\(\int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [531]

Optimal result

Integrand size = 36, antiderivative size = 224 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {(3 i A+B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2} \] Output:

1/32*((-1+3*I)*A+(1+3*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/a^ 
2/d+1/32*((-1+3*I)*A+(1+3*I)*B)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2) 
/a^2/d-1/32*((1+3*I)*A+(1-3*I)*B)*arctanh(2^(1/2)*cot(d*x+c)^(1/2)/(1+cot( 
d*x+c)))*2^(1/2)/a^2/d+1/8*(3*I*A+B)*cot(d*x+c)^(1/2)/a^2/d/(I+cot(d*x+c)) 
+1/4*(A+I*B)*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^2
 

Mathematica [A] (verified)

Time = 2.07 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-2 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (-i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\tan (c+d x)} (-3 i A-B+(A-3 i B) \tan (c+d x))\right )}{8 a^2 d (-i+\tan (c+d x))^2} \] Input:

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

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-2*(-1)^(1/4)*(I*A + B)*ArcTan[(-1 
)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c 
+ d*x)]) + (-1)^(1/4)*((-I)*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]* 
Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + Sqrt[Tan[c + d*x] 
]*((-3*I)*A - B + (A - (3*I)*B)*Tan[c + d*x])))/(8*a^2*d*(-I + Tan[c + d*x 
])^2)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.08, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4064, 3042, 4078, 27, 3042, 4079, 25, 3042, 4017, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4064

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\int -\frac {a (i A-B)-a (5 A-3 i B) \cot (c+d x)}{2 \sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)}dx}{4 a^2}+\frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\int \frac {a (i A-B)-a (5 A-3 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)} (\cot (c+d x) a+i a)}dx}{8 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\int \frac {a (i A-B)+a (5 A-3 i B) \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (i a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{8 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int -\frac {a^2 (A-3 i B)-a^2 (3 i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{2 a^2}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int \frac {a^2 (A-3 i B)-a^2 (3 i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{2 a^2}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int \frac {(A-3 i B) a^2+(3 i A+B) \tan \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {-\frac {\int -\frac {a^2 (A-3 i B-(3 i A+B) \cot (c+d x))}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{a^2 d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int \frac {a^2 (A-3 i B-(3 i A+B) \cot (c+d x))}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{a^2 d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\int \frac {A-3 i B-(3 i A+B) \cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \int \frac {\cot (c+d x)+1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {1}{2} \int \frac {1}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1+3 i) A+(1-3 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\cot (c+d x)}+1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )+\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(A+i B) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2}-\frac {\frac {\frac {1}{2} ((1-3 i) A-(1+3 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} ((1+3 i) A+(1-3 i) B) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )}{d}-\frac {(B+3 i A) \sqrt {\cot (c+d x)}}{d (\cot (c+d x)+i)}}{8 a^2}\)

Input:

Int[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^2),x]
 

Output:

((A + I*B)*Sqrt[Cot[c + d*x]])/(4*d*(I*a + a*Cot[c + d*x])^2) - (-((((3*I) 
*A + B)*Sqrt[Cot[c + d*x]])/(d*(I + Cot[c + d*x]))) + ((((1 - 3*I)*A - (1 
+ 3*I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + 
Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]))/2 + (((1 + 3*I)*A + (1 - 3*I)*B)*(-1 
/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/Sqrt[2] + Log[1 + Sq 
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2])))/2)/d)/(8*a^2)
 

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 4064

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp 
[g^(m + n)   Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d + c 
*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !Integer 
Q[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4078

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), 
 x] + Simp[1/(2*a^2*m)   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* 
x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a 
*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] 
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 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 = 0.68 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{2 \sqrt {2}-2 i \sqrt {2}}+\frac {i \left (\frac {\left (-\frac {i B}{2}+\frac {3 A}{2}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {3 B}{2}+\frac {i A}{2}\right ) \sqrt {\cot \left (d x +c \right )}}{\left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {\left (i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{4}}{a^{2} d}\)	\(146\)
default	\(\frac {\frac {i \left (-i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{2 \sqrt {2}-2 i \sqrt {2}}+\frac {i \left (\frac {\left (-\frac {i B}{2}+\frac {3 A}{2}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {3 B}{2}+\frac {i A}{2}\right ) \sqrt {\cot \left (d x +c \right )}}{\left (i+\cot \left (d x +c \right )\right )^{2}}+\frac {\left (i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{4}}{a^{2} d}\)	\(146\)

Input:

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

Output:

1/a^2/d*(1/2*I*(A-I*B)/(2^(1/2)-I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1 
/2)-I*2^(1/2)))+1/4*I*(((-1/2*I*B+3/2*A)*cot(d*x+c)^(3/2)+(3/2*B+1/2*I*A)* 
cot(d*x+c)^(1/2))/(I+cot(d*x+c))^2+(A+I*B)/(2^(1/2)+I*2^(1/2))*arctan(2*co 
t(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. 663 vs. \(2 (173) = 346\).

Time = 0.10 (sec) , antiderivative size = 663, normalized size of antiderivative = 2.96 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/32*(2*a^2*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)* 
log(-2*((I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*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^4*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^2*d 
*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((-I*a 
^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*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^4*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^2*d*sqrt((-I*A^2 
+ 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(1/8*((a^2*d*e^(2*I*d*x 
 + 2*I*c) - a^2*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^4*d^2)) + I*A - B)*e^(-2*I*d*x - 2*I 
*c)/(a^2*d)) - a^2*d*sqrt((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 
 4*I*c)*log(-1/8*((a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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^4 
*d^2)) - I*A + B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*(2*(A - I*B)*e^(4*I*d* 
x + 4*I*c) - (A - 3*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^(-4*I*d*x - 4*I*c)/(a^2*d)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

-(Integral(A/(tan(c + d*x)**2*sqrt(cot(c + d*x)) - 2*I*tan(c + d*x)*sqrt(c 
ot(c + d*x)) - sqrt(cot(c + d*x))), x) + Integral(B*tan(c + d*x)/(tan(c + 
d*x)**2*sqrt(cot(c + d*x)) - 2*I*tan(c + d*x)*sqrt(cot(c + d*x)) - sqrt(co 
t(c + d*x))), x))/a**2
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.52 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=-\frac {\sqrt {2} {\left (\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {2 \, {\left (A \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A \sqrt {\tan \left (d x + c\right )} - B \sqrt {\tan \left (d x + c\right )}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, a^{2} d} \] Input:

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

Output:

-1/16*(sqrt(2)*((I + 1)*A + (I - 1)*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(t 
an(d*x + c))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(-(1/2*I - 1/2)*sq 
rt(2)*sqrt(tan(d*x + c))) - 2*(A*tan(d*x + c)^(3/2) - 3*I*B*tan(d*x + c)^( 
3/2) - 3*I*A*sqrt(tan(d*x + c)) - B*sqrt(tan(d*x + c)))/(tan(d*x + c) - I) 
^2)/(a^2*d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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