3.12.40 \(\int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1140]

3.12.40.1 Optimal result
3.12.40.2 Mathematica [A] (verified)
3.12.40.3 Rubi [A] (verified)
3.12.40.4 Maple [B] (verified)
3.12.40.5 Fricas [B] (verification not implemented)
3.12.40.6 Sympy [F]
3.12.40.7 Maxima [F(-1)]
3.12.40.8 Giac [F(-2)]
3.12.40.9 Mupad [B] (verification not implemented)

3.12.40.1 Optimal result

Integrand size = 32, antiderivative size = 121 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f}+\frac {i \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}} \]

output
-1/2*I*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))*(c-I*d)^(1/2)/f*2^(1/2)/a^(1/2)+I*(c+d*tan(f*x+e))^(1/ 
2)/f/(a+I*a*tan(f*x+e))^(1/2)
 
3.12.40.2 Mathematica [A] (verified)

Time = 1.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {i \left (\frac {\sqrt {2} \sqrt {-a (c-i d)} \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{a}+\frac {2 \sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{2 f} \]

input
Integrate[Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + I*a*Tan[e + f*x]],x]
 
output
((I/2)*((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]])])/a + (2*Sqrt[c 
+ d*Tan[e + f*x]])/Sqrt[a + I*a*Tan[e + f*x]]))/f
 
3.12.40.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4029, 3042, 4027, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

input
Int[Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + I*a*Tan[e + f*x]],x]
 
output
((-I)*Sqrt[c - I*d]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sq 
rt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[2]*Sqrt[a]*f) + (I*Sqrt[c 
+ d*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x]])
 

3.12.40.3.1 Defintions of rubi rules used

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 4029
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^n/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 
3.12.40.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 876 vs. \(2 (95 ) = 190\).

Time = 3.48 (sec) , antiderivative size = 877, normalized size of antiderivative = 7.25

method result size
derivativedivides \(-\frac {\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (-i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d \left (\tan ^{2}\left (f x +e \right )\right )+2 i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c \tan \left (f x +e \right )-\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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c \left (\tan ^{2}\left (f x +e \right )\right )+i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d -2 \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d \tan \left (f x +e \right )+\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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c -4 i c \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, \tan \left (f x +e \right )-4 i \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, d +4 \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, d \tan \left (f x +e \right )-4 c \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\right )}{4 f a \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, \left (i c -d \right ) \left (-\tan \left (f x +e \right )+i\right )^{2}}\) \(877\)
default \(-\frac {\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (-i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d \left (\tan ^{2}\left (f x +e \right )\right )+2 i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c \tan \left (f x +e \right )-\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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c \left (\tan ^{2}\left (f x +e \right )\right )+i \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d -2 \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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d \tan \left (f x +e \right )+\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 (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c -4 i c \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, \tan \left (f x +e \right )-4 i \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, d +4 \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, d \tan \left (f x +e \right )-4 c \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\right )}{4 f a \sqrt {a \left (1+i \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )}\, \left (i c -d \right ) \left (-\tan \left (f x +e \right )+i\right )^{2}}\) \(877\)

input
int((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(1/2),x,method=_RETURNVERBOS 
E)
 
output
-1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*(-I*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*(1+ 
I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c) 
)^(1/2)*d*tan(f*x+e)^2+2*I*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2 
))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*c*tan(f*x+e)-ln((3*a*c+I*a*t 
an(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*(1+I*ta 
n(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1 
/2)*c*tan(f*x+e)^2+I*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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(ta 
n(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d-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*(1+I*tan(f*x+e))*(c+d* 
tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2)*(-a*(I*d-c))^(1/2)*d*tan(f*x+e 
)+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*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*2^(1/2 
)*(-a*(I*d-c))^(1/2)*c-4*I*c*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*t 
an(f*x+e)-4*I*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d+4*(a*(1+I*tan( 
f*x+e))*(c+d*tan(f*x+e)))^(1/2)*d*tan(f*x+e)-4*c*(a*(1+I*tan(f*x+e))*(c+d* 
tan(f*x+e)))^(1/2))/(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)/(I*c-d)/(- 
tan(f*x+e)+I)^2
 
3.12.40.5 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. 352 vs. \(2 (89) = 178\).

