\(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [742]

Optimal result

Integrand size = 26, antiderivative size = 191 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2} \] Output:

(-1/32+3/32*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a^2/d+(-1/32+3/ 
32*I)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a^2/d-(1/32+3/32*I)*arcta 
nh(2^(1/2)*cot(d*x+c)^(1/2)/(1+cot(d*x+c)))*2^(1/2)/a^2/d+3/8*I*cot(d*x+c) 
^(1/2)/a^2/d/(I+cot(d*x+c))+1/4*cot(d*x+c)^(1/2)/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 = 0.57 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {\sqrt {\cot (c+d x)} \left (\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right )+\cos (c+d x) (\cos (3 (c+d x))-i \sin (3 (c+d x)))\right )}{8 a^2 d} \] Input:

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

Output:

-1/8*(Sqrt[Cot[c + d*x]]*(Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[c + d*x 
]] - 2*Hypergeometric2F1[-1/2, 1, 1/2, I*Tan[c + d*x]] + Cos[c + d*x]*(Cos 
[3*(c + d*x)] - I*Sin[3*(c + d*x)])))/(a^2*d)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.13, 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, 4041, 27, 3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4156

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\int \frac {i a-5 a \cot (c+d x)}{\sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)}dx}{8 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\int \frac {5 \tan \left (c+d x+\frac {\pi }{2}\right ) a+i a}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (i a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{8 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int -\frac {a^2-3 i a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{2 a^2}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int \frac {a^2-3 i a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{2 a^2}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int \frac {3 i \tan \left (c+d x+\frac {\pi }{2}\right ) a^2+a^2}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int -\frac {a^2 (1-3 i \cot (c+d x))}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{a^2 d}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int \frac {a^2 (1-3 i \cot (c+d x))}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{a^2 d}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int \frac {1-3 i \cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\left (\frac {1}{2}-\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\left (\frac {1}{2}+\frac {3 i}{2}\right ) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )}{d}-\frac {3 i \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]/(4*d*(I*a + a*Cot[c + d*x])^2) - (((-3*I)*Sqrt[Cot[c + 
d*x]])/(d*(I + Cot[c + d*x])) + ((1/2 - (3*I)/2)*(-(ArcTan[1 - Sqrt[2]*Sqr 
t[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2] 
) + (1/2 + (3*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)/(8*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 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 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m)   Int[(a + 
b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta 
n[e + f*x], 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] && GtQ[n, 1] && (In 
tegerQ[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 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 = 0.45 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.63

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

Input:

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

Output:

1/a^2/d*(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)))-1/4*(-3/2*I*cot(d*x+c)^(3/2)+1/2*cot(d*x+c)^(1/2))/(I+cot(d*x+c))^ 
2+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. 509 vs. \(2 (140) = 280\).

Time = 0.13 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.66 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (4 \, a^{2} d \sqrt {\frac {i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (2 \, {\left (4 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{16 \, a^{4} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, a^{2} d \sqrt {\frac {i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, {\left (4 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{16 \, a^{4} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, a^{2} d \sqrt {-\frac {i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left (8 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{64 \, a^{4} d^{2}}} + i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 4 \, a^{2} d \sqrt {-\frac {i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (8 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{64 \, a^{4} d^{2}}} - i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (2 \, e^{\left (4 i \, d x + 4 i \, c\right )} - e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \] Input:

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

Output:

-1/16*(4*a^2*d*sqrt(1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(2*(4*(a^2*d* 
e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x 
+ 2*I*c) - 1))*sqrt(1/16*I/(a^4*d^2)) + I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x 
 - 2*I*c)) - 4*a^2*d*sqrt(1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*(4* 
(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2 
*I*d*x + 2*I*c) - 1))*sqrt(1/16*I/(a^4*d^2)) - I*e^(2*I*d*x + 2*I*c))*e^(- 
2*I*d*x - 2*I*c)) - 4*a^2*d*sqrt(-1/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*lo 
g(1/8*(8*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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^4*d^2)) + I)*e^(-2*I*d*x - 
2*I*c)/(a^2*d)) + 4*a^2*d*sqrt(-1/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log( 
-1/8*(8*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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^4*d^2)) - I)*e^(-2*I*d*x - 2 
*I*c)/(a^2*d)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1 
))*(2*e^(4*I*d*x + 4*I*c) - e^(2*I*d*x + 2*I*c) - 1))*e^(-4*I*d*x - 4*I*c) 
/(a^2*d)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {-\left (i + 1\right ) \, \sqrt {2} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \frac {2 \, {\left (\tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, \sqrt {\tan \left (d x + c\right )}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, a^{2} d} \] Input:

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

Output:

1/16*(-(I + 1)*sqrt(2)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 
(2*I - 2)*sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) + 2*(t 
an(d*x + c)^(3/2) - 3*I*sqrt(tan(d*x + c)))/(tan(d*x + c) - I)^2)/(a^2*d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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