\(\int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [373]

Optimal result

Integrand size = 27, antiderivative size = 45 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {2 i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {2 i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d} \] Input:

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

Output:

((-2*I)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{d}\)

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 728 vs. \(2 (36 ) = 72\).

Time = 0.20 (sec) , antiderivative size = 729, normalized size of antiderivative = 16.20

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-i \sqrt {a^{2}+b^{2}}-i a +b \right ) \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a +\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, b -\frac {\left (-i \sqrt {a^{2}+b^{2}}-i a +b \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\left (-2 i \sqrt {a^{2}+b^{2}}\, a^{2}-i \sqrt {a^{2}+b^{2}}\, b^{2}-2 i a^{3}-2 i a \,b^{2}+\sqrt {a^{2}+b^{2}}\, a b +a^{2} b +b^{3}\right ) \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a^{2}+i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{3}+i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \,b^{2}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a b -\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2} b -\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, b^{3}+\frac {\left (-2 i \sqrt {a^{2}+b^{2}}\, a^{2}-i \sqrt {a^{2}+b^{2}}\, b^{2}-2 i a^{3}-2 i a \,b^{2}+\sqrt {a^{2}+b^{2}}\, a b +a^{2} b +b^{3}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \left (\sqrt {a^{2}+b^{2}}\, a +a^{2}+b^{2}\right )}}{d}\)	\(729\)
default	\(\frac {\frac {\frac {\left (-i \sqrt {a^{2}+b^{2}}-i a +b \right ) \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (-i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a +\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, b -\frac {\left (-i \sqrt {a^{2}+b^{2}}-i a +b \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\left (-2 i \sqrt {a^{2}+b^{2}}\, a^{2}-i \sqrt {a^{2}+b^{2}}\, b^{2}-2 i a^{3}-2 i a \,b^{2}+\sqrt {a^{2}+b^{2}}\, a b +a^{2} b +b^{3}\right ) \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a^{2}+i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{3}+i \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \,b^{2}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a b -\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2} b -\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, b^{3}+\frac {\left (-2 i \sqrt {a^{2}+b^{2}}\, a^{2}-i \sqrt {a^{2}+b^{2}}\, b^{2}-2 i a^{3}-2 i a \,b^{2}+\sqrt {a^{2}+b^{2}}\, a b +a^{2} b +b^{3}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, \left (\sqrt {a^{2}+b^{2}}\, a +a^{2}+b^{2}\right )}}{d}\)	\(729\)
parts	\(\text {Expression too large to display}\)	\(1892\)

Input:

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

Output:

1/d*(1/(2*(a^2+b^2)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)*(1/2*(-I*(a^2+b^2)^(1 
/2)-I*a+b)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a 
)^(1/2)+(a^2+b^2)^(1/2))+2*(-I*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2)*b-1/2*(-I*(a^2+b^2)^(1/2)-I*a+b)*(2*(a^2+b^2)^(1/2)+2*a 
)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2 
*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)))+1/(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2)/(a^2+b^2)^(1/2)/((a^2+b^2)^(1/2)*a+a^2+b^2)*(-1/2*(-2*I 
*(a^2+b^2)^(1/2)*a^2-I*(a^2+b^2)^(1/2)*b^2-2*I*a^3-2*I*a*b^2+(a^2+b^2)^(1/ 
2)*a*b+a^2*b+b^3)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)- 
b*tan(d*x+c)-a-(a^2+b^2)^(1/2))+2*(I*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^ 
2)^(1/2)*a^2+I*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+I*(2*(a^2+b^2)^(1/2)+2*a) 
^(1/2)*a*b^2-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a*b-(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2)*a^2*b-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*b^3+1/2*(-2*I*(a^2+ 
b^2)^(1/2)*a^2-I*(a^2+b^2)^(1/2)*b^2-2*I*a^3-2*I*a*b^2+(a^2+b^2)^(1/2)*a*b 
+a^2*b+b^3)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*a 
rctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2 
)^(1/2)-2*a)^(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. 249 vs. \(2 (33) = 66\).

