\(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx\) [760]

Optimal result

Integrand size = 28, antiderivative size = 144 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 \sqrt [4]{-1} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 \cot ^{\frac {3}{2}}(c+d x) (i a \tan (c+d x))^{3/2} \left (\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )}{d \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(2*Cot[c + d*x]^(3/2)*(I*a*Tan[c + d*x])^(3/2)*(Sqrt[a]*ArcSinh[Sqrt[I*a*T 
an[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]] - Sqrt[2]*ArcTanh[(Sqrt[2]* 
Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d 
*x]]))/(d*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4729, 3042, 4038, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4038

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+i \int \frac {\sqrt {i \tan (c+d x) a+a} (\tan (c+d x) a+i a)}{\sqrt {\tan (c+d x)}}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+i \int \frac {\sqrt {i \tan (c+d x) a+a} (\tan (c+d x) a+i a)}{\sqrt {\tan (c+d x)}}dx\right )\)

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4038

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(c_.) + (d_.)*tan[(e_ 
.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*a   Int[Sqrt[a + b*Tan[e + f*x]]/Sqr 
t[c + d*Tan[e + f*x]], x], x] + Simp[b/a   Int[(b + a*Tan[e + f*x])*(Sqrt[a 
 + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, 
e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 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 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (114 ) = 228\).

Time = 0.40 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.32

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(334\)
default	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(334\)

Input:

int(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(I*2^ 
(1/2)*(I*a)^(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))+4*I*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)-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)-2*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)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a 
)^(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. 491 vs. \(2 (108) = 216\).

Time = 0.13 (sec) , antiderivative size = 491, normalized size of antiderivative = 3.41 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {1}{4} \, \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \log \left (-16 \, {\left (\sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{4} \, \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \log \left (16 \, {\left (\sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} - 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{4} \, \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - \frac {1}{4} \, \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) \] Input:

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/4*sqrt(-4*I*a^3/d^2)*log(-16*(sqrt(2)*(d*e^(3*I*d*x + 3*I*c) - d*e^(I*d 
*x + I*c))*sqrt(-4*I*a^3/d^2)*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*a^2*e^(2*I*d*x + 2*I 
*c) - a^2)*e^(-2*I*d*x - 2*I*c)) + 1/4*sqrt(-4*I*a^3/d^2)*log(16*(sqrt(2)* 
(d*e^(3*I*d*x + 3*I*c) - d*e^(I*d*x + I*c))*sqrt(-4*I*a^3/d^2)*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*a^2*e^(2*I*d*x + 2*I*c) + a^2)*e^(-2*I*d*x - 2*I*c)) + 1/4*sq 
rt(-32*I*a^3/d^2)*log(1/2*(sqrt(2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-3 
2*I*a^3/d^2)*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)) + 8*I*a^2*e^(I*d*x + I*c))*e^(-I*d*x - I* 
c)/a) - 1/4*sqrt(-32*I*a^3/d^2)*log(1/2*(sqrt(2)*(-I*d*e^(2*I*d*x + 2*I*c) 
 + I*d)*sqrt(-32*I*a^3/d^2)*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)) + 8*I*a^2*e^(I*d*x + I*c)) 
*e^(-I*d*x - I*c)/a)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate((I*a*tan(d*x + c) + a)^(3/2)*sqrt(cot(d*x + c)), x)
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="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 TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a*(int(sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*tan(c + d*x),x) 
*i + int(sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x)),x))