\(\int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [1105]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 30, antiderivative size = 211 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}-\frac {\left (2 c d-i \left (2 c^2+d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 (c+i d)^{3/2} f}+\frac {(2 i c-d) \sqrt {c+d \tan (e+f x)}}{8 a^2 (c+i d) f (1+i \tan (e+f x))}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \] Output:

-1/4*I*(c-I*d)^(1/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a^2/f-1 
/8*(2*c*d-I*(2*c^2+d^2))*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a^2 
/(c+I*d)^(3/2)/f+1/8*(2*I*c-d)*(c+d*tan(f*x+e))^(1/2)/a^2/(c+I*d)/f/(1+I*t 
an(f*x+e))+1/4*I*(c+d*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \left (2 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\frac {\left (2 c^2+2 i c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right ) \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+\sqrt {c+i d} (4 c+3 i d+(2 i c-d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{(c+i d)^{3/2} (-i+\tan (e+f x))^2}\right )}{8 a^2 f} \] Input:

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

Output:

((-1/8*I)*(2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] 
 + ((2*c^2 + (2*I)*c*d + d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I* 
d]]*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + Sqrt[c + I*d] 
*(4*c + (3*I)*d + ((2*I)*c - d)*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(( 
c + I*d)^(3/2)*(-I + Tan[e + f*x])^2)))/(a^2*f)
 

Rubi [F]

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4040

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 25

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

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]
 
Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.27

method result size
derivativedivides \(\frac {2 d^{3} \left (-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {\frac {\frac {d \left (2 i c^{2}-i d^{2}-3 c d \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 i c^{2}-2 i d^{2}-4 c d}-\frac {i d \left (7 i c^{2} d -3 i d^{3}+2 c^{3}-8 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (i c^{2}-i d^{2}-2 c d \right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {i \left (2 i c^{3}-i c \,d^{2}-4 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (i c^{2}-i d^{2}-2 c d \right ) \sqrt {-i d -c}}}{8 d^{3}}\right )}{f \,a^{2}}\) \(269\)
default \(\frac {2 d^{3} \left (-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}+\frac {\frac {\frac {d \left (2 i c^{2}-i d^{2}-3 c d \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 i c^{2}-2 i d^{2}-4 c d}-\frac {i d \left (7 i c^{2} d -3 i d^{3}+2 c^{3}-8 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (i c^{2}-i d^{2}-2 c d \right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {i \left (2 i c^{3}-i c \,d^{2}-4 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (i c^{2}-i d^{2}-2 c d \right ) \sqrt {-i d -c}}}{8 d^{3}}\right )}{f \,a^{2}}\) \(269\)

Input:

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

Output:

2/f/a^2*d^3*(-1/8*I*(I*d-c)^(1/2)/d^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c 
)^(1/2))+1/8/d^3*((1/2*d*(2*I*c^2-I*d^2-3*c*d)/(I*c^2-I*d^2-2*c*d)*(c+d*ta 
n(f*x+e))^(3/2)-1/2*I*d*(7*I*c^2*d-3*I*d^3+2*c^3-8*c*d^2)/(I*c^2-I*d^2-2*c 
*d)*(c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d)^2-1/2*I*(-d^3+2*I*c^3-I*c* 
d^2-4*c^2*d)/(I*c^2-I*d^2-2*c*d)/(-c-I*d)^(1/2)*arctan((c+d*tan(f*x+e))^(1 
/2)/(-c-I*d)^(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. 1086 vs. \(2 (162) = 324\).

Time = 0.25 (sec) , antiderivative size = 1086, normalized size of antiderivative = 5.15 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/32*(2*(-I*a^2*c + a^2*d)*f*sqrt(-(c - I*d)/(a^4*f^2))*e^(4*I*f*x + 4*I*e 
)*log(-2*((I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f 
*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c - I*d)/(a^4*f^2 
)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 2*(I*a^2*c 
 - a^2*d)*f*sqrt(-(c - I*d)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*((-I*a^2 
*f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c 
+ I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c - I*d)/(a^4*f^2)) - (c - I*d)*e 
^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + (-I*a^2*c + a^2*d)*f*sqrt( 
-(-4*I*c^4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a 
^4*c*d^2 - a^4*d^3)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(-2*I*c^3 + 4*c^2*d 
 + I*c*d^2 + d^3 + ((a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*e^(2*I*f*x + 2*I*e 
) + (a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I* 
e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-4*I*c^4 + 8*c^3*d + 4*c*d 
^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a^4*c*d^2 - a^4*d^3)*f^2)) + 
(-2*I*c^3 + 2*c^2*d - I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/( 
(a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f)) + (I*a^2*c - a^2*d)*f*sqrt(-(-4*I*c^ 
4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a^4*c*d^2 
- a^4*d^3)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(-2*I*c^3 + 4*c^2*d + I*c*d^ 
2 + d^3 - ((a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*e^(2*I*f*x + 2*I*e) + (a^2* 
c^2 + 2*I*a^2*c*d - a^2*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c...
 

