\(\int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [212]

Optimal result

Integrand size = 28, antiderivative size = 140 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \sqrt [4]{-1} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.27 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-4 (-1)^{3/4} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}-2 a \tan (c+d x)+i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{2 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4041, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((-4*(-1)^(1/4)*a^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + I*a*Tan[c + d*x]]])/d - ((1 - I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sq 
rt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/(2*a^2) - Sqrt[Tan[c + d 
*x]]/(d*Sqrt[a + I*a*Tan[c + d*x]])
 

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 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(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 - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta 
n[e + f*x], 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] && LtQ[m, 0] && GtQ[n, 1] && (In 
tegerQ[m] || IntegersQ[2*m, 2*n])
 

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. 576 vs. \(2 (109 ) = 218\).

Time = 1.66 (sec) , antiderivative size = 577, normalized size of antiderivative = 4.12

method	result	size
derivativedivides	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )+4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+4 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {i a}\, \sqrt {-i a}}\)	\(577\)
default	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 \sqrt {2}\, \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )+4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+4 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {i a}\, \sqrt {-i a}}\)	\(577\)

Input:

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

Output:

1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(I*(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))*a*tan(d*x+c)^2-I*(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))*a+2*2^(1/2)*(I*a)^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*ta 
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan 
(d*x+c)-8*I*(-I*a)^(1/2)*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))*a*tan(d*x+c)+4*(-I*a)^(1/2)*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))*a*tan(d*x+c)^2+4*I*(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)-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)+4* 
(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x 
+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^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. 577 vs. \(2 (102) = 204\).

Time = 0.14 (sec) , antiderivative size = 577, normalized size of antiderivative = 4.12 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*log(1/4*a*d*sqrt(-2*I/(a*d^2) 
)*e^(I*d*x + I*c) + 1/4*sqrt(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))*(e^(2*I*d*x + 2*I*c) 
+ 1)) - a*d*sqrt(-2*I/(a*d^2))*e^(I*d*x + I*c)*log(-1/4*a*d*sqrt(-2*I/(a*d 
^2))*e^(I*d*x + I*c) + 1/4*sqrt(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))*(e^(2*I*d*x + 2*I* 
c) + 1)) - a*d*sqrt(-4*I/(a*d^2))*e^(I*d*x + I*c)*log(52/605*(4*sqrt(2)*sq 
rt(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^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c)) + (3*a*d*e^(2*I 
*d*x + 2*I*c) - a*d)*sqrt(-4*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + a*d* 
sqrt(-4*I/(a*d^2))*e^(I*d*x + I*c)*log(52/605*(4*sqrt(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))*(e^(3*I*d*x + 3*I*c) + e^(I*d*x + I*c)) - (3*a*d*e^(2*I*d*x + 2*I*c) - 
 a*d)*sqrt(-4*I/(a*d^2)))/(e^(2*I*d*x + 2*I*c) + 1)) + 2*sqrt(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))*(e^(2*I*d*x + 2*I*c) + 1))*e^(-I*d*x - I*c)/(a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {a}\, \left (-5 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}+6 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -18 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}+5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d +5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{2}+1}d x \right ) d +9 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3}+\tan \left (d x +c \right )}d x \right ) \tan \left (d x +c \right )^{2} d +9 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3}+\tan \left (d x +c \right )}d x \right ) d \right )}{5 a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(2*sqrt(a)*( - 5*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)* 
*2 + 6*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 18*sqr 
t(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) + 5*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*d + 5*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)*d + 9*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*d + 9*int((sq 
rt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**3 + tan(c + d*x) 
),x)*d))/(5*a*d*(tan(c + d*x)**2 + 1))