\(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx\) [195]

Optimal result

Integrand size = 28, antiderivative size = 176 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {3 (-1)^{3/4} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.22 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a \sqrt {i a \tan (c+d x)} \left (3 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (1+i \tan (c+d x))+\sqrt {1+i \tan (c+d x)} \left ((1+i \tan (c+d x)) \sqrt {i a \tan (c+d x)}-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3042, 4039, 27, 3042, 4078, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

\(\displaystyle \frac {1}{2} a \left (\frac {2 i a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {6 i a^3 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} a \left (\frac {2 i a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {6 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a^2}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

Input:

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

Output:

-((a^2*Tan[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]])) + (a*(-(((6*(-1 
)^(3/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I* 
a*Tan[c + d*x]]])/d + ((4 + 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan 
[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a^2) + ((2*I)*a*Sqrt[Tan[c + d 
*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])))/2
 

Defintions of rubi rules used

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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*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 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. 360 vs. \(2 (143 ) = 286\).

Time = 1.54 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.05

method	result	size
derivativedivides	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+3 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(361\)
default	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+3 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a +\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(361\)

Input:

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

Output:

1/2/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*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^(1/2)*a+2*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I 
*a)^(1/2)*(-I*a)^(1/2)+3*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*t 
an(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*(-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)*a+4*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))*a*(-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. 525 vs. \(2 (134) = 268\).

Time = 0.10 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.98 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 i \, \sqrt {2} a \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}} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {9 i \, a^{3}}{d^{2}}} d \log \left (\frac {{\left (3 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} + 2 i \, \sqrt {\frac {9 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{3 \, a}\right ) + \sqrt {\frac {9 i \, a^{3}}{d^{2}}} d \log \left (\frac {{\left (3 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} - 2 i \, \sqrt {\frac {9 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{3 \, a}\right ) + \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \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}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{2 \, d} \] Input:

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

Output:

1/2*(2*I*sqrt(2)*a*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))*e^(I*d*x + I*c) - sqrt(9*I*a^3/d^2 
)*d*log(1/3*(3*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) + 
2*I*sqrt(9*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + sqrt(9*I*a^ 
3/d^2)*d*log(1/3*(3*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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 
)) - 2*I*sqrt(9*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + sqrt(8 
*I*a^3/d^2)*d*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) + I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - sqr 
t(8*I*a^3/d^2)*d*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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)) - I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/d
 

Sympy [F]

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

integrate(tan(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(tan(c + d*x)), x)
 

Maxima [F]

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

integrate(tan(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(tan(d*x + c)), x)
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*a*i*( - 2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) + int((sqrt 
(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x),x)*d + 3*int(sqrt(ta 
n(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*d))/d