\(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [135]

Optimal result

Integrand size = 36, antiderivative size = 217 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((1-3 i) A+(3+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((2+i) A+(1+4 i) B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} a d}-\frac {(A+5 i B) \sqrt {\tan (c+d x)}}{2 a d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))} \] Output:

1/8*((1-3*I)*A+(3+5*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/a/d+ 
1/8*((1-3*I)*A+(3+5*I)*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+(1+4*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+5*I*B)*tan(d*x+c)^(1/2)/a/d+1/2*(I*A-B)*tan(d*x+c 
)^(3/2)/d/(a+I*a*tan(d*x+c))
 

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.53 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 \sqrt [4]{-1} (A+2 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {i \sqrt {\tan (c+d x)} (A+5 i B-4 B \tan (c+d x))}{-i+\tan (c+d x)}}{2 a d} \] Input:

Integrate[(Tan[c + d*x]^(3/2)*(A + B*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 + (2*I)*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + (I*Sqrt[Tan[c + d* 
x]]*(A + (5*I)*B - 4*B*Tan[c + d*x]))/(-I + Tan[c + d*x]))/(2*a*d)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4078, 27, 3042, 4011, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {1}{2} \sqrt {\tan (c+d x)} (3 a (i A-B)+a (A+5 i B) \tan (c+d x))dx}{2 a^2}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 \int -\frac {a (A+5 i B-3 (i A-B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}}{4 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 \int \frac {a (A+5 i B-3 (i A-B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}}{4 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \int \frac {A+5 i B-3 (i A-B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}}{4 a^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}}{4 a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\frac {2 a (A+5 i B) \sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (\frac {1}{2} ((1-3 i) A+(3+5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) ((2+i) A+(1+4 i) B) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}}{4 a^2}\)

Input:

Int[(Tan[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x]),x]
 

Output:

-1/4*((-2*a*((((1 - 3*I)*A + (3 + 5*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 + 
(1/2 + I/2)*((2 + I)*A + (1 + 4*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]))))/d + (2*a*(A + (5*I)*B)*Sqrt[Tan[c + d*x]])/d)/a^2 + 
 ((I*A - B)*Tan[c + d*x]^(3/2))/(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 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

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]
 

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {-2 i B \sqrt {\tan \left (d x +c \right )}+\frac {i \left (-\frac {i \left (i A -B \right ) \sqrt {\tan \left (d x +c \right )}}{-i+\tan \left (d x +c \right )}-\frac {4 \left (2 i B +A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(143\)
default	\(\frac {-2 i B \sqrt {\tan \left (d x +c \right )}+\frac {i \left (-\frac {i \left (i A -B \right ) \sqrt {\tan \left (d x +c \right )}}{-i+\tan \left (d x +c \right )}-\frac {4 \left (2 i B +A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(143\)

Input:

int(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

1/d/a*(-2*I*B*tan(d*x+c)^(1/2)+1/2*I*(-I*(I*A-B)*tan(d*x+c)^(1/2)/(-I+tan( 
d*x+c))-4*(A+2*I*B)/(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. 621 vs. \(2 (160) = 320\).

Time = 0.10 (sec) , antiderivative size = 621, normalized size of antiderivative = 2.86 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(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 - 4*A*B - 4* 
I*B^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 
- 4*A*B - 4*I*B^2)/(a^2*d^2)) + I*A - 2*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2 
*a*d*sqrt((I*A^2 - 4*A*B - 4*I*B^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 - 4*A*B - 4*I*B^2)/(a^2*d^2)) - I*A + 2*B)*e 
^(-2*I*d*x - 2*I*c)/(a*d)) - 2*((A + 9*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]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=- \frac {i \left (\int \frac {A \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx + \int \frac {B \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

integrate(tan(d*x+c)**(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x)
 

Output:

-I*(Integral(A*tan(c + d*x)**(3/2)/(tan(c + d*x) - I), x) + Integral(B*tan 
(c + d*x)**(5/2)/(tan(c + d*x) - I), x))/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(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 = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.48 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A + \left (2 i + 2\right ) \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - 8 i \, B \sqrt {\tan \left (d x + c\right )} - \frac {2 \, {\left (-i \, A \sqrt {\tan \left (d x + c\right )} + B \sqrt {\tan \left (d x + c\right )}\right )}}{\tan \left (d x + c\right ) - i}}{4 \, a d} \] Input:

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x, algorith 
m="giac")
 

Output:

1/4*(2*sqrt(2)*(-(I - 1)*A + (2*I + 2)*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqr 
t(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))) - 8*I*B*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 = 7.67 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.24 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}\,1{}\mathrm {i}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,4{}\mathrm {i}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \] Input:

int((tan(c + d*x)^(3/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i),x)
 

Output:

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

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

int(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x)
 

Output:

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