\(\int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx\) [1139]

Optimal result

Integrand size = 32, antiderivative size = 151 \[ \int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx=-\frac {2 \sqrt [4]{-1} \sqrt {a} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {i \sqrt {2} \sqrt {a} \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx=\frac {i \sqrt {2} \sqrt {-a (c-i d)} \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )-\frac {2 (-1)^{3/4} \sqrt {a} \sqrt {c+i d} \sqrt {d} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {c+d \tan (e+f x)}}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4046, 3042, 4027, 221, 4082, 66, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \sqrt [4]{-1} \sqrt {a} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {i \sqrt {2} \sqrt {a} \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}\)

Input:

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

Output:

(-2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Ta 
n[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/f - (I*Sqrt[2]*Sqrt[a]*S 
qrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - 
I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f
 

Defintions of rubi rules used

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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 4027

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

rule 4046

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[(a*c - b*d)/a   Int[Sqrt[a + b*Tan[e + f 
*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Simp[d/a   Int[Sqrt[a + b*Tan[e + f 
*x]]*((b + a*Tan[e + f*x])/Sqrt[c + d*Tan[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]
 

rule 4082

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[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 865 vs. \(2 (114 ) = 228\).

Time = 0.43 (sec) , antiderivative size = 866, normalized size of antiderivative = 5.74

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {c +d \tan \left (f x +e \right )}\, a \left (-i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2} \tan \left (f x +e \right )+i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c d -\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c d \tan \left (f x +e \right )-i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c^{2}-i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d^{2}-\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2}+\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c^{2} \tan \left (f x +e \right )+\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d^{2} \tan \left (f x +e \right )\right ) \sqrt {2}}{2 f \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}\, \sqrt {-a \left (i d -c \right )}\, \left (i c -d \right ) \left (-\tan \left (f x +e \right )+i\right )}\)	\(866\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {c +d \tan \left (f x +e \right )}\, a \left (-i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2} \tan \left (f x +e \right )+i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c d -\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c d \tan \left (f x +e \right )-i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c^{2}-i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d^{2}-\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2}+\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c^{2} \tan \left (f x +e \right )+\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d^{2} \tan \left (f x +e \right )\right ) \sqrt {2}}{2 f \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}\, \sqrt {-a \left (i d -c \right )}\, \left (i c -d \right ) \left (-\tan \left (f x +e \right )+i\right )}\)	\(866\)

Input:

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

Output:

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

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. 485 vs. \(2 (109) = 218\).

Time = 0.10 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.21 \[ \int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx=\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {a c - i \, a d}{f^{2}}} \log \left ({\left (i \, \sqrt {2} f \sqrt {-\frac {a c - i \, a d}{f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{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 (-i \, f x - i \, e\right )}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {a c - i \, a d}{f^{2}}} \log \left ({\left (-i \, \sqrt {2} f \sqrt {-\frac {a c - i \, a d}{f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{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 (-i \, f x - i \, e\right )}\right ) - \frac {1}{2} \, \sqrt {\frac {4 i \, a d}{f^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} + i \, f \sqrt {\frac {4 i \, a d}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {4 i \, a d}{f^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} - i \, f \sqrt {\frac {4 i \, a d}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sqrt(I*a*tan(f*x + e) + a)*sqrt(d*tan(f*x + e) + c), x)
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 17.49 (sec) , antiderivative size = 2101, normalized size of antiderivative = 13.91 \[ \int \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2^(1/2)*a^(1/2)*d^(1/2)*log((2*d^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2) - 2* 
a^(1/2)*d^(1/2) - 2^(1/2)*a^(1/2)*c^(1/2)*(1 + 1i) + 2^(1/2)*a^(1/2)*(c + 
d*tan(e + f*x))^(1/2)*(1 + 1i))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)))*(1 
 + 1i))/f - (a^(1/2)*atan((4*(29*(1i/2)^(1/2)*c^3*(c*1i + d)^(1/2)*(a + a* 
tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(1/2) - 29*(1i/2)^(1/2)*c^(7/2 
)*(c*1i + d)^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2) - 6*(1i/2)^(1/2)*a^(1/2)* 
c^(7/2)*(c*1i + d)^(1/2) + (1i/2)^(1/2)*d^3*(c*1i + d)^(1/2)*(a + a*tan(e 
+ f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(1/2)*1i + (1i/2)^(1/2)*c^(5/2)*d*(c 
*1i + d)^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2)*1i - 35*(1i/2)^(1/2)*a^(1/2)* 
c^(5/2)*(c*1i + d)^(1/2)*(c + d*tan(e + f*x)) + (1i/2)^(1/2)*c^(1/2)*d^3*( 
c*1i + d)^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2)*5i - 25*(1i/2)^(1/2)*c^(3/2) 
*d^2*(c*1i + d)^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2) + 41*(1i/2)^(1/2)*a^(1 
/2)*c^3*(c*1i + d)^(1/2)*(c + d*tan(e + f*x))^(1/2) - (1i/2)^(1/2)*a^(1/2) 
*d^3*(c*1i + d)^(1/2)*(c + d*tan(e + f*x))^(1/2)*1i + (1i/2)^(1/2)*a^(1/2) 
*c^(5/2)*d*(c*1i + d)^(1/2)*25i - (1i/2)^(1/2)*a^(1/2)*c^(1/2)*d^3*(c*1i + 
 d)^(1/2)*5i + 28*(1i/2)^(1/2)*a^(1/2)*c^(3/2)*d^2*(c*1i + d)^(1/2) - 35*( 
1i/2)^(1/2)*a^(1/2)*c^(5/2)*d*tan(e + f*x)*(c*1i + d)^(1/2) + 3*(1i/2)^(1/ 
2)*a^(1/2)*c^(1/2)*d^3*tan(e + f*x)*(c*1i + d)^(1/2) + (1i/2)^(1/2)*a^(1/2 
)*c^(3/2)*d^2*tan(e + f*x)*(c*1i + d)^(1/2)*26i - 27*(1i/2)^(1/2)*c*d^2*(c 
*1i + d)^(1/2)*(a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(1/2)...
 

Reduce [F]

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

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

Output:

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