\(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx\) [739]

Optimal result

Integrand size = 26, antiderivative size = 179 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {3}{4}+\frac {5 i}{4}\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {3}{4}-\frac {5 i}{4}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {5 i}{2 a d \sqrt {\cot (c+d x)}}-\frac {1}{2 d \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \] Output:

(-3/8-5/8*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a/d-(3/8+5/8*I)*a 
rctan(1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a/d+(-3/8+5/8*I)*arctanh(2^(1/2) 
*cot(d*x+c)^(1/2)/(1+cot(d*x+c)))*2^(1/2)/a/d-5/2*I/a/d/cot(d*x+c)^(1/2)-1 
/2/d/cot(d*x+c)^(1/2)/(I*a+a*cot(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 1.87 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {-\frac {8}{i+\cot (c+d x)}-\frac {5 i \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+8 \sqrt {\tan (c+d x)}\right )}{\sqrt {\tan (c+d x)}}+8 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)}{16 a d \sqrt {\cot (c+d x)}} \] Input:

Integrate[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])),x]
 

Output:

(-8/(I + Cot[c + d*x]) - ((5*I)*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + 
 d*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Sqrt[2]*Log[1 
 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sq 
rt[Tan[c + d*x]] + Tan[c + d*x]] + 8*Sqrt[Tan[c + d*x]]))/Sqrt[Tan[c + d*x 
]] + 8*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2]*Tan[c + d*x])/(16*a 
*d*Sqrt[Cot[c + d*x]])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.15, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 4156, 3042, 4035, 27, 3042, 4012, 25, 3042, 4017, 25, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot (c+d x)^{5/2} (a+i a \tan (c+d x))}dx\)

\(\Big \downarrow \) 4156

\(\displaystyle \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4035

\(\displaystyle -\frac {\int \frac {5 i a-3 a \cot (c+d x)}{2 \cot ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {5 i a-3 a \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)}dx}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {3 \tan \left (c+d x+\frac {\pi }{2}\right ) a+5 i a}{\left (-\tan \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 4012

\(\displaystyle -\frac {\int -\frac {5 i \cot (c+d x) a+3 a}{\sqrt {\cot (c+d x)}}dx+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {10 i a}{d \sqrt {\cot (c+d x)}}-\int \frac {5 i \cot (c+d x) a+3 a}{\sqrt {\cot (c+d x)}}dx}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {10 i a}{d \sqrt {\cot (c+d x)}}-\int \frac {3 a-5 i a \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 4017

\(\displaystyle -\frac {\frac {10 i a}{d \sqrt {\cot (c+d x)}}-\frac {2 \int -\frac {a (5 i \cot (c+d x)+3)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {2 \int \frac {a (5 i \cot (c+d x)+3)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {2 a \int \frac {5 i \cot (c+d x)+3}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 1482

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \int \frac {\cot (c+d x)+1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}-\frac {5 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\cot (c+d x)}+1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )+\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {\frac {2 a \left (\left (\frac {3}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\left (\frac {3}{2}-\frac {5 i}{2}\right ) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {10 i a}{d \sqrt {\cot (c+d x)}}}{4 a^2}-\frac {1}{2 d \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)}\)

Input:

Int[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])),x]
 

Output:

-1/2*1/(d*Sqrt[Cot[c + d*x]]*(I*a + a*Cot[c + d*x])) - (((10*I)*a)/(d*Sqrt 
[Cot[c + d*x]]) + (2*a*((3/2 + (5*I)/2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + 
 d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) + (3/2 
- (5*I)/2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/Sqrt[2 
] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2]))))/d)/( 
4*a^2)
 

