\(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [133]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \] Output:

5*I*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*I*arctanh(1/2* 
(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2)/d+1/5*cot(d*x+c) 
/d/(a+I*a*tan(d*x+c))^(5/2)+19/30*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(3/2)+ 
41/12*cot(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-21/4*cot(d*x+c)*(a+I*a*tan 
(d*x+c))^(1/2)/a^3/d
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.74 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {600 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {15 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{5/2}}+\frac {2 \cot (c+d x) \left (315 i+740 \cot (c+d x)-497 i \cot ^2(c+d x)-60 \cot ^3(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{a^3 (i+\cot (c+d x))^3}}{120 d} \] Input:

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

Output:

(((600*I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/a^(5/2) + ((15*I)*S 
qrt[2]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(5/2) + (2 
*Cot[c + d*x]*(315*I + 740*Cot[c + d*x] - (497*I)*Cot[c + d*x]^2 - 60*Cot[ 
c + d*x]^3)*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*(I + Cot[c + d*x])^3))/(120*d 
)
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.10, 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, 4079, 27, 3042, 4081, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (12 a-7 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (12 a-7 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {12 a-7 i a \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {5 \cot ^2(c+d x) \left (22 a^2-19 i a^2 \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 \int \frac {\cot ^2(c+d x) \left (22 a^2-19 i a^2 \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \int \frac {22 a^2-19 i a^2 \tan (c+d x)}{\tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {5 \left (\frac {\int \frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (42 a^3-41 i a^3 \tan (c+d x)\right )dx}{a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 \left (\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (42 a^3-41 i a^3 \tan (c+d x)\right )dx}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (42 a^3-41 i a^3 \tan (c+d x)\right )}{\tan (c+d x)^2}dx}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (\frac {\int -\cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^4+20 i a^4\right )dx}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^4+20 i a^4\right )dx}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (21 \tan (c+d x) a^4+20 i a^4\right )}{\tan (c+d x)}dx}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {a^4 \int \sqrt {i \tan (c+d x) a+a}dx+20 i a^3 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {a^4 \int \sqrt {i \tan (c+d x) a+a}dx+20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^5 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {20 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\frac {20 i a^5 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\frac {40 a^4 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {5 \left (\frac {41 a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {3 \left (-\frac {-\frac {40 i a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {42 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}\right )}{6 a^2}+\frac {19 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

Cot[c + d*x]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((19*a*Cot[c + d*x])/(3* 
d*(a + I*a*Tan[c + d*x])^(3/2)) + (5*((41*a^2*Cot[c + d*x])/(d*Sqrt[a + I* 
a*Tan[c + d*x]]) + (3*(-((((-40*I)*a^(9/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d* 
x]]/Sqrt[a]])/d - (I*Sqrt[2]*a^(9/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(S 
qrt[2]*Sqrt[a])])/d)/a) - (42*a^3*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) 
/d))/(2*a^2)))/(6*a^2))/(10*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 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 = 1.54 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{5}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {11}{2}}}-\frac {17}{8 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5}{12 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{10 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\)	\(156\)
default	\(\frac {2 i a^{3} \left (\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{5}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {11}{2}}}-\frac {17}{8 a^{5} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5}{12 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{10 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\)	\(156\)

Input:

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

Output:

2*I/d*a^3*(1/a^5*(1/2*I*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+5/2/a^(1/2)* 
arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2)))+1/16/a^(11/2)*2^(1/2)*arctanh(1 
/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-17/8/a^5/(a+I*a*tan(d*x+c))^( 
1/2)-5/12/a^4/(a+I*a*tan(d*x+c))^(3/2)-1/10/a^3/(a+I*a*tan(d*x+c))^(5/2))
 

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. 612 vs. \(2 (165) = 330\).

Time = 0.10 (sec) , antiderivative size = 612, normalized size of antiderivative = 2.86 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/120*(15*sqrt(1/2)*(-I*a^3*d*e^(7*I*d*x + 7*I*c) + I*a^3*d*e^(5*I*d*x + 
5*I*c))*sqrt(1/(a^5*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I 
*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I* 
d*x + I*c))*e^(-I*d*x - I*c)) + 15*sqrt(1/2)*(I*a^3*d*e^(7*I*d*x + 7*I*c) 
- I*a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(-4*(sqrt(2)*sqrt(1/2) 
*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 150*(-I*a^3*d*e^(7 
*I*d*x + 7*I*c) + I*a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(16*(3 
*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^ 
(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a^2)* 
e^(-2*I*d*x - 2*I*c)) + 150*(I*a^3*d*e^(7*I*d*x + 7*I*c) - I*a^3*d*e^(5*I* 
d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*sqrt 
(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x 
 + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(2)*s 
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-403*I*e^(8*I*d*x + 8*I*c) - 151*I*e^(6* 
I*d*x + 6*I*c) + 280*I*e^(4*I*d*x + 4*I*c) + 31*I*e^(2*I*d*x + 2*I*c) + 3* 
I))/(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, a {\left (\frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 205 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 38 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 12 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}} + \frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {7}{2}}} + \frac {600 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}}{240 \, d} \] Input:

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

Output:

-1/240*I*a*(4*(315*(I*a*tan(d*x + c) + a)^3 - 205*(I*a*tan(d*x + c) + a)^2 
*a - 38*(I*a*tan(d*x + c) + a)*a^2 - 12*a^3)/((I*a*tan(d*x + c) + a)^(7/2) 
*a^3 - (I*a*tan(d*x + c) + a)^(5/2)*a^4) + 15*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) + 600*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*t 
an(d*x + c) + a) + sqrt(a)))/a^(7/2))/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),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 = 1.17 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,19{}\mathrm {i}}{30\,d}+\frac {a\,1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,41{}\mathrm {i}}{12\,a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3\,21{}\mathrm {i}}{4\,a^2\,d}}{a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,5{}\mathrm {i}}{a^5\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,1{}\mathrm {i}}{8\,a^5\,d} \] Input:

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

Output:

- (((a + a*tan(c + d*x)*1i)*19i)/(30*d) + (a*1i)/(5*d) + ((a + a*tan(c + d 
*x)*1i)^2*41i)/(12*a*d) - ((a + a*tan(c + d*x)*1i)^3*21i)/(4*a^2*d))/(a*(a 
 + a*tan(c + d*x)*1i)^(5/2) - (a + a*tan(c + d*x)*1i)^(7/2)) - (atan(((-a^ 
5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^3)*(-a^5)^(1/2)*5i)/(a^5*d) - (2 
^(1/2)*atan((2^(1/2)*(-a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^3))* 
(-a^5)^(1/2)*1i)/(8*a^5*d)
 

Reduce [F]

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

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

Output:

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