Sympy [F]

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

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

Output:

-Integral(sqrt(c + d*tan(e + f*x))/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1 
), x)/a**2
 

Maxima [F(-2)]

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

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

Output:

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (162) = 324\).

Time = 0.53 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.17 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {2 \, \sqrt {2} {\left (i \, c + d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {4 \, \sqrt {2} {\left (2 \, c^{2} + 2 i \, c d + d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{-4 \, {\left (-i \, c + d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} d + i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{2} - 5 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d^{2} + 3 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} {\left (c + i \, d\right )}}}{8 \, a^{2} f} \] Input:

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

Output:

1/8*(2*sqrt(2)*(I*c + d)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + 
 d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I* 
sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + s 
qrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) 
+ 1)) + 4*sqrt(2)*(2*c^2 + 2*I*c*d + d^2)*arctan(2*(sqrt(d*tan(f*x + e) + 
c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt 
(c^2 + d^2)) + I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + 
 d^2)*sqrt(-c + sqrt(c^2 + d^2))))/((4*I*c - 4*d)*sqrt(-c + sqrt(c^2 + d^2 
))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + (2*(d*tan(f*x + e) + c)^(3/2)*c*d - 
2*sqrt(d*tan(f*x + e) + c)*c^2*d + I*(d*tan(f*x + e) + c)^(3/2)*d^2 - 5*I* 
sqrt(d*tan(f*x + e) + c)*c*d^2 + 3*sqrt(d*tan(f*x + e) + c)*d^3)/((d*tan(f 
*x + e) - I*d)^2*(c + I*d)))/(a^2*f)
 

Mupad [B] (verification not implemented)

Time = 4.96 (sec) , antiderivative size = 14675, normalized size of antiderivative = 69.55 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

log(((-(3*d^9 - c*d^8*9i + 12*c^2*d^7 - c^3*d^6*16i + 8*c^4*d^5 - c^5*d^4* 
8i - a^4*c^2*f^2*((((3*d^11 + 27*c^2*d^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^ 
4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24 
*c^5*d^6 + 8*c^7*d^4)/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 
 4*(256*d^6 + 256*c^2*d^4)*((((5*c^3*d^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 
 + (c^7*d^5)/8)*1i)/(a^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^1 
2/64 - (11*c^2*d^10)/32 - (11*c^4*d^8)/64 + (c^6*d^6)/8 + (c^8*d^4)/16)/(a 
^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4)))^(1/2)*1i + a^4*d^2*f^2*((( 
(3*d^11 + 27*c^2*d^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^4*c^4*f^2 + a^4*d^4* 
f^2 + 2*a^4*c^2*d^2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24*c^5*d^6 + 8*c^7*d^4 
)/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 4*(256*d^6 + 256*c^ 
2*d^4)*((((5*c^3*d^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/ 
(a^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^12/64 - (11*c^2*d^10) 
/32 - (11*c^4*d^8)/64 + (c^6*d^6)/8 + (c^8*d^4)/16)/(a^8*c^4*f^4 + a^8*d^4 
*f^4 + 2*a^8*c^2*d^2*f^4)))^(1/2)*1i + 2*a^4*c*d*f^2*((((3*d^11 + 27*c^2*d 
^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^ 
2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24*c^5*d^6 + 8*c^7*d^4)/(a^4*c^4*f^2 + a 
^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 4*(256*d^6 + 256*c^2*d^4)*((((5*c^3*d 
^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/(a^8*c^4*f^4 + a^8 
*d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^12/64 - (11*c^2*d^10)/32 - (11*c^4*d...
 

Reduce [F]

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

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

Output:

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