*x + c)^3 - (a^6*b + a^5*b^2 - 3*a^4*b^3 - 5*a^3*b^4 - 2*a^2*b^5)*cos(d*x + c)^2 + (a^7 + a^6*b - 3*a^5*b^2 - 5*a^4*b^3 - 2*a^3* b^4)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^6*b - a^5*b^2 - 3*a^4* b^3 + 5*a^3*b^4 - 2*a^2*b^5 - (a^7 - a^6*b - 3*a^5*b^2 + 5*a^4*b^3 - 2*a^3 *b^4)*cos(d*x + c)^3 - (a^6*b - a^5*b^2 - 3*a^4*b^3 + 5*a^3*b^4 - 2*a^2*b^ 5)*cos(d*x + c)^2 + (a^7 - a^6*b - 3*a^5*b^2 + 5*a^4*b^3 - 2*a^3*b^4)*cos( d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^ 3*b^6)*d*cos(d*x + c)^3 + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*cos( d*x + c)^2 - (a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c) - (a^8 *b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d)
\[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(cot(d*x+c)**3/(a+b*sec(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**3/(a + b*sec(c + d*x))**2, x)
Time = 0.07 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.54 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} + \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {a^{4} b + a^{2} b^{3} + 2 \, b^{5} - 2 \, {\left (a^{4} b + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )}}{2 \, d} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
Output:
-1/2*(2*(5*a^2*b^4 - b^6)*log(a*cos(d*x + c) + b)/(a^8 - 3*a^6*b^2 + 3*a^4 *b^4 - a^2*b^6) + (a - 2*b)*log(cos(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (a + 2*b)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (a^4*b + a^2*b^3 + 2*b^5 - 2*(a^4*b + b^5)*cos(d*x + c)^2 + (a^5 - a^3*b ^2)*cos(d*x + c))/(a^6*b - 2*a^4*b^3 + a^2*b^5 - (a^7 - 2*a^5*b^2 + a^3*b^ 4)*cos(d*x + c)^3 - (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)))/d
Time = 0.17 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.34 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {{\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | a \cos \left (d x + c\right ) + b \right |}\right )}{a^{8} d - 3 \, a^{6} b^{2} d + 3 \, a^{4} b^{4} d - a^{2} b^{6} d} - \frac {{\left (a - 2 \, b\right )} \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | -\cos \left (d x + c\right ) + 1 \right |}\right )}{2 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {a^{4} b + a^{2} b^{3} + 2 \, b^{5} - 2 \, {\left (a^{4} b + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right ) + b\right )} {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} a^{2} d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
-(5*a^2*b^4 - b^6)*log(abs(a*cos(d*x + c) + b))/(a^8*d - 3*a^6*b^2*d + 3*a ^4*b^4*d - a^2*b^6*d) - 1/2*(a - 2*b)*log(abs(cos(d*x + c) + 1))/(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) - 1/2*(a + 2*b)*log(abs(-cos(d*x + c) + 1)) /(a^3*d + 3*a^2*b*d + 3*a*b^2*d + b^3*d) + 1/2*(a^4*b + a^2*b^3 + 2*b^5 - 2*(a^4*b + b^5)*cos(d*x + c)^2 + (a^5 - a^3*b^2)*cos(d*x + c))/((a*cos(d*x + c) + b)*(a + b)^2*(a - b)^2*a^2*d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1) )
Time = 11.62 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.59 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {a^2-2\,a\,b+b^2}{2\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-4\,a^4\,b+6\,a^3\,b^2-4\,a^2\,b^3+a\,b^4-16\,b^5\right )}{2\,a\,{\left (a+b\right )}^2\,\left (a-b\right )}}{d\,\left (\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-4\,a^3+4\,a^2\,b+4\,a\,b^2-4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^2}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+2\,b\right )}{d\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (5\,a^2-b^2\right )}{a^2\,d\,{\left (a^2-b^2\right )}^3} \] Input:
int(cot(c + d*x)^3/(a + b/cos(c + d*x))^2,x)
Output:
((a^2 - 2*a*b + b^2)/(2*(a + b)) - (tan(c/2 + (d*x)/2)^2*(a*b^4 - 4*a^4*b + a^5 - 16*b^5 - 4*a^2*b^3 + 6*a^3*b^2))/(2*a*(a + b)^2*(a - b)))/(d*(tan( c/2 + (d*x)/2)^2*(4*a*b^2 + 4*a^2*b - 4*a^3 - 4*b^3) + tan(c/2 + (d*x)/2)^ 4*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3))) - tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^2) + log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - (log(tan(c/2 + (d*x)/2))* (a + 2*b))/(d*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - (b^4*log(a + b - a*tan(c/ 2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(5*a^2 - b^2))/(a^2*d*(a^2 - b^2) ^3)
Time = 0.16 (sec) , antiderivative size = 1143, normalized size of antiderivative = 5.80 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x)
Output:
(8*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**7 - 24*cos (c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**5*b**2 + 24*cos( c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**3*b**4 - 8*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a*b**6 - 40*cos(c + d* x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x) **2*a**3*b**4 + 8*cos(c + d*x)*log(tan((c + d*x)/2)**2*a - tan((c + d*x)/2 )**2*b - a - b)*sin(c + d*x)**2*a*b**6 - 8*cos(c + d*x)*log(tan((c + d*x)/ 2))*sin(c + d*x)**2*a**7 + 8*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d* x)**2*a**6*b + 24*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**5* b**2 - 40*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**4*b**3 + 1 6*cos(c + d*x)*log(tan((c + d*x)/2))*sin(c + d*x)**2*a**3*b**4 + cos(c + d *x)*sin(c + d*x)**2*a**7 + 4*cos(c + d*x)*sin(c + d*x)**2*a**6*b - 6*cos(c + d*x)*sin(c + d*x)**2*a**5*b**2 + 4*cos(c + d*x)*sin(c + d*x)**2*a**4*b* *3 - 3*cos(c + d*x)*sin(c + d*x)**2*a**3*b**4 + 8*cos(c + d*x)*sin(c + d*x )**2*a**2*b**5 - 8*cos(c + d*x)*sin(c + d*x)**2*a*b**6 - 4*cos(c + d*x)*a* *7 + 8*cos(c + d*x)*a**5*b**2 - 4*cos(c + d*x)*a**3*b**4 + 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**6*b - 24*log(tan((c + d*x)/2)**2 + 1)*s in(c + d*x)**2*a**4*b**3 + 24*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2 *a**2*b**5 - 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*b**7 - 40*log( tan((c + d*x)/2)**2*a - tan((c + d*x)/2)**2*b - a - b)*sin(c + d*x)**2*...