Time = 0.26 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.91 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {{\left (\sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (i \, \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-i \, \sqrt {2} a f \sqrt {-\frac {c - i \, d}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) - 2 \, \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-i \, e^{\left (2 i \, f x + 2 i \, e\right )} - i\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a f} \]

input
integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fr 
icas")
 
output
1/4*(sqrt(2)*a*f*sqrt(-(c - I*d)/(a*f^2))*e^(I*f*x + I*e)*log(I*sqrt(2)*a* 
f*sqrt(-(c - I*d)/(a*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)) - sqrt(2)*a*f*sqrt(-(c - I*d)/(a*f 
^2))*e^(I*f*x + I*e)*log(-I*sqrt(2)*a*f*sqrt(-(c - I*d)/(a*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)) - 2*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))*(-I*e^(2*I*f*x + 2*I*e) 
- I))*e^(-I*f*x - I*e)/(a*f)
 
3.12.40.6 Sympy [F]

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

input
integrate((c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(1/2),x)
 
output
Integral(sqrt(c + d*tan(e + f*x))/sqrt(I*a*(tan(e + f*x) - I)), x)
 
3.12.40.7 Maxima [F(-1)]

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

input
integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="ma 
xima")
 
output
Timed out
 
3.12.40.8 Giac [F(-2)]

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

input
integrate((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="gi 
ac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0,0] was discard 
ed and replaced randomly by 0=[-63,1]Warning, replacing -63 by 7, a substi 
tution va
 
3.12.40.9 Mupad [B] (verification not implemented)

Time = 21.84 (sec) , antiderivative size = 1724, normalized size of antiderivative = 14.25 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\text {Too large to display} \]

input
int((c + d*tan(e + f*x))^(1/2)/(a + a*tan(e + f*x)*1i)^(1/2),x)
 
output
(2*(c + d*1i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(d*f*((c + d*tan( 
e + f*x))^(1/2) - c^(1/2))*((a*1i)/d + ((a + a*tan(e + f*x)*1i)^(1/2) - a^ 
(1/2))^2/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 - (a^(1/2)*c^(1/2)*((a + 
 a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*2i)/(d*((c + d*tan(e + f*x))^(1/2) - 
c^(1/2))))) - (2^(1/2)*atan(((2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(4*a^(3/2)*c 
^(3/2)*f - a^(3/2)*c^(1/2)*d*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^(1/2 
) - a^(1/2))*(a*d^2*f - a*c^2*f + a*c*d*f*2i))/((c + d*tan(e + f*x))^(1/2) 
 - c^(1/2)) - (4*d^8*(a^(1/2)*c^(3/2)*f*4i + 4*a^(1/2)*c^(1/2)*d*f)*((a + 
a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/ 
2))^2 - (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(a^2*c*f^2*4i - 4*a^2*d*f^2) - (1 
6*d^7*(a^(3/2)*c^(3/2)*f^2*2i + 6*a^(3/2)*c^(1/2)*d*f^2)*((a + a*tan(e + f 
*x)*1i)^(1/2) - a^(1/2)))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) + (4*d^8* 
(20*a*c*f^2 - a*d*f^2*12i)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2)/(( 
c + d*tan(e + f*x))^(1/2) - c^(1/2))^2))/(4*a^(1/2)*f))*1i)/(4*a^(1/2)*f) 
+ (2^(1/2)*(d*1i - c)^(1/2)*(4*d^7*(4*a^(3/2)*c^(3/2)*f - a^(3/2)*c^(1/2)* 
d*f*4i) + (16*d^7*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))*(a*d^2*f - a*c 
^2*f + a*c*d*f*2i))/((c + d*tan(e + f*x))^(1/2) - c^(1/2)) - (4*d^8*(a^(1/ 
2)*c^(3/2)*f*4i + 4*a^(1/2)*c^(1/2)*d*f)*((a + a*tan(e + f*x)*1i)^(1/2) - 
a^(1/2))^2)/((c + d*tan(e + f*x))^(1/2) - c^(1/2))^2 + (2^(1/2)*(d*1i - c) 
^(1/2)*(4*d^7*(a^2*c*f^2*4i - 4*a^2*d*f^2) - (16*d^7*(a^(3/2)*c^(3/2)*f...