\(\int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [179]

Optimal result

Integrand size = 38, antiderivative size = 205 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(-1)^{3/4} (2 i A-B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.72 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {(-1)^{3/4} (2 A+i B) \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} (-A-2 i B+B \tan (c+d x))}{d \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4078, 27, 3042, 4080, 25, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((I*A - B)*Tan[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (-(((-2*(- 
1)^(1/4)*a^(5/2)*(2*A + I*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]] 
)/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((1 - I)*a^(5/2)*(A - I*B)*ArcTanh[((1 
+ I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a) + (2*a 
*(A + (2*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d)/(2*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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[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 4080

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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

rule 4082

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

rule 4084

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (165 ) = 330\).

Time = 0.36 (sec) , antiderivative size = 1135, normalized size of antiderivative = 5.54

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1135\)
default	\(\text {Expression too large to display}\)	\(1135\)
parts	\(\text {Expression too large to display}\)	\(1199\)

Input:

int(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x,method=_R 
ETURNVERBOSE)
 

Output:

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

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. 807 vs. \(2 (155) = 310\).

Time = 0.20 (sec) , antiderivative size = 807, normalized size of antiderivative = 3.94 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, al 
gorithm="fricas")
 

Output:

-1/4*(sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c)*l 
og((I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c) + 
 sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2* 
I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/( 
4*I*A + 4*B)) - sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d* 
x + I*c)*log((-I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d 
*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2 
*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* 
c) + 1)))/(4*I*A + 4*B)) - a*d*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2))*e^ 
(I*d*x + I*c)*log(104/605*(2*sqrt(2)*((2*I*A - B)*e^(3*I*d*x + 3*I*c) + (2 
*I*A - B)*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2 
*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + (3*I*a*d*e^(2*I*d*x + 2* 
I*c) - I*a*d)*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2)))/((2*I*A - B)*e^(2* 
I*d*x + 2*I*c) + 2*I*A - B)) + a*d*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2) 
)*e^(I*d*x + I*c)*log(104/605*(2*sqrt(2)*((2*I*A - B)*e^(3*I*d*x + 3*I*c) 
+ (2*I*A - B)*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I* 
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + (-3*I*a*d*e^(2*I*d*x 
 + 2*I*c) + I*a*d)*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2)))/((2*I*A - B)* 
e^(2*I*d*x + 2*I*c) + 2*I*A - B)) + 2*sqrt(2)*((A + 3*I*B)*e^(2*I*d*x + 2* 
I*c) + A + I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x +...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, al 
gorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, al 
gorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar 
gument Ty
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(a)*( - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 
*a - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2*b*i - 2 
*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*b - 6*sqrt(tan(c 
 + d*x))*sqrt(tan(c + d*x)*i + 1)*a - 6*sqrt(tan(c + d*x))*sqrt(tan(c + d* 
x)*i + 1)*b*i + 2*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + 
 d*x)**3)/(tan(c + d*x)**2 + 1),x)*tan(c + d*x)**2*a*d + int((sqrt(tan(c + 
 d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3)/(tan(c + d*x)**2 + 1),x)* 
tan(c + d*x)**2*b*d*i + 2*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) 
*tan(c + d*x)**3)/(tan(c + d*x)**2 + 1),x)*a*d + int((sqrt(tan(c + d*x))*s 
qrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3)/(tan(c + d*x)**2 + 1),x)*b*d*i + 
3*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**3 + tan 
(c + d*x)),x)*tan(c + d*x)**2*a*d + 3*int((sqrt(tan(c + d*x))*sqrt(tan(c + 
 d*x)*i + 1))/(tan(c + d*x)**3 + tan(c + d*x)),x)*tan(c + d*x)**2*b*d*i + 
3*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**3 + tan 
(c + d*x)),x)*a*d + 3*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(t 
an(c + d*x)**3 + tan(c + d*x)),x)*b*d*i + 3*int((sqrt(tan(c + d*x))*sqrt(t 
an(c + d*x)*i + 1))/(tan(c + d*x)**2 + 1),x)*tan(c + d*x)**2*a*d*i + 3*int 
((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**2 + 1),x)*a* 
d*i))/(a*d*(tan(c + d*x)**2 + 1))