\(\int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx\) [1122]

Optimal result

Integrand size = 30, antiderivative size = 155 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a \sqrt {c-i d} f}+\frac {(i c-2 d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{3/2} f}-\frac {\sqrt {c+d \tan (e+f x)}}{2 (i c-d) f (a+i a \tan (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \left (\frac {(c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}-\frac {(c+2 i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {i \sqrt {c+d \tan (e+f x)}}{-i+\tan (e+f x)}\right )}{2 a (c+i d) f} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.76 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4035, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4035

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[1/((a + I*a*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]),x]
 

Output:

((2*a*(I*c - d)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + (2 
*a*(I*c - 2*d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f))/(4*a 
^2*(I*c - d)) - Sqrt[c + d*Tan[e + f*x]]/(2*(I*c - d)*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 4035

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* 
c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d))   Int[(c + d 
*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\)	\(150\)
default	\(\frac {2 d^{2} \left (\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\)	\(150\)

Input:

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

Output:

2/f/a*d^2*(1/4*I/d^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^( 
1/2))+1/4/d^2*(-1/(c+I*d)*d*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+I*d)-(I* 
c-2*d)/(c+I*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. 990 vs. \(2 (119) = 238\).

Time = 0.17 (sec) , antiderivative size = 990, normalized size of antiderivative = 6.39 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),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. 370 vs. \(2 (119) = 238\).

Time = 0.51 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.39 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} {\left (c + 2 i \, d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} + i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (c + i \, d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {\sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {4 \, \sqrt {d \tan \left (f x + e\right ) + c} d}{-4 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )} {\left (i \, c - d\right )}}\right )}}{2 \, a f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.42 (sec) , antiderivative size = 12379, normalized size of antiderivative = 79.86 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(a*d^6*f*1i - ((-(c*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c 
^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^ 
2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f 
^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4 
)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5) 
*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d 
^4))^(1/2)*1i + a^2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2 
*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 3 
2*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 
+ 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ( 
(4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4)) 
*(256*d^6 + 256*c^2*d^4))^(1/2)*1i + 2*a^2*c*d*f^2*((((48*c^2*d^7 - 48*d^9 
 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144* 
c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d 
^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 
 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^ 
4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2))/(512*(d^6 + c^2*d^ 
4)*(a^2*d^2*f^2*1i - a^2*c^2*f^2*1i + 2*a^2*c*d*f^2)))^(1/2)*(24*a^3*d^7*f 
^3 + a^3*c*d^6*f^3*16i + 32*a^3*c^2*d^5*f^3 + a^3*c^3*d^4*f^3*16i + 8*a^3* 
c^4*d^3*f^3 - 2*(c + d*tan(e + f*x))^(1/2)*(a^2*c^2*d^3*f^2*64i - 32*a^...
 

Reduce [F]

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

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

Output:

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