\(\int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx\) [1104]

Optimal result

Integrand size = 30, antiderivative size = 140 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {i c \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a \sqrt {c+i d} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {c \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right ) (1+i \tan (e+f x))+\sqrt {c-i d} (-i c+d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right ) (-i+\tan (e+f x))+(c+i d) \sqrt {c+d \tan (e+f x)}}{2 a (c+i d) f (-i+\tan (e+f x))} \] Input:

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

Output:

(c*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]*(1 + I*Ta 
n[e + f*x]) + Sqrt[c - I*d]*((-I)*c + d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/ 
Sqrt[c - I*d]]*(-I + Tan[e + f*x]) + (c + I*d)*Sqrt[c + d*Tan[e + f*x]])/( 
2*a*(c + I*d)*f*(-I + Tan[e + f*x]))
 

Rubi [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4032, 27, 3042, 4022, 3042, 4020, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4032

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {2 i a \left (c^2+d^2\right ) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a c (-d+i c) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{4 a^2 (-d+i c)}+\frac {i \sqrt {c+d \tan (e+f x)}}{2 f (a+i a \tan (e+f x))}\)

Input:

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

Output:

(((2*I)*a*(c^2 + d^2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f 
) + (2*a*c*(I*c - d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) 
)/(4*a^2*(I*c - d)) + ((I/2)*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + 
 f*x]))
 

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 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 4020

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

rule 4022

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

rule 4032

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{2 f a}-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{2 f a \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {i c \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 f a \sqrt {-i d -c}}\)	\(126\)
default	\(-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{2 f a}-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{2 f a \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {i c \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 f a \sqrt {-i d -c}}\)	\(126\)

Input:

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

Output:

-1/2*I/f/a*(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))-1/2/ 
f/a*d*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+I*d)-1/2*I/f/a*c/(-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. 686 vs. \(2 (104) = 208\).

Time = 0.14 (sec) , antiderivative size = 686, normalized size of antiderivative = 4.90 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e)),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. 355 vs. \(2 (104) = 208\).

Time = 0.53 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.54 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} c \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {\sqrt {2} {\left (c - i \, d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {i \, \sqrt {d \tan \left (f x + e\right ) + c} d}{d \tan \left (f x + e\right ) - i \, d}\right )}}{2 \, a f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 4.49 (sec) , antiderivative size = 763, normalized size of antiderivative = 5.45 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

- 2*atanh((4*a^2*d^4*f^2*((d*1i)/(16*a^2*f^2) - c/(16*a^2*f^2))^(1/2)*(c + 
 d*tan(e + f*x))^(1/2))/(a*d^5*f + a*c*d^4*f*1i))*((d*1i)/(16*a^2*f^2) - c 
/(16*a^2*f^2))^(1/2) - atan((a^2*d^4*f^2*(c + d*tan(e + f*x))^(1/2)*(-(c^2 
*1i)/(16*(a^2*c*f^2*1i - a^2*d*f^2)))^(1/2)*4i)/(a*c*d^4*f*1i - a*c^2*d^3* 
f + (a^3*c^2*d^4*f^3)/(a^2*c*f^2*1i - a^2*d*f^2) + (a^3*c^3*d^3*f^3*1i)/(a 
^2*c*f^2*1i - a^2*d*f^2)) + (8*a^4*c^3*d^2*f^4*(c + d*tan(e + f*x))^(1/2)* 
(-(c^2*1i)/(16*(a^2*c*f^2*1i - a^2*d*f^2)))^(1/2))/(a^3*c*d^5*f^3*1i + a^3 
*c^3*d^3*f^3*1i + (a^5*c^2*d^5*f^5)/(a^2*c*f^2*1i - a^2*d*f^2) + (a^5*c^4* 
d^3*f^5)/(a^2*c*f^2*1i - a^2*d*f^2)) - (8*a^2*c*d^3*f^2*(c + d*tan(e + f*x 
))^(1/2)*(-(c^2*1i)/(16*(a^2*c*f^2*1i - a^2*d*f^2)))^(1/2))/(a*c*d^4*f*1i 
- a*c^2*d^3*f + (a^3*c^2*d^4*f^3)/(a^2*c*f^2*1i - a^2*d*f^2) + (a^3*c^3*d^ 
3*f^3*1i)/(a^2*c*f^2*1i - a^2*d*f^2)) - (a^2*c^2*d^2*f^2*(c + d*tan(e + f* 
x))^(1/2)*(-(c^2*1i)/(16*(a^2*c*f^2*1i - a^2*d*f^2)))^(1/2)*8i)/(a*c*d^4*f 
*1i - a*c^2*d^3*f + (a^3*c^2*d^4*f^3)/(a^2*c*f^2*1i - a^2*d*f^2) + (a^3*c^ 
3*d^3*f^3*1i)/(a^2*c*f^2*1i - a^2*d*f^2)))*(-(c^2*1i)/(16*(a^2*c*f^2*1i - 
a^2*d*f^2)))^(1/2)*2i - (d*(c + d*tan(e + f*x))^(1/2))/(2*(a*d*f*1i - a*d* 
f*tan(e + f*x)))
 

Reduce [F]

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

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

Output:

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