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 217

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

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

rule 4012

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4035

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* 
c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d))   Int[(c + d 
*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
&& NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]
 

rule 4156

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Cot[e + f*x])^(m - n*p 
)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-\frac {2 i}{\sqrt {\cot \left (d x +c \right )}}-\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {i \sqrt {\cot \left (d x +c \right )}}{2 \left (i+\cot \left (d x +c \right )\right )}-\frac {4 i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{i \sqrt {2}+\sqrt {2}}\right )}{i \sqrt {2}+\sqrt {2}}}{d a}\)	\(118\)
default	\(\frac {-\frac {2 i}{\sqrt {\cot \left (d x +c \right )}}-\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {i \sqrt {\cot \left (d x +c \right )}}{2 \left (i+\cot \left (d x +c \right )\right )}-\frac {4 i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{i \sqrt {2}+\sqrt {2}}\right )}{i \sqrt {2}+\sqrt {2}}}{d a}\)	\(118\)

Input:

int(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d/a*(-2*I/cot(d*x+c)^(1/2)-I/(2^(1/2)-I*2^(1/2))*arctan(2*cot(d*x+c)^(1/ 
2)/(2^(1/2)-I*2^(1/2)))-1/2*I*cot(d*x+c)^(1/2)/(I+cot(d*x+c))-4*I/(I*2^(1/ 
2)+2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(I*2^(1/2)+2^(1/2))))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (130) = 260\).

Time = 0.09 (sec) , antiderivative size = 548, normalized size of antiderivative = 3.06 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx =\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
 

Output:

1/4*((a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(1/4*I/(a^2*d 
^2))*log(2*(2*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) 
+ I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/4*I/(a^2*d^2)) + I*e^(2*I*d*x + 2*I 
*c))*e^(-2*I*d*x - 2*I*c)) - (a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2 
*I*c))*sqrt(1/4*I/(a^2*d^2))*log(-2*(2*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqr 
t((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/4*I/(a^2*d 
^2)) - I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - (a*d*e^(4*I*d*x + 4* 
I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-4*I/(a^2*d^2))*log(-((a*d*e^(2*I*d*x 
 + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1 
))*sqrt(-4*I/(a^2*d^2)) + 2*I)*e^(-2*I*d*x - 2*I*c)/(a*d)) + (a*d*e^(4*I*d 
*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt(-4*I/(a^2*d^2))*log(((a*d*e^(2 
*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* 
c) - 1))*sqrt(-4*I/(a^2*d^2)) - 2*I)*e^(-2*I*d*x - 2*I*c)/(a*d)) - sqrt((I 
*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(9*e^(4*I*d*x + 4*I*c 
) - 8*e^(2*I*d*x + 2*I*c) - 1))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x 
+ 2*I*c))
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=- \frac {i \int \frac {1}{\tan {\left (c + d x \right )} \cot ^{\frac {5}{2}}{\left (c + d x \right )} - i \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a} \] Input:

integrate(1/cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*Integral(1/(tan(c + d*x)*cot(c + d*x)**(5/2) - I*cot(c + d*x)**(5/2)), 
x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {i \, {\left (-\left (4 i - 4\right ) \, \sqrt {2} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \left (i + 1\right ) \, \sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - 8 \, \sqrt {\tan \left (d x + c\right )} + \frac {2 i \, \sqrt {\tan \left (d x + c\right )}}{\tan \left (d x + c\right ) - i}\right )}}{4 \, a d} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
 

Output:

1/4*I*(-(4*I - 4)*sqrt(2)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c))) 
 + (I + 1)*sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 8*s 
qrt(tan(d*x + c)) + 2*I*sqrt(tan(d*x + c))/(tan(d*x + c) - I))/(a*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:

int(1/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)),x)
 

Output:

int(1/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)), x)
 

Reduce [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\int \frac {\sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{3} \tan \left (d x +c \right ) i +\cot \left (d x +c \right )^{3}}d x}{a} \] Input:

int(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x)
 

Output:

int(sqrt(cot(c + d*x))/(cot(c + d*x)**3*tan(c + d*x)*i + cot(c + d*x)**3), 
x)/a