Integrand size = 23, antiderivative size = 260 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{4 (a-b)^{3/2} d}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{4 (a+b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d}+\frac {\sqrt {a+b \sec (c+d x)}}{4 (a+b) d (1-\sec (c+d x))}+\frac {\sqrt {a+b \sec (c+d x)}}{4 (a-b) d (1+\sec (c+d x))} \]
-1/4*b*arctanh((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(3/2)/d+1/4*b*arc tanh((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(3/2)/d-2*arctanh((a+b*sec( d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)+arctanh((a+b*sec(d*x+c))^(1/2)/(a-b)^(1/2 ))/d/(a-b)^(1/2)+arctanh((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2))/d/(a+b)^(1/2) +1/4*(a+b*sec(d*x+c))^(1/2)/(a+b)/d/(1-sec(d*x+c))+1/4*(a+b*sec(d*x+c))^(1 /2)/(a-b)/d/(1+sec(d*x+c))
Time = 2.38 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {-\frac {b^2 \arctan \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {-a+b}}\right )}{(-a+b)^{3/2}}-\frac {8 b \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {4 b \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {4 a \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+4 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )-\frac {b \sqrt {a+b \sec (c+d x)}}{(a+b) (-1+\sec (c+d x))}+\frac {b \sqrt {a+b \sec (c+d x)}}{(a-b) (1+\sec (c+d x))}}{4 b d} \]
(-((b^2*ArcTan[Sqrt[a + b*Sec[c + d*x]]/Sqrt[-a + b]])/(-a + b)^(3/2)) - ( 8*b*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (4*b*ArcTanh[Sqrt [a + b*Sec[c + d*x]]/Sqrt[a - b]])/Sqrt[a - b] + (b^2*ArcTanh[Sqrt[a + b*S ec[c + d*x]]/Sqrt[a + b]])/(a + b)^(3/2) - (4*a*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/Sqrt[a + b] + 4*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]] - (b*Sqrt[a + b*Sec[c + d*x]])/((a + b)*(-1 + Sec[c + d*x])) + (b*Sqrt[a + b*Sec[c + d*x]])/((a - b)*(1 + Sec[c + d*x])))/(4*b* d)
Time = 0.47 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4373, 561, 25, 1567, 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.
| \(\displaystyle \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \sqrt {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle \frac {b^4 \int \frac {\cos (c+d x)}{b \sqrt {a+b \sec (c+d x)} \left (b^2-b^2 \sec ^2(c+d x)\right )^2}d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 561 |
| \(\displaystyle \frac {2 b^4 \int -\frac {1}{\left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \sec (c+d x)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 b^4 \int \frac {1}{\left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}d\sqrt {a+b \sec (c+d x)}}{d}\) |
| \(\Big \downarrow \) 1567 |
| \(\displaystyle -\frac {2 b^4 \int \left (-\frac {1}{2 b^4 \left (-b^2 \sec ^2(c+d x)+a+b\right )}+\frac {1}{2 b^4 \left (b^2 \sec ^2(c+d x)-a+b\right )}-\frac {1}{4 b^3 \left (-b^2 \sec ^2(c+d x)+a+b\right )^2}+\frac {1}{4 b^3 \left (b^2 \sec ^2(c+d x)-a+b\right )^2}+\frac {1}{b^4 \left (a-b^2 \sec ^2(c+d x)\right )}\right )d\sqrt {a+b \sec (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b^4 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} b^4}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{2 b^4 \sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{2 b^4 \sqrt {a+b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{8 b^3 (a-b)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{8 b^3 (a+b)^{3/2}}-\frac {\sec (c+d x)}{8 b^2 (a-b) \left (a-b^2 \sec ^2(c+d x)-b\right )}+\frac {\sec (c+d x)}{8 b^2 (a+b) \left (a-b^2 \sec ^2(c+d x)+b\right )}\right )}{d}\) |
(2*b^4*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^4)) + ArcTa nh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^4) - ArcTanh[Sqr t[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(8*(a - b)^(3/2)*b^3) + ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(8*b^3*(a + b)^(3/2)) + ArcTanh[Sqrt[a + b* Sec[c + d*x]]/Sqrt[a + b]]/(2*b^4*Sqrt[a + b]) - Sec[c + d*x]/(8*(a - b)*b ^2*(a - b - b^2*Sec[c + d*x]^2)) + Sec[c + d*x]/(8*b^2*(a + b)*(a + b - b^ 2*Sec[c + d*x]^2))))/d
3.4.30.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0] || IGtQ[q, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(3106\) vs. \(2(218)=436\).
