\(\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [768]

Optimal result

Integrand size = 28, antiderivative size = 181 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d} \] Output:

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

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.66 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {-14+4 i \cot (c+d x)-4 \cot ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}}}{6 d \sqrt {\cot (c+d x)}} \] Input:

Integrate[Cot[c + d*x]^(5/2)/Sqrt[a + I*a*Tan[c + d*x]],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]] + (-14 + (4*I)*Cot[c + d*x] - 4*Cot[c + d* 
x]^2)/Sqrt[a + I*a*Tan[c + d*x]])/(6*d*Sqrt[Cot[c + d*x]])
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4729, 3042, 4042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (10 \tan (c+d x) a^2+7 i a^2\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (10 \tan (c+d x) a^2+7 i a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (10 \tan (c+d x) a^2+7 i a^2\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {\frac {2 \int \frac {3 a^3 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {3 a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {3 a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 4027

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {-\frac {6 i a^4 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {-\frac {\frac {(3-3 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {14 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {10 a \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {1}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(1/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I* 
a*Tan[c + d*x]]) + ((-10*a*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^( 
3/2)) - (((3 - 3*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/S 
qrt[a + I*a*Tan[c + d*x]]])/d - ((14*I)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d 
*Sqrt[Tan[c + d*x]]))/(3*a))/(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 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 [A] (verified)

Time = 2.03 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.29

method	result	size
default	\(\frac {\left (-\frac {1}{6}-\frac {i}{6}\right ) \left (-3 \sin \left (d x +c \right )^{3} \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right ) \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}+i \left (3 \cos \left (d x +c \right )+3\right ) \sqrt {2}\, \sin \left (d x +c \right )^{2} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right ) \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}+i \left (\left (-2 \cos \left (d x +c \right )-5 \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2}-2 \sin \left (d x +c \right )\right )+\left (-2 \cos \left (d x +c \right )+5 \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )\right ) \cot \left (d x +c \right )^{\frac {5}{2}} \sec \left (d x +c \right )^{3}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(233\)

Input:

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

Output:

(-1/6-1/6*I)/d*(-3*sin(d*x+c)^3*2^(1/2)*arctan((1/2+1/2*I)*2^(1/2)*(-csc(d 
*x+c)+cot(d*x+c))^(1/2))*(-csc(d*x+c)+cot(d*x+c))^(1/2)+I*(3*cos(d*x+c)+3) 
*2^(1/2)*sin(d*x+c)^2*arctan((1/2+1/2*I)*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))^ 
(1/2))*(-csc(d*x+c)+cot(d*x+c))^(1/2)+I*((-2*cos(d*x+c)-5*sin(d*x+c))*sin( 
d*x+c)^2-2*sin(d*x+c))+(-2*cos(d*x+c)+5*sin(d*x+c))*sin(d*x+c)^2+2*sin(d*x 
+c))*cot(d*x+c)^(5/2)/(a*(1+I*tan(d*x+c)))^(1/2)*sec(d*x+c)^3
 

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. 388 vs. \(2 (135) = 270\).

Time = 0.12 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.14 \[ \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (-7 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )} - 3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (-2 \, {\left (\sqrt {2} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left (-2 \, {\left (\sqrt {2} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:

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

Output:

-1/12*(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))*(-7*I*e^(4*I*d*x + 4*I*c) + 18*I*e^(2 
*I*d*x + 2*I*c) - 3*I) - 3*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c)) 
*sqrt(-2*I/(a*d^2))*log(-2*(sqrt(2)*(I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*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))*sqrt(-2*I/(a*d^2)) - 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - 
I*c)) + 3*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqrt(-2*I/(a*d^2 
))*log(-2*(sqrt(2)*(-I*a*d*e^(2*I*d*x + 2*I*c) + I*a*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(-2*I/(a*d^2)) - 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a*d*e^(3 
*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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