\(\int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx\) [164]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {43 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))} \] Output:

2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-43/32*2^(1/2)*a^(5/2)* 
arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d-1/4*a^2*(a+a*sec(d*x 
+c))^(1/2)/d/(1-sec(d*x+c))^2-11/16*a^2*(a+a*sec(d*x+c))^(1/2)/d/(1-sec(d* 
x+c))
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.74 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {(a (1+\sec (c+d x)))^{5/2} \left (64 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\frac {2 \sqrt {1+\sec (c+d x)} (-15+11 \sec (c+d x))}{(-1+\sec (c+d x))^2}\right )}{32 d (1+\sec (c+d x))^{5/2}} \] Input:

Integrate[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

((a*(1 + Sec[c + d*x]))^(5/2)*(64*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 43*Sqr 
t[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + (2*Sqrt[1 + Sec[c + d*x]]*( 
-15 + 11*Sec[c + d*x]))/(-1 + Sec[c + d*x])^2))/(32*d*(1 + Sec[c + d*x])^( 
5/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 25, 4368, 25, 27, 114, 27, 168, 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.

\(\displaystyle \int \cot ^5(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{5/2}}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}dx\)

\(\Big \downarrow \) 4368

\(\displaystyle \frac {a^6 \int -\frac {\cos (c+d x)}{a^3 (1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a^6 \int \frac {\cos (c+d x)}{a^3 (1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^3 \int \frac {\cos (c+d x)}{(1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\)

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}-\frac {\int -\frac {a \cos (c+d x) (11 \sec (c+d x)+32)}{2 (1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{2 a}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \int \frac {\cos (c+d x) (11 \sec (c+d x)+32)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

\(\Big \downarrow \) 174

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (43 \int \frac {1}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+32 \int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {86 \int \frac {1}{2-\frac {\sec (c+d x) a+a}{a}}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {64 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\)

Input:

Int[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(5/2),x]
 

Output:

-((a^3*(Sqrt[a + a*Sec[c + d*x]]/(4*a*(1 - Sec[c + d*x])^2) + (((-64*ArcTa 
nh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (43*Sqrt[2]*ArcTanh[Sqrt[a 
 + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a])/4 + (11*Sqrt[a + a*Sec[c + 
 d*x]])/(2*a*(1 - Sec[c + d*x])))/8))/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 168

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] && ILtQ[m, -1]
 

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. \(301\) vs. \(2(122)=244\).

Time = 205.93 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.05

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (134 \cos \left (d x +c \right )^{4}+996 \cos \left (d x +c \right )^{3}+776 \cos \left (d x +c \right )^{2}-516 \cos \left (d x +c \right )-430\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cot \left (d x +c \right ) \csc \left (d x +c \right )^{3}+\left (-480 \cos \left (d x +c \right )-480\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right )+\left (-645 \cos \left (d x +c \right )-645\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+\left (-182 \cos \left (d x +c \right )^{4}+704 \cos \left (d x +c \right )^{3}-532 \cos \left (d x +c \right )^{2}-320 \cos \left (d x +c \right )+330\right ) \cot \left (d x +c \right ) \csc \left (d x +c \right )^{3}\right )}{480 d \left (1+\cos \left (d x +c \right )\right )}\)	\(302\)

Input:

int(cot(d*x+c)^5*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/480/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((134*cos(d*x+c)^4+996 
*cos(d*x+c)^3+776*cos(d*x+c)^2-516*cos(d*x+c)-430)*2^(1/2)*(-cos(d*x+c)/(1 
+cos(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)^3+(-480*cos(d*x+c)-480)*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))+(-645*cos(d*x+c)- 
645)*(-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))+(-182*cos(d*x+c)^4+704*cos(d*x+c)^3-532*cos(d*x+c)^ 
2-320*cos(d*x+c)+330)*cot(d*x+c)*csc(d*x+c)^3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (118) = 236\).

Time = 0.10 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.33 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {43 \, \sqrt {\frac {1}{2}} {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (-\frac {4 \, \sqrt {\frac {1}{2}} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + 32 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right ) - 2 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{32 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}, \frac {43 \, \sqrt {\frac {1}{2}} {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 32 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate(cot(d*x+c)^5*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[1/32*(43*sqrt(1/2)*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(a 
)*log(-(4*sqrt(1/2)*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)) + 32*(a^2*cos(d*x + c)^ 
2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(a)*log(-2*a*cos(d*x + c) - 2*sqrt(a)*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - a) - 2*(15*a^2*cos(d* 
x + c)^2 - 11*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/( 
d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d), 1/16*(43*sqrt(1/2)*(a^2*cos(d*x 
+ c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(-a)*arctan(2*sqrt(1/2)*sqrt(-a)*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 
 32*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(-a)*arctan(sqrt(- 
a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + 
a)) - (15*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c)))/(d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d)]
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**5*(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)^5*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.20 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {\frac {43 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - \frac {64 \, a^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - \frac {\sqrt {2} {\left (13 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 11 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{32 \, d} \] Input:

integrate(cot(d*x+c)^5*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

1/32*(43*sqrt(2)*a^3*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))* 
sgn(cos(d*x + c))/sqrt(-a) - 64*a^3*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x 
 + 1/2*c)^2 + a)/sqrt(-a))*sgn(cos(d*x + c))/sqrt(-a) - sqrt(2)*(13*(-a*ta 
n(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^3*sgn(cos(d*x + c)) - 11*sqrt(-a*tan(1/2 
*d*x + 1/2*c)^2 + a)*a^4*sgn(cos(d*x + c)))/(a^2*tan(1/2*d*x + 1/2*c)^4))/ 
d
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^5*(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)^5*(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{5} \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{5} \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{5}d x \right ) \] Input:

int(cot(d*x+c)^5*(a+a*sec(d*x+c))^(5/2),x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**5*sec(c + d*x)**2,x 
) + 2*int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**5*sec(c + d*x),x) + int(sqr 
t(sec(c + d*x) + 1)*cot(c + d*x)**5,x))