Time = 2.01 (sec) , antiderivative size = 3107, normalized size of antiderivative = 11.95
1/8/d/(a+b)^2/(a-b)^(5/2)/a*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x +c))^2*csc(d*x+c)^2-a-b)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos( d*x+c))^2*csc(d*x+c)^2-1)*(8*ln(2*(-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+b*(1 -cos(d*x+c))^2*csc(d*x+c)^2+2*a^(1/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1- cos(d*x+c))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)+ 2*a+b)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))*a^(7/2)*(a-b)^(3/2)*(1-cos(d*x+c ))^2*csc(d*x+c)^2+8*ln(2*(-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+b*(1-cos(d*x+ c))^2*csc(d*x+c)^2+2*a^(1/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c ))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)+2*a+b)/(( 1-cos(d*x+c))^2*csc(d*x+c)^2+1))*a^(5/2)*(a-b)^(3/2)*b*(1-cos(d*x+c))^2*cs c(d*x+c)^2-8*ln(2*(-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+b*(1-cos(d*x+c))^2*c sc(d*x+c)^2+2*a^(1/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4*b* csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)+2*a+b)/((1-cos(d *x+c))^2*csc(d*x+c)^2+1))*a^(3/2)*(a-b)^(3/2)*b^2*(1-cos(d*x+c))^2*csc(d*x +c)^2-(a-b)^(3/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4*b*csc( d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)*a^3*(1-cos(d*x+c))^4 *csc(d*x+c)^4+2*(a-b)^(3/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c) )^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)*a^2*b*(1-c os(d*x+c))^4*csc(d*x+c)^4-(a-b)^(3/2)*((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1- cos(d*x+c))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/...
Leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (216) = 432\).
Time = 25.89 (sec) , antiderivative size = 4336, normalized size of antiderivative = 16.68 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Too large to display} \]
[-1/16*(8*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2) *sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 + 4*(2*a*cos (d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + (4*a^4 + 3*a^3*b - 6*a^2*b^2 - 5*a*b^3 - (4*a^4 + 3*a^3*b - 6*a^2*b ^2 - 5*a*b^3)*cos(d*x + c)^2)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos( d*x + c)^2 + b^2 + 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c ))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + (4*a^4 - 3*a^3*b - 6*a^2*b^2 + 5*a*b^3 - (4*a^4 - 3*a^3*b - 6*a^2*b^2 + 5*a*b^3)*cos(d*x + c)^2)*sqrt(a + b)*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b^2 + 4*((2*a + b)*cos(d *x + c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)) - 8*((a^4 - a^2*b^2)*cos(d*x + c)^2 - (a^3*b - a*b^3)*cos(d*x + c))*s qrt((a*cos(d*x + c) + b)/cos(d*x + c)))/((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), -1/16*(16*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt ((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) + (4*a^4 + 3*a^3*b - 6*a^2*b^2 - 5*a*b^3 - (4*a^4 + 3*a^3*b - 6*a^2*b^2 - 5 *a*b^3)*cos(d*x + c)^2)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 + 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*...
\[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {\cot \left (d x + c\right )^{3}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (216) = 432\).
Time = 0.92 (sec) , antiderivative size = 533, normalized size of antiderivative = 2.05 \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {\frac {16 \, \arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, {\left (4 \, a + 5 \, b\right )} \arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}}{\sqrt {-a - b}}\right )}{{\left (a + b\right )} \sqrt {-a - b}} + \frac {{\left (4 \, a - 5 \, b\right )} \log \left ({\left | {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )} {\left (a - b\right )} - \sqrt {a - b} a \right |}\right )}{{\left (a - b\right )}^{\frac {3}{2}}} + \frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}}{a - b} - \frac {2 \, {\left ({\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )} a - {\left (a + b\right )} \sqrt {a - b}\right )}}{{\left ({\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{2} - a - b\right )} {\left (a + b\right )}}}{8 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]
1/8*(16*arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d *x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + sqrt(a - b))/sqrt(-a))/sqrt(-a) - 2*(4*a + 5*b)*arctan(-(sqrt(a - b )*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))/sqrt(-a - b))/((a + b)*sq rt(-a - b)) + (4*a - 5*b)*log(abs((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sq rt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))*(a - b) - sqrt(a - b)*a))/(a - b)^(3/2) + sqrt(a*tan(1 /2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)/(a - b) - 2*((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2 *d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))*a - (a + b)*sqrt(a - b))/(((sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - s qrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^2 - a - b)*(a + b)))/(d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {\cot ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]