\(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [784]

Optimal result

Integrand size = 28, antiderivative size = 188 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] 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))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/a^(5/2)/d+1/5*I/d/cot(d*x+c)^(1/2)/( 
a+I*a*tan(d*x+c))^(5/2)+1/10*I/a/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/ 
2)-1/20*I/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \left (\frac {5 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}-\frac {2 \left (a+5 a \cot ^2(c+d x)\right )}{(i+\cot (c+d x))^2 \sqrt {a+i a \tan (c+d x)}}\right )}{40 a^3 d \sqrt {\cot (c+d x)}} \] Input:

Integrate[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

((-1/40*I)*((5*Sqrt[2]*a*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*(a + 5*a*Cot[c + d*x]^2)) 
/((I + Cot[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])))/(a^3*d*Sqrt[Cot[c + d 
*x]])
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 199, 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, 4729, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4040

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

Int[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((I/5)*Sqrt[Tan[c + d*x]])/(d*(a + 
I*a*Tan[c + d*x])^(5/2)) - (((-I)*a*Sqrt[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[T 
an[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]
 

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. 449 vs. \(2 (147 ) = 294\).

Time = 0.43 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.39

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-5 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-5 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +15 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-20 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{80 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, a^{4} \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(450\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-5 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-5 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +15 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}-20 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{80 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, a^{4} \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(450\)

Input:

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

Output:

-1/80/d*(a*(1+I*tan(d*x+c)))^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*( 
15*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)^2-5*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)^3-5*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+15*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*(-I*a)^(1/2)* 
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-20*(-I*a)^(1/2)*(a*tan( 
d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(1/tan(d*x+c))^(1/2)/a^4/(1+I*tan(d*x+c))/ 
tan(d*x+c)/(-tan(d*x+c)+I)^3/(-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. 366 vs. \(2 (136) = 272\).

Time = 0.10 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (2 \, \sqrt {2} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{8 \, a^{5} d^{2}}} + i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (2 \, \sqrt {2} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{8 \, a^{5} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \] Input:

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

Sympy [F]

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

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

Output:

Integral(1/((I*a*(tan(c + d*x) - I))**(5/2)*sqrt(cot(c + d*x))), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/cot(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. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

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

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

Output:

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