\(\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [774]

Optimal result

Integrand size = 28, antiderivative size = 221 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 \cot ^{\frac {3}{2}}(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {13 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{2 a^2 d}-\frac {7 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a^2 d} \] Output:

(-1/4+1/4*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^(3/2)/d+1/3*cot(d*x+c)^(3/2)/d/(a+ 
I*a*tan(d*x+c))^(3/2)+5/2*cot(d*x+c)^(3/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)+13 
/2*I*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d-7/2*cot(d*x+c)^(3/2)* 
(a+I*a*tan(d*x+c))^(1/2)/a^2/d
 

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.67 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}-\frac {2 \left (39 i+57 \cot (c+d x)-12 i \cot ^2(c+d x)+4 \cot ^3(c+d x)\right )}{(i+\cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}}{12 a d \sqrt {\cot (c+d x)}} \] Input:

Integrate[Cot[c + d*x]^(5/2)/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-3*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*(39*I + 57*Cot[c + d*x] - (12*I)*Cot[ 
c + d*x]^2 + 4*Cot[c + d*x]^3))/((I + Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d 
*x]]))/(12*a*d*Sqrt[Cot[c + d*x]])
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4729, 3042, 4042, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (21 a^2-20 i a^2 \tan (c+d x)\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx}{a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (21 a^2-20 i a^2 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (21 a^2-20 i a^2 \tan (c+d x)\right )}{\tan (c+d x)^{5/2}}dx}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {2 \int -\frac {3 \sqrt {i \tan (c+d x) a+a} \left (14 \tan (c+d x) a^3+13 i a^3\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (14 \tan (c+d x) a^3+13 i a^3\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (14 \tan (c+d x) a^3+13 i a^3\right )}{\tan (c+d x)^{3/2}}dx}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {\frac {2 \int \frac {a^4 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {26 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 4027

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {-\frac {2 i a^5 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {26 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {-\frac {14 a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {26 i a^3 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{a}}{2 a^2}+\frac {5 a}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {1}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}\right )\)

Input:

Int[Cot[c + d*x]^(5/2)/(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(1/(3*d*Tan[c + d*x]^(3/2)*(a + I*a* 
Tan[c + d*x])^(3/2)) + ((5*a)/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d 
*x]]) + ((-14*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) - ((( 
1 - I)*a^(7/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*T 
an[c + d*x]]])/d - ((26*I)*a^3*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + 
 d*x]]))/a)/(2*a^2))/(2*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 4042

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

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 4081

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

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. 552 vs. \(2 (175 ) = 350\).

Time = 0.46 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.50

method	result	size
derivativedivides	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{5}-9 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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+9 \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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+384 \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}+156 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 )^{4}-3 \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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-32 \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}-276 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}-16 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \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 )^{3} \sqrt {-i a}}\)	\(553\)
default	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{5}-9 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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}+9 \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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+384 \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}+156 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 )^{4}-3 \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 )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-32 \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}-276 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}-16 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \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 )^{3} \sqrt {-i a}}\)	\(553\)

Input:

int(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24/d*(1/tan(d*x+c))^(5/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(3* 
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) 
+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan(d*x+c)^5-9*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)+3*a*tan(d*x+c)-I*a) 
/(tan(d*x+c)+I))*a*tan(d*x+c)^3+9*2^(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)+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan 
(d*x+c)^4+384*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c 
)^3+156*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4- 
3*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) 
+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan(d*x+c)^2-32*tan(d*x+c)*(a*tan(d 
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)-276*I*(a*tan(d*x+c)*(1+I*tan(d* 
x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-16*I*(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 
)^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. 418 vs. \(2 (165) = 330\).

Time = 0.11 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.89 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (52 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 87 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} + 3 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \] Input:

integrate(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/12*(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))*(52*I*e^(6*I*d*x + 6*I*c) - 87*I*e^(4*I* 
d*x + 4*I*c) + 18*I*e^(2*I*d*x + 2*I*c) + I) + 3*(a^2*d*e^(5*I*d*x + 5*I*c 
) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(-1/2*I/(a^3*d^2))*log(-4*(sqrt(2)*(I*a 
^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*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(-1/2*I/(a^3*d 
^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*(a^2*d*e^(5*I*d*x + 5*I* 
c) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(-1/2*I/(a^3*d^2))*log(-4*(sqrt(2)*(-I 
*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq 
rt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/2*I/(a^3 
*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a^2*d*e^(5*I*d*x + 5*I*c 
) - a^2*d*e^(3*I*d*x + 3*I*c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(5/2)/(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. \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{5/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:

int(cot(c + d*x)^(5/2)/(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int(cot(c + d*x)^(5/2)/(a + a*tan(c + d*x)*1i)^(3/2), x)
 

Reduce [F]

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

int(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

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