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

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

3.12.4.2 Mathematica [A] (verified)

Time = 0.80 (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]))
 

3.12.4.3 Rubi [A] (warning: unable to verify)

Time = 0.77 (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]))
 

3.12.4.3.1 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]
 

3.12.4.4 Maple [A] (verified)

Time = 1.78 (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/(-I*d-c)^(1/2 
)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))
 

3.12.4.5 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. 686 vs. \(2 (104) = 208\).

Time = 0.28 (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=-\frac {{\left (a f \sqrt {-\frac {c - i \, d}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left ({\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \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 {c - i \, d}{a^{2} f^{2}}} - {\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - c\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a f \sqrt {-\frac {c - i \, d}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left ({\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \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 {c - i \, d}{a^{2} f^{2}}} - {\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - c\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 2 \, a f \sqrt {-\frac {i \, c^{2}}{4 \, {\left (i \, a^{2} c - a^{2} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} + i \, c d - 2 \, {\left ({\left (i \, a c - a d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, a c - a d\right )} f\right )} \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 {i \, c^{2}}{4 \, {\left (i \, a^{2} c - a^{2} d\right )} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (i \, a c - a d\right )} f}\right ) + 2 \, a f \sqrt {-\frac {i \, c^{2}}{4 \, {\left (i \, a^{2} c - a^{2} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2} + i \, c d - 2 \, {\left ({\left (-i \, a c + a d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, a c + a d\right )} f\right )} \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 {i \, c^{2}}{4 \, {\left (i \, a^{2} c - a^{2} d\right )} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (i \, a c - a d\right )} f}\right ) - 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}} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a f} \]

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)
 

3.12.4.6 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
 

3.12.4.7 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.
 

3.12.4.8 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. 372 vs. \(2 (104) = 208\).

Time = 0.45 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.66 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {1}{2} \, d^{2} {\left (\frac {\sqrt {d \tan \left (f x + e\right ) + c}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d f} - \frac {2 i \, c \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{2} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 \, {\left (-i \, c - d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{2} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \]

input

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

output

1/2*d^2*(sqrt(d*tan(f*x + e) + c)/((d*tan(f*x + e) - I*d)*a*d*f) - 2*I*c*a 
rctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) 
+ c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2)) 
*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a*sqrt(-2*c + 2*sqr 
t(c^2 + d^2))*d^2*f*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2*(-I*c - d)*arctan 
(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c)) 
/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - 
sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a*sqrt(-2*c + 2*sqrt(c^2 
 + d^2))*d^2*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))
 

3.12.4.9 Mupad [B] (verification not implemented)

Time = 8.05 (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=-2\,\mathrm {atanh}\left (\frac {4\,a^2\,d^4\,f^2\,\sqrt {-\frac {c}{16\,a^2\,f^2}+\frac {d\,1{}\mathrm {i}}{16\,a^2\,f^2}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{a\,f\,d^5+1{}\mathrm {i}\,a\,c\,f\,d^4}\right )\,\sqrt {-\frac {c}{16\,a^2\,f^2}+\frac {d\,1{}\mathrm {i}}{16\,a^2\,f^2}}-\mathrm {atan}\left (\frac {a^2\,d^4\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^2\,1{}\mathrm {i}}{16\,\left (-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}\right )}}\,4{}\mathrm {i}}{a\,c\,d^4\,f\,1{}\mathrm {i}-a\,c^2\,d^3\,f+\frac {a^3\,c^2\,d^4\,f^3}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}+\frac {a^3\,c^3\,d^3\,f^3\,1{}\mathrm {i}}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}}+\frac {8\,a^4\,c^3\,d^2\,f^4\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^2\,1{}\mathrm {i}}{16\,\left (-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}\right )}}}{a^3\,c\,d^5\,f^3\,1{}\mathrm {i}+a^3\,c^3\,d^3\,f^3\,1{}\mathrm {i}+\frac {a^5\,c^2\,d^5\,f^5}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}+\frac {a^5\,c^4\,d^3\,f^5}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}}-\frac {8\,a^2\,c\,d^3\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^2\,1{}\mathrm {i}}{16\,\left (-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}\right )}}}{a\,c\,d^4\,f\,1{}\mathrm {i}-a\,c^2\,d^3\,f+\frac {a^3\,c^2\,d^4\,f^3}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}+\frac {a^3\,c^3\,d^3\,f^3\,1{}\mathrm {i}}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}}-\frac {a^2\,c^2\,d^2\,f^2\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {c^2\,1{}\mathrm {i}}{16\,\left (-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}\right )}}\,8{}\mathrm {i}}{a\,c\,d^4\,f\,1{}\mathrm {i}-a\,c^2\,d^3\,f+\frac {a^3\,c^2\,d^4\,f^3}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}+\frac {a^3\,c^3\,d^3\,f^3\,1{}\mathrm {i}}{-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}}}\right )\,\sqrt {-\frac {c^2\,1{}\mathrm {i}}{16\,\left (-a^2\,d\,f^2+a^2\,c\,f^2\,1{}\mathrm {i}\right )}}\,2{}\mathrm {i}-\frac {d\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\left (-a\,d\,f\,\mathrm {tan}\left (e+f\,x\right )+a\,d\,f\,1{}\mathrm {i}\right )} \]

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)))