\(\int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [125]

Optimal result

Integrand size = 26, antiderivative size = 220 \[ \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {23 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d} \] Output:

23/4*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d-1/4*arctanh(1/2*( 
a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(3/2)/d+1/3*cot(d*x+c)^ 
2/d/(a+I*a*tan(d*x+c))^(3/2)+17/6*cot(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(1/2 
)+21/4*I*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d-11/3*cot(d*x+c)^2*(a+I* 
a*tan(d*x+c))^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.66 \[ \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {69 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-\frac {\sqrt {a} \left (63 i+82 \cot (c+d x)-9 i \cot ^2(c+d x)+6 \cot ^3(c+d x)\right )}{(i+\cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}}{12 a^{3/2} d} \] Input:

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

Output:

(69*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - 3*Sqrt[2]*ArcTanh[Sqrt[a 
 + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] - (Sqrt[a]*(63*I + 82*Cot[c + d*x] 
 - (9*I)*Cot[c + d*x]^2 + 6*Cot[c + d*x]^3))/((I + Cot[c + d*x])*Sqrt[a + 
I*a*Tan[c + d*x]]))/(12*a^(3/2)*d)
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.08, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {\cot ^3(c+d x) (10 a-7 i a \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cot ^3(c+d x) (10 a-7 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {10 a-7 i a \tan (c+d x)}{\tan (c+d x)^3 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {1}{2} \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (88 a^2-85 i a^2 \tan (c+d x)\right )dx}{a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (88 a^2-85 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (88 a^2-85 i a^2 \tan (c+d x)\right )}{\tan (c+d x)^3}dx}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {\frac {\int -6 \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (22 \tan (c+d x) a^3+21 i a^3\right )dx}{2 a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (22 \tan (c+d x) a^3+21 i a^3\right )dx}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (22 \tan (c+d x) a^3+21 i a^3\right )}{\tan (c+d x)^2}dx}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (23 a^4-21 i a^4 \tan (c+d x)\right )dx}{a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (23 a^4-21 i a^4 \tan (c+d x)\right )dx}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (23 a^4-21 i a^4 \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {2 i a^4 \int \sqrt {i \tan (c+d x) a+a}dx+23 a^3 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {2 i a^4 \int \sqrt {i \tan (c+d x) a+a}dx+23 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\frac {4 a^5 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+23 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {23 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {2 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\frac {23 a^5 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {2 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {-\frac {3 \left (\frac {\frac {2 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {46 i a^4 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {44 a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {3 \left (\frac {\frac {2 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {46 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a}-\frac {21 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{a}}{2 a^2}+\frac {17 a \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

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

Output:

Cot[c + d*x]^2/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((17*a*Cot[c + d*x]^2) 
/(d*Sqrt[a + I*a*Tan[c + d*x]]) + ((-44*a^2*Cot[c + d*x]^2*Sqrt[a + I*a*Ta 
n[c + d*x]])/d - (3*(((-46*a^(9/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt 
[a]])/d + (2*Sqrt[2]*a^(9/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*S 
qrt[a])])/d)/(2*a) - ((21*I)*a^3*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/ 
d))/a)/(2*a^2))/(6*a^2)
 

Defintions of rubi rules used

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 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 221

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

rule 3042

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

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(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))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*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] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

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 4081

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 

rule 4082

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[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4083

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

Maple [A] (verified)

Time = 1.56 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {2 a^{4} \left (\frac {-\frac {-\frac {7 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {23 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{5}}-\frac {7}{4 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {11}{2}}}\right )}{d}\)	\(155\)
default	\(\frac {2 a^{4} \left (\frac {-\frac {-\frac {7 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {23 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{5}}-\frac {7}{4 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {11}{2}}}\right )}{d}\)	\(155\)

Input:

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

Output:

2/d*a^4*(1/a^5*(-(-7/8*(a+I*a*tan(d*x+c))^(3/2)+9/8*a*(a+I*a*tan(d*x+c))^( 
1/2))/a^2/tan(d*x+c)^2+23/8/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/ 
2)))-7/4/a^5/(a+I*a*tan(d*x+c))^(1/2)-1/6/a^4/(a+I*a*tan(d*x+c))^(3/2)-1/8 
/a^(11/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(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. 678 vs. \(2 (171) = 342\).

Time = 0.09 (sec) , antiderivative size = 678, normalized size of antiderivative = 3.08 \[ \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/48*(12*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x + 5*I* 
c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2) 
*(a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 12*sqrt(1/2)*(a^2* 
d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3 
*I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I 
*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - a*e^(I* 
d*x + I*c))*e^(-I*d*x - I*c)) - 69*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^ 
(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(16*(3 
*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^ 
(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a^2)* 
e^(-2*I*d*x - 2*I*c)) + 69*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x 
 + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(16*(3*a^2*e^( 
2*I*d*x + 2*I*c) - 2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + 
 I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a^2)*e^(-2*I* 
d*x - 2*I*c)) + 4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(37*e^(8*I*d*x 
 + 8*I*c) - 33*e^(6*I*d*x + 6*I*c) - 50*e^(4*I*d*x + 4*I*c) + 21*e^(2*I*d* 
x + 2*I*c) + 1))/(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x + 5*I*c) 
+ a^2*d*e^(3*I*d*x + 3*I*c))
 

Sympy [F]

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

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

Output:

Integral(cot(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00 \[ \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {a^{2} {\left (\frac {2 \, {\left (63 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 107 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 34 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 4 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5}} - \frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {7}{2}}} + \frac {69 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}}{24 \, d} \] Input:

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

Output:

-1/24*a^2*(2*(63*(I*a*tan(d*x + c) + a)^3 - 107*(I*a*tan(d*x + c) + a)^2*a 
 + 34*(I*a*tan(d*x + c) + a)*a^2 + 4*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^ 
3 - 2*(I*a*tan(d*x + c) + a)^(5/2)*a^4 + (I*a*tan(d*x + c) + a)^(3/2)*a^5) 
 - 3*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)* 
sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(7/2) + 69*log((sqrt(I*a*tan(d*x 
+ c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(7/2))/d
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {a^2}{3}+\frac {21\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{4\,a}+\frac {17\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}-\frac {107\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{12}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-2\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {23\,\mathrm {atanh}\left (\frac {a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}}\right )}{4\,d\,\sqrt {a^3}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a^3}}\right )}{4\,d\,\sqrt {a^3}} \] Input:

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

Output:

(23*atanh((a*(a + a*tan(c + d*x)*1i)^(1/2))/(a^3)^(1/2)))/(4*d*(a^3)^(1/2) 
) - ((21*(a + a*tan(c + d*x)*1i)^3)/(4*a) - (107*(a + a*tan(c + d*x)*1i)^2 
)/12 + (17*a*(a + a*tan(c + d*x)*1i))/6 + a^2/3)/(d*(a + a*tan(c + d*x)*1i 
)^(7/2) - 2*a*d*(a + a*tan(c + d*x)*1i)^(5/2) + a^2*d*(a + a*tan(c + d*x)* 
1i)^(3/2)) - (2^(1/2)*atanh((2^(1/2)*a*(a + a*tan(c + d*x)*1i)^(1/2))/(2*( 
a^3)^(1/2))))/(4*d*(a^3)^(1/2))
 

Reduce [F]

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

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

Output:

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