\(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [749]

Optimal result

Integrand size = 26, antiderivative size = 183 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2} \] Output:

1/16*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a^3/d+1/16*arctan(1+2^(1/ 
2)*cot(d*x+c)^(1/2))*2^(1/2)/a^3/d+1/16*arctanh(2^(1/2)*cot(d*x+c)^(1/2)/( 
1+cot(d*x+c)))*2^(1/2)/a^3/d+1/6*I*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^3 
+1/12*cot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\frac {\cot ^2(c+d x) (3 i+\cot (c+d x))}{(i+\cot (c+d x))^3}-\operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right )}{12 a^3 d \cot ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]
 

Output:

((Cot[c + d*x]^2*(3*I + Cot[c + d*x]))/(I + Cot[c + d*x])^3 - Hypergeometr 
ic2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])/(12*a^3*d*Cot[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.17, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.731, Rules used = {3042, 4156, 3042, 4040, 3042, 4079, 27, 2011, 3042, 3957, 266, 755, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4156

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4040

\(\displaystyle \frac {\int \frac {a-5 i a \cot (c+d x)}{\sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)^2}dx}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 i \tan \left (c+d x+\frac {\pi }{2}\right ) a+a}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (i a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int -\frac {6 \left (\cot (c+d x) a^2+i a^2\right )}{\sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)}dx}{4 a^2}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}-\frac {3 \int \frac {\cot (c+d x) a^2+i a^2}{\sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)}dx}{2 a^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 2011

\(\displaystyle \frac {-\frac {3 \int \frac {1}{\sqrt {\cot (c+d x)}}dx}{2 a}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 \int \frac {1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {3 \int \frac {1}{\sqrt {\cot (c+d x)} \left (\cot ^2(c+d x)+1\right )}d\cot (c+d x)}{2 a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {3 \int \frac {1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {\cot (c+d x)+1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\cot (c+d x)}+1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{a d}+\frac {a \sqrt {\cot (c+d x)}}{d (a \cot (c+d x)+i a)^2}}{12 a^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}\)

Input:

Int[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]
 

Output:

((I/6)*Sqrt[Cot[c + d*x]])/(d*(I*a + a*Cot[c + d*x])^3) + ((a*Sqrt[Cot[c + 
 d*x]])/(d*(I*a + a*Cot[c + d*x])^2) + (3*((-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[ 
c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2])/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]))/2))/(a*d))/(12*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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4040

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*(a + b*Tan[e + f*x])^m*(Sqrt[c + d 
*Tan[e + f*x]]/(2*a*f*m)), x] + Simp[1/(4*a^2*m)   Int[(a + b*Tan[e + f*x]) 
^(m + 1)*(Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x]/Sqrt[c + d*Ta 
n[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] & 
& EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && IntegersQ[2*m]
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 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 = 0.44 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {\frac {2 \cot \left (d x +c \right )^{\frac {3}{2}}}{3}+2 i \sqrt {\cot \left (d x +c \right )}}{8 \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{i \sqrt {2}+\sqrt {2}}\right )}{4 i \sqrt {2}+4 \sqrt {2}}}{d \,a^{3}}\)	\(120\)
default	\(\frac {-\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {\frac {2 \cot \left (d x +c \right )^{\frac {3}{2}}}{3}+2 i \sqrt {\cot \left (d x +c \right )}}{8 \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{i \sqrt {2}+\sqrt {2}}\right )}{4 i \sqrt {2}+4 \sqrt {2}}}{d \,a^{3}}\)	\(120\)

Input:

int(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d/a^3*(-1/4*I/(2^(1/2)-I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)-I*2 
^(1/2)))+1/8*(2/3*cot(d*x+c)^(3/2)+2*I*cot(d*x+c)^(1/2))/(I+cot(d*x+c))^3+ 
1/4*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. 520 vs. \(2 (144) = 288\).

Time = 0.09 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.84 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/48*(12*a^3*d*sqrt(1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(2*(8*(a^3*d* 
e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x 
+ 2*I*c) - 1))*sqrt(1/64*I/(a^6*d^2)) + I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x 
 - 2*I*c)) - 12*a^3*d*sqrt(1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8 
*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^( 
2*I*d*x + 2*I*c) - 1))*sqrt(1/64*I/(a^6*d^2)) - I*e^(2*I*d*x + 2*I*c))*e^( 
-2*I*d*x - 2*I*c)) + 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)* 
log(1/8*(8*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) 
 + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) + I)*e^(-2*I*d*x 
- 2*I*c)/(a^3*d)) - 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*l 
og(-1/8*(8*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) 
 + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) - I)*e^(-2*I*d*x 
- 2*I*c)/(a^3*d)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) 
- 1))*(2*e^(6*I*d*x + 6*I*c) - 5*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x + 2*I* 
c) - 1))*e^(-6*I*d*x - 6*I*c)/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(1/cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {i \, {\left (-\left (3 i - 3\right ) \, \sqrt {2} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \left (3 i + 3\right ) \, \sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {4 \, {\left (3 i \, \tan \left (d x + c\right )^{\frac {5}{2}} + \tan \left (d x + c\right )^{\frac {3}{2}}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{3}}\right )}}{48 \, a^{3} d} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

-1/48*I*(-(3*I - 3)*sqrt(2)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c) 
)) - (3*I + 3)*sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 
 4*(3*I*tan(d*x + c)^(5/2) + tan(d*x + c)^(3/2))/(tan(d*x + c) - I)^3)/(a^ 
3*d)
 

Mupad [F(-1)]

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

int(1/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^3),x)
 

Output:

int(1/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^3), x)
 

Reduce [F]

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

int(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,x)
 

Output:

( - int(sqrt(cot(c + d*x))/(cot(c + d*x)**3*tan(c + d*x)**3*i + 3*cot(c + 
d*x)**3*tan(c + d*x)**2 - 3*cot(c + d*x)**3*tan(c + d*x)*i - cot(c + d*x)* 
*3),x))/a**3