\(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx\) [168]

Optimal result

Integrand size = 28, antiderivative size = 202 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx=-\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{4}+\frac {i}{4}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))} \] Output:

(-3/8+1/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a/d^(1 
/2)/f+(3/8-1/8*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a 
/d^(1/2)/f+(3/8+1/8*I)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/2)/(d^(1/2)+d^(1/ 
2)*tan(f*x+e)))*2^(1/2)/a/d^(1/2)/f+1/2*(d*tan(f*x+e))^(1/2)/d/f/(a+I*a*ta 
n(f*x+e))
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx=\frac {-\frac {\sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a}-\frac {2 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a}+\frac {\sqrt {d \tan (e+f x)}}{a+i a \tan (e+f x)}}{2 d f} \] Input:

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

Output:

(-(((-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/ 
a) - (2*(-1)^(1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[ 
d]])/a + Sqrt[d*Tan[e + f*x]]/(a + I*a*Tan[e + f*x]))/(2*d*f)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.19, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4035, 27, 3042, 4017, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4035

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\left (\frac {3}{2}+\frac {i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {3}{2}-\frac {i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )}{2 a f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\left (\frac {3}{2}+\frac {i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {3}{2}-\frac {i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )}{2 a f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (\frac {3}{2}+\frac {i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {3}{2}-\frac {i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )}{2 a f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\left (\frac {3}{2}-\frac {i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {3}{2}+\frac {i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )}{2 a f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\)

Input:

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

Output:

((3/2 - I/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2 
]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]* 
Sqrt[d])) + (3/2 + I/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]*Sqr 
t[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2]*Sq 
rt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d])))/(2*a*f) + Sqrt[d*Tan[e + 
 f*x]]/(2*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 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 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 = 1.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {\sqrt {d \tan \left (f x +e \right )}}{2 f a \left (i d \tan \left (f x +e \right )+d \right )}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{f a \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}}\)	\(100\)
default	\(\frac {\sqrt {d \tan \left (f x +e \right )}}{2 f a \left (i d \tan \left (f x +e \right )+d \right )}-\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{f a \sqrt {-i d}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{2 f a \sqrt {i d}}\)	\(100\)

Input:

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

Output:

1/2/f/a*(d*tan(f*x+e))^(1/2)/(I*d*tan(f*x+e)+d)-I/f/a/(-I*d)^(1/2)*arctan( 
(d*tan(f*x+e))^(1/2)/(-I*d)^(1/2))+1/2*I/f/a/(I*d)^(1/2)*arctan((d*tan(f*x 
+e))^(1/2)/(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. 515 vs. \(2 (146) = 292\).

Time = 0.09 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.55 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx=-\frac {{\left (a d f \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (2 \, {\left (2 \, {\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt {\frac {i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{a^{2} d f^{2}}} + i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + a d f \sqrt {\frac {i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{a^{2} d f^{2}}} - i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a d f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(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 [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx=\frac {i \, {\left (\frac {\sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} - \frac {2 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} - \frac {\sqrt {d \tan \left (f x + e\right )}}{d \tan \left (f x + e\right ) - i \, d}\right )}}{2 \, a f} \] Input:

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

Output:

1/2*I*(sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3/2) + s 
qrt(2)*sqrt(d)*abs(d)))/(sqrt(d)*(I*d/abs(d) + 1)) - 2*sqrt(2)*arctan(2*sq 
rt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/( 
sqrt(d)*(-I*d/abs(d) + 1)) - sqrt(d*tan(f*x + e))/(d*tan(f*x + e) - I*d))/ 
(a*f)
 

Mupad [B] (verification not implemented)

Time = 2.93 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx=-\mathrm {atan}\left (2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{4\,a^2\,d\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{4\,a^2\,d\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d\,f^2}}\,2{}\mathrm {i}+\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \] Input:

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

Output:

atan(4*a*f*(d*tan(e + f*x))^(1/2)*(-1i/(16*a^2*d*f^2))^(1/2))*(-1i/(16*a^2 
*d*f^2))^(1/2)*2i - atan(2*a*f*(d*tan(e + f*x))^(1/2)*(1i/(4*a^2*d*f^2))^( 
1/2))*(1i/(4*a^2*d*f^2))^(1/2)*2i + ((d*tan(e + f*x))^(1/2)*1i)/(2*a*f*(d* 
1i - d*tan(e + f*x)))
 

Reduce [F]

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

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

Output:

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