\(\int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx\) [137]

Optimal result

Integrand size = 36, antiderivative size = 184 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((2+i) A+B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((2+i) A+B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {((3+i) A-(1+i) B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \] Output:

(1/8-1/8*I)*((2+I)*A+B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/a/d+(1 
/8-1/8*I)*((2+I)*A+B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/a/d+1/8*( 
(3+I)*A-(1+I)*B)*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/ 
a/d+1/2*(A+I*B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.53 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {-\sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 \sqrt [4]{-1} A \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {(-i A+B) \sqrt {\tan (c+d x)}}{-i+\tan (c+d x)}}{2 a d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\int \frac {a (3 A-i B-(i A-B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{2 a^2 d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 A-i B-(i A-B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{2 a d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}\)

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((1/2 - I/2)*((2 + I)*A + B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqr 
t[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + (((3 + I)*A - (1 
 + I)*B)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] 
+ Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2)/(2*a 
*d) + ((A + I*B)*Sqrt[Tan[c + d*x]])/(2*d*(a + I*a*Tan[c + d*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 4079

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-\frac {i \left (i B +A \right ) \sqrt {\tan \left (d x +c \right )}}{2 \left (-i+\tan \left (d x +c \right )\right )}-\frac {2 i A \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {4 \left (\frac {i A}{4}+\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(121\)
default	\(\frac {-\frac {i \left (i B +A \right ) \sqrt {\tan \left (d x +c \right )}}{2 \left (-i+\tan \left (d x +c \right )\right )}-\frac {2 i A \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {4 \left (\frac {i A}{4}+\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(121\)

Input:

int((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

1/d/a*(-1/2*I*(A+I*B)*tan(d*x+c)^(1/2)/(-I+tan(d*x+c))-2*I*A/(2^(1/2)-I*2^ 
(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))+4*(1/4*I*A+1/4*B)/(2 
^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(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. 571 vs. \(2 (139) = 278\).

Time = 0.10 (sec) , antiderivative size = 571, normalized size of antiderivative = 3.10 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx =\text {Too large to display} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorith 
m="fricas")
 

Output:

-1/8*(a*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log 
(-2*((I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I) 
/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A 
- I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - a*d*sqrt((-I 
*A^2 - 2*A*B + I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(-2*((-I*a*d*e^(2* 
I*d*x + 2*I*c) - I*a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2* 
I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x 
 + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 2*a*d*sqrt(I*A^2/(a^2*d^2))*e 
^(2*I*d*x + 2*I*c)*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d* 
x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(I*A^2/(a^2*d^2)) + I*A)*e^ 
(-2*I*d*x - 2*I*c)/(a*d)) + 2*a*d*sqrt(I*A^2/(a^2*d^2))*e^(2*I*d*x + 2*I*c 
)*log(-((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/ 
(e^(2*I*d*x + 2*I*c) + 1))*sqrt(I*A^2/(a^2*d^2)) - I*A)*e^(-2*I*d*x - 2*I* 
c)/(a*d)) - 2*((A + I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*sqrt((-I*e^(2*I*d* 
x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(a*d)
 

Sympy [F(-2)]

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

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

Output:

Exception raised: TypeError >> Invalid comparison of non-real -I
 

Maxima [F(-2)]

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

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorith 
m="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.48 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx=-\frac {\left (2 i - 2\right ) \, \sqrt {2} A \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \frac {2 \, {\left (i \, A \sqrt {\tan \left (d x + c\right )} - B \sqrt {\tan \left (d x + c\right )}\right )}}{\tan \left (d x + c\right ) - i}}{4 \, a d} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorith 
m="giac")
 

Output:

-1/4*((2*I - 2)*sqrt(2)*A*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c))) 
 + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan 
(d*x + c))) + 2*(I*A*sqrt(tan(d*x + c)) - B*sqrt(tan(d*x + c)))/(tan(d*x + 
 c) - I))/(a*d)
 

Mupad [B] (verification not implemented)

Time = 4.54 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \, dx=-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {2\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,B\,\mathrm {atanh}\left (4\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{a\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \] Input:

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

Output:

atan((4*a*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(16*a^2*d^2))^(1/2))/A)*(-(A^2*1 
i)/(16*a^2*d^2))^(1/2)*2i - atan((2*a*d*tan(c + d*x)^(1/2)*((A^2*1i)/(4*a^ 
2*d^2))^(1/2))/A)*((A^2*1i)/(4*a^2*d^2))^(1/2)*2i - (2*(1i/16)^(1/2)*B*ata 
nh(4*(1i/16)^(1/2)*tan(c + d*x)^(1/2)))/(a*d) + (A*tan(c + d*x)^(1/2))/(2* 
a*d*(tan(c + d*x)*1i + 1)) + (B*tan(c + d*x)^(1/2)*1i)/(2*a*d*(tan(c + d*x 
)*1i + 1))
 

Reduce [F]

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

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

Output:

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