Time = 0.10 (sec) , antiderivative size = 249, normalized size of antiderivative = 5.53 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {1}{4} \, \sqrt {-\frac {4 i}{{\left (i \, a + b\right )} d^{2}}} \log \left ({\left ({\left ({\left (i \, a + b\right )} d e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, a + b\right )} d\right )} \sqrt {\frac {{\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {4 i}{{\left (i \, a + b\right )} d^{2}}} + 2 \, {\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \frac {1}{4} \, \sqrt {-\frac {4 i}{{\left (i \, a + b\right )} d^{2}}} \log \left ({\left ({\left ({\left (-i \, a - b\right )} d e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, a - b\right )} d\right )} \sqrt {\frac {{\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + a + i \, b}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {4 i}{{\left (i \, a + b\right )} d^{2}}} + 2 \, {\left (a - i \, b\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

I*(Integral(-I/sqrt(a + b*tan(c + d*x)), x) + Integral(tan(c + d*x)/sqrt(a 
 + b*tan(c + d*x)), x))
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 6480 vs. \(2 (33) = 66\).

Time = 0.35 (sec) , antiderivative size = 6480, normalized size of antiderivative = 144.00 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((1+I*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

-1/4*(sqrt(2*a^2 + 2*b^2)*sqrt(a + sqrt(a^2 + b^2))*(2*arctan2((b^2*cos(2* 
d*x + 2*c) - a*b*sin(2*d*x + 2*c) + ((a^4 + 2*a^2*b^2 + b^4)*cos(2*d*x + 2 
*c)^4 + (a^4 + 2*a^2*b^2 + b^4)*sin(2*d*x + 2*c)^4 + a^4 + 2*a^2*b^2 + b^4 
 + 4*(a^4 + a^2*b^2)*cos(2*d*x + 2*c)^3 + 4*(a^3*b + a*b^3)*sin(2*d*x + 2* 
c)^3 + 2*(3*a^4 + 2*a^2*b^2 - b^4)*cos(2*d*x + 2*c)^2 + 2*(a^4 + 2*a^2*b^2 
 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cos(2*d*x + 2*c)^2 + 2*(a^4 + a^2*b^2)*co 
s(2*d*x + 2*c))*sin(2*d*x + 2*c)^2 + 4*(a^4 + a^2*b^2)*cos(2*d*x + 2*c) + 
4*(a^3*b + a*b^3 + (a^3*b + a*b^3)*cos(2*d*x + 2*c)^2 + 2*(a^3*b + a*b^3)* 
cos(2*d*x + 2*c))*sin(2*d*x + 2*c))^(1/4)*b*sin(1/2*arctan2(-2*(a*b*cos(2* 
d*x + 2*c)^2 - a*b*sin(2*d*x + 2*c)^2 + a*b*cos(2*d*x + 2*c) - (a^2 + (a^2 
 - b^2)*cos(2*d*x + 2*c))*sin(2*d*x + 2*c))/b^2, (2*a^2*cos(2*d*x + 2*c) + 
 (a^2 - b^2)*cos(2*d*x + 2*c)^2 - (a^2 - b^2)*sin(2*d*x + 2*c)^2 + a^2 + b 
^2 + 2*(2*a*b*cos(2*d*x + 2*c) + a*b)*sin(2*d*x + 2*c))/b^2)))/b^2, -(a*co 
s(2*d*x + 2*c) + b*sin(2*d*x + 2*c) - ((a^4 + 2*a^2*b^2 + b^4)*cos(2*d*x + 
 2*c)^4 + (a^4 + 2*a^2*b^2 + b^4)*sin(2*d*x + 2*c)^4 + a^4 + 2*a^2*b^2 + b 
^4 + 4*(a^4 + a^2*b^2)*cos(2*d*x + 2*c)^3 + 4*(a^3*b + a*b^3)*sin(2*d*x + 
2*c)^3 + 2*(3*a^4 + 2*a^2*b^2 - b^4)*cos(2*d*x + 2*c)^2 + 2*(a^4 + 2*a^2*b 
^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cos(2*d*x + 2*c)^2 + 2*(a^4 + a^2*b^2)* 
cos(2*d*x + 2*c))*sin(2*d*x + 2*c)^2 + 4*(a^4 + a^2*b^2)*cos(2*d*x + 2*c) 
+ 4*(a^3*b + a*b^3 + (a^3*b + a*b^3)*cos(2*d*x + 2*c)^2 + 2*(a^3*b + a*...
 

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. 159 vs. \(2 (33) = 66\).

