3.2.42 \(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\) [142]

3.2.42.1 Optimal result

Integrand size = 36, antiderivative size = 279 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((-1+3 i) A+(1+3 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(1-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1+3 i) A+(1-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {(i A+3 B) \sqrt {\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]

output

-1/32*((-1+3*I)*A+(1+3*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d*2^( 
1/2)-1/32*((-1+3*I)*A+(1+3*I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^2/d* 
2^(1/2)+1/64*((1+3*I)*A+(1-3*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c 
))/a^2/d*2^(1/2)-1/64*((1+3*I)*A+(1-3*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+ 
tan(d*x+c))/a^2/d*2^(1/2)+1/8*(I*A+3*B)*tan(d*x+c)^(1/2)/a^2/d/(1+I*tan(d* 
x+c))+1/4*(I*A-B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^2
 

3.2.42.2 Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {-2 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (-i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt {\tan (c+d x)} (-3 i A-B+(A-3 i B) \tan (c+d x))}{8 a^2 d (-i+\tan (c+d x))^2} \]

input

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

output

(-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)*ArcT 
anh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Si 
n[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)
 

3.2.42.3 Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.88, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4078, 27, 3042, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

input

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

output

-1/8*(((((-1 + 3*I)*A + (1 + 3*I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d* 
x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + (((1 
+ 3*I)*A + (1 - 3*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*Sqr 
t[2])))/2)/d - ((I*A + 3*B)*Sqrt[Tan[c + d*x]])/(d*(1 + I*Tan[c + d*x])))/ 
a^2 + ((I*A - B)*Sqrt[Tan[c + d*x]])/(4*d*(a + I*a*Tan[c + d*x])^2)
 

3.2.42.3.1 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 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 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]
 

3.2.42.4 Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.52

method	result	size
derivativedivides	\(\frac {\frac {\left (-\frac {3 i B}{2}+\frac {A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {B}{2}-\frac {3 i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\)	\(144\)
default	\(\frac {\frac {\left (-\frac {3 i B}{2}+\frac {A}{2}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-\frac {B}{2}-\frac {3 i A}{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{4 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\left (i B +A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2 \left (\sqrt {2}+i \sqrt {2}\right )}}{d \,a^{2}}\)	\(144\)

input

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

output

1/d/a^2*(1/4*((-3/2*I*B+1/2*A)*tan(d*x+c)^(3/2)+(-1/2*B-3/2*I*A)*tan(d*x+c 
)^(1/2))/(tan(d*x+c)-I)^2-1/4*(A+I*B)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x 
+c)^(1/2)/(2^(1/2)-I*2^(1/2)))-1/2*I*(I*A+B)/(2^(1/2)+I*2^(1/2))*arctan(2* 
tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2))))
 

3.2.42.5 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. 658 vs. \(2 (212) = 424\).

Time = 0.28 (sec) , antiderivative size = 658, normalized size of antiderivative = 2.36 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, a^{2} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{4} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a^{2} d \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} + A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + a^{2} d \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} + 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} - A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left (2 \, {\left (-i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (3 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]

input

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(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*((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)) + ( 
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*sq 
rt((I*A^2 + 2*A*B - I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((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)) - (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))*s 
qrt((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2)) + A + I*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) 
) - A - I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*(2*(-I*A - B)*e^(4*I*d*x + 
4*I*c) - (3*I*A + B)*e^(2*I*d*x + 2*I*c) - I*A + 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)
 

3.2.42.6 Sympy [F]

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

input

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

output

-(Integral(A*sqrt(tan(c + d*x))/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), 
x) + Integral(B*tan(c + d*x)**(3/2)/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 
1), x))/a**2
 

3.2.42.7 Maxima [F(-2)]

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

input

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(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.
 

3.2.42.8 Giac [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (A + i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} + \frac {A \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 3 i \, A \sqrt {\tan \left (d x + c\right )} - B \sqrt {\tan \left (d x + c\right )}}{8 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \]

input

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

output

-(1/16*I + 1/16)*sqrt(2)*(A + I*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d 
*x + c)))/(a^2*d) - (1/8*I - 1/8)*sqrt(2)*(A - I*B)*arctan(-(1/2*I - 1/2)* 
sqrt(2)*sqrt(tan(d*x + c)))/(a^2*d) + 1/8*(A*tan(d*x + c)^(3/2) - 3*I*B*ta 
n(d*x + c)^(3/2) - 3*I*A*sqrt(tan(d*x + c)) - B*sqrt(tan(d*x + c)))/(a^2*d 
*(tan(d*x + c) - I)^2)
 

3.2.42.9 Mupad [B] (verification not implemented)

Time = 11.29 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {3\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^2\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-\frac {-\frac {3\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{8\,a^2\,d}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{8\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}-2\,\mathrm {atanh}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}-\mathrm {atan}\left (\frac {8\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{64\,a^4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^4\,d^2}}\,2{}\mathrm {i} \]

input

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

output

((3*A*tan(c + d*x)^(1/2))/(8*a^2*d) + (A*tan(c + d*x)^(3/2)*1i)/(8*a^2*d)) 
/(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i) - ((B*tan(c + d*x)^(1/2)*1i)/(8 
*a^2*d) - (3*B*tan(c + d*x)^(3/2))/(8*a^2*d))/(2*tan(c + d*x) + tan(c + d* 
x)^2*1i - 1i) - 2*atanh((8*a^2*d*tan(c + d*x)^(1/2)*((A^2*1i)/(64*a^4*d^2) 
)^(1/2))/A)*((A^2*1i)/(64*a^4*d^2))^(1/2) + 2*atanh((16*a^2*d*tan(c + d*x) 
^(1/2)*(-(A^2*1i)/(256*a^4*d^2))^(1/2))/A)*(-(A^2*1i)/(256*a^4*d^2))^(1/2) 
 - atan((8*a^2*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(64*a^4*d^2))^(1/2))/B)*(-( 
B^2*1i)/(64*a^4*d^2))^(1/2)*2i - atan((16*a^2*d*tan(c + d*x)^(1/2)*((B^2*1 
i)/(256*a^4*d^2))^(1/2))/B)*((B^2*1i)/(256*a^4*d^2))^(1/2)*2i