\(\int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [229]

Optimal result

Integrand size = 28, antiderivative size = 168 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i \sqrt {\tan (c+d x)}}{10 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \sqrt {\tan (c+d x)}}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

(-1/8-1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/ 
2))/a^(5/2)/d+1/5*I*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(5/2)+1/10*I*tan 
(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)-1/20*I*tan(d*x+c)^(1/2)/a^2/d/( 
a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}-\frac {2 \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (5+\tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^3}}{40 a^3 d \sqrt {\tan (c+d x)}} \] Input:

Integrate[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(-5*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]] - (2*Tan[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]] 
*(5 + Tan[c + d*x]^2))/(-I + Tan[c + d*x])^3)/(40*a^3*d*Sqrt[Tan[c + d*x]] 
)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4040, 3042, 4079, 27, 3042, 4079, 27, 3042, 4027, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4040

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {i a-4 a \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {i a-4 a \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {3 \left (3 i a^2-2 a^2 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {3 i a^2-2 a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {3 i a^2-2 a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\int \frac {5 i a^3 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {5}{2} i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {5}{2} i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {5 a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {i \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\left (\frac {5}{2}+\frac {5 i}{2}\right ) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {i a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}-\frac {i a \sqrt {\tan (c+d x)}}{d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

Input:

Int[Sqrt[Tan[c + d*x]]/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((I/5)*Sqrt[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) - (((-I)*a*Sqr 
t[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((5/2 + (5*I)/2)*a^(3 
/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x] 
]])/d + (I*a^2*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a^2) 
)/(10*a^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 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 4040

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*(a + b*Tan[e + f*x])^m*(Sqrt[c + d 
*Tan[e + f*x]]/(2*a*f*m)), x] + Simp[1/(4*a^2*m)   Int[(a + b*Tan[e + f*x]) 
^(m + 1)*(Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x]/Sqrt[c + d*Ta 
n[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] && LtQ[m, 0] && IntegersQ[2*m]
 

rule 4079

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 569 vs. \(2 (131 ) = 262\).

Time = 1.58 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.39

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 (20 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 ) a \tan \left (d x +c \right )^{3}-5 \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 ) a \tan \left (d x +c \right )^{4}-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}-20 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 ) a \tan \left (d x +c \right )+30 \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 ) a \tan \left (d x +c \right )^{2}+4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-5 \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 ) a +20 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-20 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{80 d \,a^{3} \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 )^{4} \sqrt {-i a}}\)	\(570\)
default	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (20 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 ) a \tan \left (d x +c \right )^{3}-5 \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 ) a \tan \left (d x +c \right )^{4}-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3}-20 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 ) a \tan \left (d x +c \right )+30 \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 ) a \tan \left (d x +c \right )^{2}+4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-5 \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 ) a +20 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-20 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{80 d \,a^{3} \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 )^{4} \sqrt {-i a}}\)	\(570\)

Input:

int(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/80/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(20*I*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)^3-5*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)^4-4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d* 
x+c)^3-20*I*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)+30*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+4*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I 
*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-5*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*t 
an(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+20 
*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-20*tan(d*x+c)*(a*tan 
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c 
)))^(1/2)/(-tan(d*x+c)+I)^4/(-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. 331 vs. \(2 (120) = 240\).

Time = 0.13 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/40*(10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(I*a^3*d*sqrt( 
1/8*I/(a^5*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)) - 10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)* 
log(-I*a^3*d*sqrt(1/8*I/(a^5*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)) + 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))*(I*e 
^(6*I*d*x + 6*I*c) + 4*I*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I 
))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

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

integrate(tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),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. \[ \int \frac {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^(1/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int(tan(c + d*x)^(1/2)/(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

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

int(tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(sqrt(tan(c + d*x))/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*s 
qrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sq 
rt(a)*a**2)