Time = 0.53 (sec) , antiderivative size = 159, normalized size of antiderivative = 3.53 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {2 i \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {b \tan \left (d x + c\right ) + a} a - \sqrt {a^{2} + b^{2}} \sqrt {b \tan \left (d x + c\right ) + a}\right )}}{\sqrt {2} a \sqrt {-a + \sqrt {a^{2} + b^{2}}} - i \, \sqrt {2} \sqrt {-a + \sqrt {a^{2} + b^{2}}} b - \sqrt {2} \sqrt {a^{2} + b^{2}} \sqrt {-a + \sqrt {a^{2} + b^{2}}}}\right )}{\sqrt {-a + \sqrt {a^{2} + b^{2}}} d {\left (-\frac {i \, b}{a - \sqrt {a^{2} + b^{2}}} + 1\right )}} \] Input:

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

Output:

2*I*sqrt(2)*arctan(2*(sqrt(b*tan(d*x + c) + a)*a - sqrt(a^2 + b^2)*sqrt(b* 
tan(d*x + c) + a))/(sqrt(2)*a*sqrt(-a + sqrt(a^2 + b^2)) - I*sqrt(2)*sqrt( 
-a + sqrt(a^2 + b^2))*b - sqrt(2)*sqrt(a^2 + b^2)*sqrt(-a + sqrt(a^2 + b^2 
))))/(sqrt(-a + sqrt(a^2 + b^2))*d*(-I*b/(a - sqrt(a^2 + b^2)) + 1))
 

Mupad [B] (verification not implemented)

Time = 5.68 (sec) , antiderivative size = 1410, normalized size of antiderivative = 31.33 \[ \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

2*atanh((32*b^2*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2*d^2 
))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((a^2*b^2*d^2*64i)/(4*a^2*d^3 + 4*b^2 
*d^3) - (b^2*16i)/d + (64*a*b^3*d^2)/(4*a^2*d^3 + 4*b^2*d^3)) - (128*a^2*b 
^2*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + 
 b*tan(c + d*x))^(1/2))/((a^2*b^4*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (a^2 
*b^2*64i)/d - (b^4*64i)/d + (256*a^3*b^3*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a 
^4*b^2*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) + (256*a*b^5*d^2)/(4*a^2*d^3 + 4* 
b^2*d^3)) + (a*b^3*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2* 
d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*128i)/((a^2*b^4*d^2*256i)/(4*a^2*d^ 
3 + 4*b^2*d^3) - (a^2*b^2*64i)/d - (b^4*64i)/d + (256*a^3*b^3*d^2)/(4*a^2* 
d^3 + 4*b^2*d^3) + (a^4*b^2*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) + (256*a*b^5 
*d^2)/(4*a^2*d^3 + 4*b^2*d^3)))*(-(a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2 
) + (log(d*(-1/(d^2*(a - b*1i)))^(1/2)*(a + b*tan(c + d*x))^(1/2)*1i + 1)* 
(-1/(a*d^2 - b*d^2*1i))^(1/2))/2 - log(d*(-1/(d^2*(a - b*1i)))^(1/2)*(a + 
b*tan(c + d*x))^(1/2) + 1i)*(-1/(4*(a*d^2 - b*d^2*1i)))^(1/2) + (log(16*b^ 
2*(a + b*tan(c + d*x))^(1/2) + 16*b^3*d*(-1/(d^2*(a - b*1i)))^(1/2) - (16* 
a*b^2*(a + b*tan(c + d*x))^(1/2))/(a - b*1i))*(-1/(a*d^2 - b*d^2*1i))^(1/2 
))/2 - log(16*b^3*d*(-1/(d^2*(a - b*1i)))^(1/2) - 16*b^2*(a + b*tan(c + d* 
x))^(1/2) + (16*a*b^2*(a + b*tan(c + d*x))^(1/2))/(a - b*1i))*(-1/(4*(a*d^ 
2 - b*d^2*1i)))^(1/2) + 2*atanh((32*b^2*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2)...
 

Reduce [F]

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

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

Output:

(2*sqrt(tan(c + d*x)*b + a) - int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)** 
2)/(tan(c + d*x)*b + a),x)*b*d + int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x 
))/(tan(c + d*x)*b + a),x)*b*d*i)/(b*d)