\(\int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [187]

Optimal result

Integrand size = 23, antiderivative size = 176 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {11 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} a^{3/2} d}-\frac {3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac {5}{24 d (a+a \sec (c+d x))^{3/2}}+\frac {21}{16 a d \sqrt {a+a \sec (c+d x)}} \] Output:

-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+11/32*arctanh(1/2*(a+ 
a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(3/2)/d-3/20*a/d/(a+a*sec(d 
*x+c))^(5/2)+1/2*a/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(5/2)+5/24/d/(a+a*sec 
(d*x+c))^(3/2)+21/16/a/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.51 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {a \left (-10-11 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1+\sec (c+d x))\right ) (-1+\sec (c+d x))+8 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},1+\sec (c+d x)\right ) (-1+\sec (c+d x))\right )}{20 d (-1+\sec (c+d x)) (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

(a*(-10 - 11*Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sec[c + d*x])/2]*(-1 + 
Sec[c + d*x]) + 8*Hypergeometric2F1[-5/2, 1, -3/2, 1 + Sec[c + d*x]]*(-1 + 
 Sec[c + d*x])))/(20*d*(-1 + Sec[c + d*x])*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 25, 4368, 27, 114, 27, 169, 27, 169, 27, 169, 27, 174, 73, 219, 220}

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.

\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4368

\(\displaystyle \frac {a^4 \int \frac {\cos (c+d x)}{a^2 (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{7/2}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \int \frac {\cos (c+d x)}{(1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{7/2}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 114

\(\displaystyle \frac {a^2 \left (\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}-\frac {\int -\frac {a \cos (c+d x) (7 \sec (c+d x)+4)}{2 (1-\sec (c+d x)) (\sec (c+d x) a+a)^{7/2}}d\sec (c+d x)}{2 a}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \int \frac {\cos (c+d x) (7 \sec (c+d x)+4)}{(1-\sec (c+d x)) (\sec (c+d x) a+a)^{7/2}}d\sec (c+d x)+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\int \frac {5 a \cos (c+d x) (3 \sec (c+d x)+8)}{2 (1-\sec (c+d x)) (\sec (c+d x) a+a)^{5/2}}d\sec (c+d x)}{5 a^2}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\int \frac {\cos (c+d x) (3 \sec (c+d x)+8)}{(1-\sec (c+d x)) (\sec (c+d x) a+a)^{5/2}}d\sec (c+d x)}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\int \frac {3 a \cos (c+d x) (16-5 \sec (c+d x))}{2 (1-\sec (c+d x)) (\sec (c+d x) a+a)^{3/2}}d\sec (c+d x)}{3 a^2}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\int \frac {\cos (c+d x) (16-5 \sec (c+d x))}{(1-\sec (c+d x)) (\sec (c+d x) a+a)^{3/2}}d\sec (c+d x)}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {\int \frac {a \cos (c+d x) (32-21 \sec (c+d x))}{2 (1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{a^2}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {\int \frac {\cos (c+d x) (32-21 \sec (c+d x))}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{2 a}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {11 \int \frac {1}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+32 \int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{2 a}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {\frac {22 \int \frac {1}{2-\frac {\sec (c+d x) a+a}{a}}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}}{2 a}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {11 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}}{2 a}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {a^2 \left (\frac {1}{4} \left (\frac {\frac {\frac {\frac {11 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {64 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 a}+\frac {21}{a \sqrt {a \sec (c+d x)+a}}}{2 a}+\frac {5}{3 a (a \sec (c+d x)+a)^{3/2}}}{2 a}-\frac {3}{5 a (a \sec (c+d x)+a)^{5/2}}\right )+\frac {1}{2 a (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

Input:

Int[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^(3/2),x]
 

Output:

(a^2*(1/(2*a*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(5/2)) + (-3/(5*a*(a 
+ a*Sec[c + d*x])^(5/2)) + (5/(3*a*(a + a*Sec[c + d*x])^(3/2)) + (((-64*Ar 
cTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (11*Sqrt[2]*ArcTanh[Sqr 
t[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a])/(2*a) + 21/(a*Sqrt[a + 
a*Sec[c + d*x]]))/(2*a))/(2*a))/4))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4368

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1)   Subst[Int[(-a + b*x)^((m - 1)/2 
)*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(407\) vs. \(2(143)=286\).

Time = 1.06 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.32

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (5122 \cos \left (d x +c \right )^{5}+6354 \cos \left (d x +c \right )^{4}-5500 \cos \left (d x +c \right )^{3}-14124 \cos \left (d x +c \right )^{2}-9702 \cos \left (d x +c \right )-2310\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cot \left (d x +c \right ) \csc \left (d x +c \right )+\left (-10080 \cos \left (d x +c \right )^{4}-40320 \cos \left (d x +c \right )^{3}-60480 \cos \left (d x +c \right )^{2}-40320 \cos \left (d x +c \right )-10080\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right )+\left (-3465 \cos \left (d x +c \right )^{4}-13860 \cos \left (d x +c \right )^{3}-20790 \cos \left (d x +c \right )^{2}-13860 \cos \left (d x +c \right )-3465\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\left (29102 \cos \left (d x +c \right )^{5}+54922 \cos \left (d x +c \right )^{4}+19548 \cos \left (d x +c \right )^{3}-43932 \cos \left (d x +c \right )^{2}-46410 \cos \left (d x +c \right )-13230\right ) \cot \left (d x +c \right ) \csc \left (d x +c \right )\right )}{10080 d \,a^{2} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\)	\(408\)

Input:

int(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-1/10080/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)^3+3*cos 
(d*x+c)^2+3*cos(d*x+c)+1)*((5122*cos(d*x+c)^5+6354*cos(d*x+c)^4-5500*cos(d 
*x+c)^3-14124*cos(d*x+c)^2-9702*cos(d*x+c)-2310)*2^(1/2)*(-cos(d*x+c)/(1+c 
os(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cot(d*x+c)*csc(d*x+ 
c)+(-10080*cos(d*x+c)^4-40320*cos(d*x+c)^3-60480*cos(d*x+c)^2-40320*cos(d* 
x+c)-10080)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2 
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+(-3465*cos(d*x+c)^4-13860*cos(d*x+ 
c)^3-20790*cos(d*x+c)^2-13860*cos(d*x+c)-3465)*(-2*cos(d*x+c)/(1+cos(d*x+c 
)))^(1/2)*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+(29102*co 
s(d*x+c)^5+54922*cos(d*x+c)^4+19548*cos(d*x+c)^3-43932*cos(d*x+c)^2-46410* 
cos(d*x+c)-13230)*cot(d*x+c)*csc(d*x+c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (141) = 282\).

Time = 0.18 (sec) , antiderivative size = 592, normalized size of antiderivative = 3.36 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

[1/960*(165*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c) - 
1)*sqrt(a)*log((2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* 
cos(d*x + c) + 3*a*cos(d*x + c) + a)/(cos(d*x + c) - 1)) + 480*(cos(d*x + 
c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c) - 1)*sqrt(a)*log(-8*a*cos(d*x + c 
)^2 + 4*(2*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 4*(449*cos(d*x + c)^4 + 351*cos( 
d*x + c)^3 - 365*cos(d*x + c)^2 - 315*cos(d*x + c))*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 - 2*a^2* 
d*cos(d*x + c) - a^2*d), -1/480*(165*sqrt(2)*(cos(d*x + c)^4 + 2*cos(d*x + 
 c)^3 - 2*cos(d*x + c) - 1)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d 
*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 480*(cos(d 
*x + c)^4 + 2*cos(d*x + c)^3 - 2*cos(d*x + c) - 1)*sqrt(-a)*arctan(2*sqrt( 
-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) 
 + a)) - 2*(449*cos(d*x + c)^4 + 351*cos(d*x + c)^3 - 365*cos(d*x + c)^2 - 
 315*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^2*d*cos(d*x 
 + c)^4 + 2*a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c) - a^2*d)]
 

Sympy [F]

\[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(cot(d*x+c)**3/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Integral(cot(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\cot \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(cot(d*x + c)^3/(a*sec(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.44 \[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {165 \, \sqrt {2} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {960 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {15 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {2 \, \sqrt {2} {\left (3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{16} + 20 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{17} + 165 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{18}\right )}}{a^{20} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{480 \, d} \] Input:

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

-1/480*(165*sqrt(2)*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/( 
sqrt(-a)*a*sgn(cos(d*x + c))) - 960*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x 
 + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a*sgn(cos(d*x + c))) + 15*sqrt(2)*sqr 
t(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a^2*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2* 
c)^2) - 2*sqrt(2)*(3*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x 
+ 1/2*c)^2 + a)*a^16 + 20*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^17 + 165 
*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^18)/(a^20*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(cot(c + d*x)^3/(a + a/cos(c + d*x))^(3/2),x)
 

Output:

int(cot(c + d*x)^3/(a + a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

int(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*cot(c + d*x)**3)/(sec(c + d*x)**2 + 2 
*sec(c + d*x) + 1),x))/a**2